CBSE Class 11 Chemistry Notes For Chapter 2 Structure Of Atom

CBSE Class 11 Chemistry Notes For Chapter 2 Structure Of Atom Discovery Of Fundamental Particles

The atomic theory of matter was first proposed by Sir John Dalton (an English scientist) in 1808. His theory, called Dalton’s atomic theory was a landmark in the history of chemistry. According to this theory, the atom is the smallest, indivisible, discrete particle of matter, which takes part in chemical reactions.

However, the researches done by eminent scientists like J. J. Thomson, Goldstein, Rutherford, Chadwick, Bohr, and others towards the end of the 19th century and in the beginning of the 20th century have conclusively proved that atoms were no longer the smallest indivisible particle. Rather, atoms are -3 composed of several smaller particles called subatomic particles. At present J. scientists have identified about 35 different subatomic particles, some of L which are stable while the others are unstable. These particles may be divided under three heads which is shown in the adjacent table.

The three subatomic particles namely electrons, protons, and neutrons are the main constituents of an atom and are regarded as the fundamental particles

Cathode rays: Discovery of electron 

William Crookes, in 1878, studied the conduction of electricity through gases. A discharge tube was filled with a gas at very low pressure (0.01 mm Hg) and the electrodes were connected to a source of high voltage \(\approx 10^4 \mathrm{~V}\).

It was observed that the glass wall behind the anode began to glow with a faint greenish light called fluorescence.

Further investigations revealed that this fluorescence was due to the bombardment of the glass wall by certain invisible rays which were emitted from the cathode surface and moved towards the anode with tremendous speed.

The rays originated from the cathode and were called the cathode ray.

Nature of cathode rays: J.J Thomson (in 1897) and others characterized these rays based on different experimental findings.

1. Cathode rays are emitted perpendicularly from the cathode surface and travel towards the anode.
2. They cast a sharp shadow on any opaque object placed in their path. So, like ordinary light, cathode rays also travel in straight lines.
3. When a light paddle wheel (made of mica) is placed in their path, the wheel begins to rotate. This indicates that cathode rays are composed of material particles and possess momentum.
4. As cathode rays possess momentum, they can penetrate thin foil metals like aluminum.

1. Cathode rays are deflected by an electric field (towards the positive plate) as well as by a magnetic field, suggesting that cathode rays consist of negatively charged particles.
2. When these rays strike a metal foil, the foil gets heated indicating that cathode rays produce a heating effect.
3. Cathode rays ionize the gas through which they pass.
4. Like ordinary light, cathode rays affect photographic plates. This is called fogging.
5. Cathode rays produce fluorescence on the glass walls of a tube or a screen coated with zinc sulfide 1 ‘(ZnS) or barium platinocyanide Ba[Pt(CN)6].
6. Cathode rays produce X-rays when they strike against the surface of hard metals like tungsten, molybdenum, etc.
7. The characteristics of cathode rays are independent of the material of the cathode nature of the gas used in the discharge tube.
8. Considering the various properties of cathode rays, J. J. Thomson concluded that cathode rays are made of material particles, cathode rays carry a negative charge.
9. He named these negatively charged particles as negatron. Later, these particles were named electrons by G. J. Stoney (in 1874).

J. J. Thomson (1897) used discharge tubes fitted with electrodes made of different metals and filled different gases in the tube. Every time he found that the ratio of charge to mass of electrons (e/m) was the same.

e/m of electron = \(\frac{\text { charge of cathode ray particle }}{\text { mass of cathode ray particle }}\)

= \(1.76 \times 10^8 \mathrm{C} \cdot \mathrm{g}^{-1}\)

= \(1.76 \times 10^{11} \mathrm{C} \cdot \mathrm{kg}^{-1}\)

R. A. Millikan (1917) with the help of his oil drop experiment, determined the charge of an electron. Charge ofan electron (e) = -1.602 x 10^–19 C (or, -4.8 x 10^–1°esu)

No other fundamental particle is known to contain a charge lower than this. This is the minimum measurable quantity of negative charge.

The quantity of electrical charge carried by all negatively charged panicles is an integral multiple of this charge.

Hence, the electronic charge is considered to be die fundamental unit of electricity and is called one unit. The mass of an electron can be calculated from the values of e and e/m.

Mass of an electron = \(\frac{e}{e / m}=\frac{1.602 \times 10^{-19} \mathrm{C}}{1.76 \times 10^8 \mathrm{C} \cdot \mathrm{g}^{-1}} \)

= \(9.11 \times 10^{-28} \mathrm{~g}=9.11 \times 10^{-31} \mathrm{~kg}\)

= 0.000548

1 u = \(1.66 \times 10^{-24}\)

Mass of an electron = \(\frac{1}{1837}\) x mass of a hydrogen atom Thus, a hydrogen atom is 1837 times heavier than an electron. The mass of an electron being very small may be considered as negligible for all practical involving chemical calculations.

Therefore, an electron may be defined as a subatomic particle having a unit negative charge (1.602 × 10^-9C) & negligible mass (9.11 × 10^-28g).

The radius of an electron = 2.8 × 10^-13cm

Electrons are universal constituents of matter:

The elm ratio of negatively charged particles constituting the cathode rays was found to be the same irrespective of the nature of the cathode and the nature of the gas used in the discharge tube, thereby showing that the electrons are the basic constituents of all atoms.

Other experiments showing the existence of electrons:

The following experiments show that the same charge-to-mass ratio exists for the electrons emitted—

1. Spontaneously from radioactive substances in the form of X-rays.
2. when ultraviolet rays are incident on the surface of active metals (Example: Na, K, etc.).
3. When certain metal filaments are heated to a very high temperature.
4. When any form of matter is exposed to X-rays.

Origin of cathode rays in discharge tube:

1. On applying high voltage, electrons are first emitted from the surface of the cathode which travel in straight lines with high speed.
2. During their passage through the gas inside the tube, more electrons are ejected due to the bombardment of the gas molecules by the high-speed electrons.
3. On increasing the voltage in the discharge tube, the speed of the electrons increases, and the electron density in the cathode ray increases.

Applications of cathode ray tube:

In picture tubes of televisions:

The picture tube of a television is a cathode ray tube. Due to fluorescence, a picture is formed on the television screen when an electron beam strikes the screen coated with a fluorescent or phosphorescent material.

In fluorescent tubes:

These types of tubes are filled with gases like argon, nitrogen, etc. along with a small amount of mercury vapor under very low pressure.

• They are cathode ray tubes with their inner walls coated with a suitable fluorescent material.
• On passing electric current, electric discharge occurs inside the tube.
• As a result, electrons from the cathode are transferred to the higher energy state due to collision with the atoms and molecules of the gases or the mercury vapor.
• On their return to the ground state photons are emitted in the form of UV rays. These rays hit the inner walls of the tube and produce visible light.

Anode rays: Discovery of proton

• Since negatively charged electrons are the essential constituents of all atoms and the atom as a whole is electrically neutral, it was thought that some positively charged particles must also be present in the atom.
• Goldstein, in 1886, performed the discharge tube experiment using a perforated cathode. On passing electric discharge at low pressure, he observed that some luminous rays were emitted from the side of the anode which passed through the holes in the cathode and produced fluorescence on the glass wall coated with zinc sulfide.
• These rays were originally called canal rays as they passed through the holes in the cathode.
• As they travel from the side of the anode towards the cathode, these rays are also called anode rays.
• These were named positive rays (since they were found to carry a positive charge) by J. J. Thomson (1907).

Characteristics of anode rays:

• Anode rays travel in straight lines but their speed is much less than that ofthe cathode rays.
• They consist of material particles.
• They are positively charged as indicated by the direction of their deflection in the electric and magnetic fields.
• From the extent of deflection in the electric field, it was proved that the particles constituting the anode rays arc are much heavier than electrons.
• The E/M value of the particles in the anode rays is much smaller than that of the cathode ray particles.
• Furthermore, the e/m value of the particles depends on the nature ofthe gas taken in the discharge tube.
• The E/M value of anode ray particles is maximum when hydrogen gas (the lightest element) is used in the discharge tube.
• The mass of the particle with this maximum e/m value is almost the same as that of an H-atom and its charge is equal to that of an electron but opposite in sign.
• It may therefore be concluded that such anode ray particles with maximum e/m value are none other than H-atoms devoid of electrons. These were called protons by Rutherford (1911).

Anode rays produced using H₂ gas in the discharge tube consist of positively charged particles:

H⁺ [highest e/m ], D⁺, H⁺, HD⁺, and D⁺ [lowest e/m].

Origin of anode rays:

• On applying high voltage between the electrodes in the discharge tube, cathode ray particles (i.e., electrons) move at a high speed toward the anode.
• In the course of their motion, they collide with the gas molecules or atoms and knock out one or more electrons to produce positively charged ions, thereby constituting anode rays.

⇒ \(\text { A (neutral gaseous atom) } \stackrel{\text { ionised }}{\longrightarrow} \mathrm{A}^{+}+e\)

• Thus, the anode rays are not emitted from the anode but produced from the gaseous substance present in the discharge tube.
• So it is clear that the e/m ratio of the ray particles depends on the gaseous substance in the discharge tube. For example, the use of hydrogen to the formation of H+ ions which constitute the orhunode rays.

⇒ \(\mathbf{H} \text { (atom) }-\boldsymbol{e} \stackrel{+}{\longrightarrow} \mathbf{H}^{+} \text {(proton) }\)

E/m of proton = +9.58× 10⁴C.g^-1

Charge of proton = + 1.602×10^-19 C (Or, +4.8 × 10^-10esu)

Mass of proton (m) = \(\frac{\text { charge of a proton }}{c / m \text { of a proton }}\)

= \(\frac{1.602 \times 10^{-19}}{9.58 \times 10^4}\)

= \(1.6725 \times 10^{-24} \mathrm{~g}\)

This mass is nearly the same as that of an H-atom. A proton Is 1836 times heavier than an electron. So, a proton carries 1 unit of +ve charge and it is 1 836 times heavier than an electron.

Radius of a proton \(\approx 1.2 \times 10^{-13} \mathrm{~cm}\)

Proton is a fundamental constituent of all atoms:

• Taking different gases in the discharge tube experiment, it was shown that the mass of anode ray particles is minimal when hydrogen is used as the gaseous substance.
• Masses of other anode ray particles produced in different experiments using different gases are always integral multiples of the mass of a proton.
• Thus a proton, like an electron, is also a fundamental constituent of all atoms.
• Apart from the electrical discharge in gases under low pressure, protons are also emitted in certain nuclear reactions

Example: The bombardment of aluminum metal with a particle

⇒ \(\left({ }_2^4 \mathrm{He}\right)\) or bombardment of nitrogen gas with neutron \(\left({ }_0^1 n\right)\)

⇒ \({ }_{13}^{27} \mathrm{Al}+{ }_2^4 \mathrm{He} \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_1^1 \mathrm{H} ; \quad{ }_7^{14} \mathrm{~N}+{ }_0^1 n \rightarrow{ }_6^{14} \mathrm{C}+{ }_1^1 \mathrm{H}\)

Differences Between Cathode And Anode Rays

Radioactivity:

After the discovery of electrons and protons, it is well established that the atoms are divisible into sub-atomic particles.

• This was further supported by the phenomenon of radioactivity, discovered by Becquerel in 1896.
• The phenomenon of spontaneous emission of active radiations by certain elements like uranium, radium, etc.
• Is called radioactivity, and the elements emitting such radiation are called radioactive elements.

On placing a sample of uranium mineral in a cavity made in a block of lead and allowing the emitted rays to pass through strong electric or magnetic fields, the radiation is resolved in three directions.

• The rays which are deflected slightly towards the negative plate and hence carry + ve charge are called α rays. The particles present in a -rays are called particles, α- rays are called α – particles, and each particle carries 2 units +ve charge and has a mass of 4u. So these are helium atoms with two units +ve charge and are represented as ₂⁴He.
• The rays that are deflected towards the positive plate to a larger extent must carry a -ve charge. These are called α-rays. The particles present in β -rays are called β -P-particles. These particles have the same charge and mass as that of the electrons and are represented as _-1⁰e.
• The rays that remain undeflected are called γ-rays. These are purely electromagnetic radiations.

Discovery Of Neutron

• According to the atomic mass scale, a proton has a mass of one unit while an electron has a negligible mass.
• If an atom comprises only electrons and protons then the mass of an atom will be almost equal to the sum of the masses of the protons present in it.
• In practice, it is found that except for ordinary hydrogen, all other atoms have masses much greater than the sum of the masses of the protons contained by diem.
• This led the scientists to search for a neutral particle having considerable mass. Rutherford, in 1920, predicted the presence of a fundamental particle within an atom, having no charge but one unit mass. This particle was termed a neutron.
• James Chadwick (a student of Rutherford), in 1932, performed some scattering experiments in which he bombarded beryllium metal with fast-moving a -particles emitted from radioactive polonium.
• He observed that new types of particles were emitted which was not deflected by the electric or magnetic field. i.e., those were neutral.
• The mass of such a particle was found to be nearly equal to that of a proton or a hydrogen atom.
• These neutral particles having a unit mass must be the same particle i.e., neutron as predicted by Rutherford. Being neutral, neutrons are more penetrating than electrons or protons.

Mass of neutron = 1.675 × 10^-24 g

⇒ \(\approx\)mass of a proton

= 1837 x mass ofan electron

e/m of neutron = [because neutron has no charge]

Atoms of all elements except hydrogen constitute three fundamental particles:

1. Electron,
2. Proton and
3. Neutron.

In each atom, the number of protons is equal to the number of electrons because an atom as a whole is electrically neutral. Ordinary hydrogen contains no neutron, it only contains one proton and one electron.

Subatomic particles other than an electron, proton, and neutron, are also known

Example: Positron \(\left({ }_{+1}^0 e\right),\); 71 -meson, neutrino (v), photon, etc.

Structure Of Atom Numerical Examples

Question 1. An atom of an element contains 2 electrons In the first shell (n – 1 ), 8 electrons in the second shell (n = 2), and 5 electrons in the third shell (n = 3).

There are 16 neutrons in the nucleus of the atom. From these data, find—

1. The atomic no. of the element,
2. the no. of s and -electrons in the atom,
3. Mass no. of the element.

Answer:

1. Atomic number = number of protons = number of electrons =2+8 + 5

= 15

2. \(\begin{array}{|c|c|c|c|}
\hline \text { Shell No. } & \begin{array}{c}
\text { Total no. of } \\
\text { electrons }
\end{array} & \begin{array}{c}
\text { No. of } \\
s \text {-electrons }
\end{array} & \begin{array}{c}
\text { No. of } \\
p \text {-electrons }
\end{array} \\
\hline 1 & 2 & 2 & 0 \\
\hline 2 & 8 & 2 & (8-2)=6 \\
\hline 3 & 5 & 2 & (5-2)=3 \\
\hline
\end{array}\)

∴ Total number of s- electron – 2 + 2+2 =6

Total number of p -electron 6 + 3 =9

3. The mass number of protons + number of neutrons = 15+16

= 31.

Question 2. Determine the number of protons present In 5.6 I16. of n sample of oxygen gas at STP, containing \({ }^{16} 0\) isotope only.
Answer:

The number of oxygen molecules present in 22.4 L of the gas = 6.022 × 10²³.

∴ 5.6 L of oxygen at STP contains = \(\frac{5.6 \times 6.022 \times 10^{23}}{22.4}\)

= 1.50055  × 10²³ Molecules

Number of O-atoms in the given volume of the gas

⇒ \(2 \times 1.50055 \times 10^{23}=3.011 \times 10^{23}\)

since oxygen is diatomic]

Now, each ¹⁶O atom contains 8 protons.

∴ Number of protons in 3.011 × 10²³ atoms of oxygen

= 8 × 3.01 1 × 10²³

= 2.4088 × 10²⁴

Question 3. Find the number of protons required to fill u sphere of 10cm3 volume. What Is the muss of those number of protons?
Answer:

The radius of a proton = 1.2 × 10¹³ cm (approx.)

∴ The approximate volume of a proton \(=\frac{4}{3} \pi r^3\)

⇒ \(\frac{4}{3} \times 3.14 \times\left(1.2 \times 10^{-13}\right)^3 \mathrm{~cm}^3=7.235 \times 10^{-39} \mathrm{~cm}^3\)

∴ The number of protons which can be accommodated in the sphere of capacity 10 cm3

⇒ \(\frac{10}{7.235 \times 10^{-39}}=1.382 \times 10^{39}\)

∴ The mass of that number of protons

1.382 × 10³⁹ × 1.672 ×10^-24 g

= 2.311 × 10¹⁵g

= 2.311 × 10¹² kg

Question 4. Determine the number of neutrons and their mass, present in 7 mg of \({ }^{14}{ }_6^{14} \mathrm{C}\) Assume that the mass of I neutron = muss of 1 H-atom.
Answer:

No. of neutrons present In a \({ }_6^{14} \mathrm{C} \text {-atom }\) atom = (14-6)= 8

Number of atoms in 1 gram morn or 14 g of carbon-containing only \({ }_6^{14} \mathrm{C}\) isotope 6.022 x 10²³

∴ In 7 mg or 0.007 g of carbon, the number of atoms \(=\frac{6.022 \times 10^{23} \times 0.007}{14}\)

∴ In 7 mg of carbon, the number of neutrons

∴ \(=\frac{6.022 \times 10^{23} \times 0.007 \times 8}{14}=24.088 \times 10^{20}\)

Again, mass of 6.022 × 10²³ atoms of hydrogen = 1.008 g

∴ Mass of 1 atom of hydrogen \(=\frac{1.008}{6.022 \times 10^{23}} \mathrm{~g}\)

= mass of 1 neutron [according to the given condition]

∴ Mass of 24.088 × 10²⁰ neutrons

⇒ \(\frac{1.008 \times 24.088 \times 10^{20}}{6.022 \times 10^{23}}=4.032 \times 10^{-3} \mathrm{~g}\)

Question 5. How many different types of HCl molecule can be produced from two natural isotopes of hydrogen ¹H = 99% and ²H = 1%) and two natural isotopes of chlorine (³CI = 76% and 37CI = 24%). Arrange the molecules obtained in the decreasing order of their availability.
Answer:

Four different types of HCl molecules can be produced.

They are

⇒ \({ }^1 \mathrm{H}^{35} \mathrm{Cl},{ }^1 \mathrm{H}^{37} \mathrm{Cl},{ }^2 \mathrm{H}^{35} \mathrm{Cl},{ }^2 \mathrm{H}^{37} \mathrm{Cl} \text {. }\)

Since the availability of XH and 35C1 are comparatively 2H and 37C1, the decreasing order of their availability

⇒ \({ }^1 \mathrm{H}^{35} \mathrm{Cl}>{ }^1 \mathrm{H}^{37} \mathrm{Cl}>{ }^2 \mathrm{H}^{35} \mathrm{Cl}>{ }^2 \mathrm{H}^{37} \mathrm{Cl}\)

Question 6. A sample of oxygen contains the isotope, 180. How many neutrons are present in 11.2L of the gas at STP?
Answer:

Atomic Number of oxygen

= 8 Number of neutrons in a 180 isotope

= 18-8

= 10

∴ Number of atoms in 22.4 L of 18O isotope at STP

= 2 × 6.022 × 10²³

∴ Number of neutrons in 22.4L of 180 isotopes at STP

= 10 × 2 × 6.022 ×10²³ [since oxygen molecule is diatomic]

∴ Number of neutrons in 11.2Lof 180 isotopes at STP

=  \(\frac{10 \times 2 \times 6.022 \times 10^{23} \times 11.2}{22.4}\)

= \(6.022 \times 10^{24}\)

Question 7. A sample contains two isotopes, 160 and 180. How many protons are present in 11.2L of the sample at STP? What would be the difference in the no. of protons, if the sample contains only one isotope?
Answer:

Each of the atoms of 160 and 180 contains 8 protons (atomic number of oxygen = 8 ) Total number of atoms is 22.41. oxygen ( 160 and 180 ) at SIT = 6.022 × 10²³×2

[since oxygen molecule is diatomic]

Number of protons in 22.4L oxygen sample at STP

= \(\frac{8 \times 2 \times 6.022 \times 10^{23} \times 11.2}{22.4}\)

∴ Number of protons in 11.2 L oxygen sample at STP

= \(4.8176 \times 10^{24}\)

If the sample contains only one isotope, then there will be no difference in the number of protons because both 16O and 180 isotopes contain 8 protons in their atoms.

Electromagnetic Radiation: Origin Of Atomic Spectra

We have already seen that Rutherford’s atomic model failed to explain the atomic spectra and stability of an atom.

• In order to investigate the reasons behind the failure of the Rutherford model, scientists felt the need to explain the nature and origin of atomic spectra.
• Atomic spectra result from the electromagnetic radiations emitted by the excited atoms.
• These electromagnetic radiations can pass through a vacuum also. According to Newton and other physicists, light is nothing but a stream of very small particles. The phenomena of reflection, refraction, etc.
• Can be successfully explained using Newton’s theory but those like dispersion, interference, etc. cannot. Thus, the particle nature of light was replaced by the wave nature of light.

Wave nature of electromagnetic radiation: Maxwell’s theory

James Maxwell (1865) studied the nature of light and concluded that light is transmitted in the form of electromagnetic waves, which are associated with oscillating electric and magnetic fields.

• The oscillating electric and magnetic fields are perpendicular to each other and are both perpendicular to the direction of propagation ofthe wave.
• The absorption or emission of radiation by a body occurs continuously.
• The radiations possessing wave character, travel with the velocity of light (3 x 108m-s_1 in vacuum).
• The color of the radiation depends on its wavelength.
• Electromagnetic waves, unlike sound waves or water waves, do not need a material medium for propagation, Electromagnetic waves can travel through a vacuum.

Some quantities related to electromagnetic radiation

Electromagnetic spectrum:

Different types of electromagnetic radiation differ only in their frequencies and hence in their wavelengths.

The order of increasing wavelengths is:

Cosmic rays <γ -rays < X-rays < UV rays (150-3800Å) < visible light (3800-7600Å) < IR-rays (7600-6 × 106Å) < Microwaves < Radio waves.

• The complete spectrum obtained by arranging these electromagnetic radiations in order of their increasing wavelengths or decreasing frequencies is called electromagnetic spectrum.
• Various types of electromagnetic radiation have different energies and are being used for different applications.

Some applications of electromagnetic waves are listed below.

Limitations of electromagnetic wave theory:

This theory successfully explains the properties of light such as interference and diffraction.

However, it failed to explain the phenomena such as

1. Black body radiation
2. Photoelectric effect line spectra of atoms
3. Variation of heat capacity of solids as a function of temperature.

Black body radiation:

An ideal body, which emits and absorbs radiations of all frequencies is called a black body and the radiation emitted by such a body is called black body radiation.

• When a substance having a high melting point is heated, viz an iron bar, it first turns red, then yellow, and then glows with a white light and finally with a blue light.
• According to electromagnetic wave theory, since the emission and absorption of energy occur continuously, the energy of an electromagnetic wave is supposed to be proportional to its intensity and independent of its frequency or wavelength.
• So according to wave theory, a body should emit radiation ofthe same colour although its intensity may vary as the heating is continued.
• We have already seen that, on applying heat, the color of a solid changes with the temperature rise.
• The color change indicates that the frequency of radiation emitted increases with the rise in temperature (since red light indicates low frequency and blue light indicates high).
• Hence, electromagnetic wave theory can’t explain black body radiation.

The variation of intensity with wavelength of radiation emitted by a black body at two different temperatures ( T₁ and T₂ ).

The intensity of the radiation emitted depends only on temperature.

At a given temperature, the intensity of radiation emitted increases with a decrease in wavelength reaches a maximum value at a certain wavelength, and then starts decreasing with a further decrease in wavelength. (Observe the variation in the plot from right to left).

Photoelectric effect:

The phenomenon of ejection of electrons from the surface of a metal when the light of a suitable wavelength falls on it, is called the photoelectric effect and the ejected electrons are called photoelectrons.

• The apparatus used for studying the photoelectric effect. In the figure, Tis an evacuated glass tube fitted with two quartz windows.
• Two metal plates, A and B, are fitted at two ends of the tube. The plate A is photosensitive and acts as the cathode.
• The plate B is the electron detector and acts as the anode. The plate electrodes are connected to a battery via a milliammeter.
• When the light of a suitable wavelength strikes plate A, photoelectrons are ejected from it The electrons are attracted by detector B and so they move toward it
• These electrons are responsible for the flow of current through the external circuit which can be measured using the milliamm.
• By measuring the strength of the current flowing through the circuit, it is possible to determine the kinetic energy ofthe electrons.

The following are a few important observations about the Velocity 3 x 108 photoelectric effect: 

• For any given metal, there exists a minimum frequency called threshold frequency (VG).
• At frequencies below the threshold frequency, no electrons are emitted, no matter how large the intensity of the incident light or how long the irradiation occurs.
• The number of photoelectrons emitted per second is proportional to the intensity ofthe incident light.
• The kinetic energy of the photoelectrons is direct radiation of a frequency of 97.8 MHz. proportional to the frequency of the incident radiation.
• With an increase in the frequency of light, the kinetic energy of the electrons increases.
• The kinetic energy ofthe photoelectrons are independent of the intensity ofthe incident radiation.

The observations cannot be explained based on electromagnetic wave theory. For instance:

• According to this theory, the absorption or emission of radiation occurs continuously. Hence, the energy of light is expected to depend on the intensity of the incident light.
• Thus, light of any frequency can be made to have sufficient energy to cause the emission of electrons, merely by increasing its intensity. However, this is not true in practice.
• According to this theory, the energy of the ejected electrons should be proportional to the intensity of the incident light, which is not the case in reality.

Numerical examples

Question 1. Yellow light emitted from a sodium lamp has a wavelength of 580 nm. Calculate the frequency and wave number of yellow light.

Answer:

⇒ \(\lambda=580 \mathrm{~nm}=580 \times 10^{-7} \mathrm{~cm}\)

Since 1nm = 10^-7 cm

⇒ \(v=\frac{c}{\lambda}=\frac{3 \times 10^{10}}{580 \times 10^{-7}}=5.17 \times 10^{14} \mathrm{~s}^{-1}\)

⇒ \(\vec{v}=\frac{1}{\lambda}=\frac{1}{580 \times 10^{-7}}=17241.37 \mathrm{~cm}^{-1}\)

Question 2. A radio station broadcasts at a frequency of 100 MH_z. How long would it take to reach a receiving system at a distance of 300 km? Calculate the wavelength and wave number of these radiations.
Answer:

Required time \(=\frac{\text { Distance }}{\text { Velocity }}=\frac{300 \times 100}{3 \times 10^8}=3 \times 10^{-3} \mathrm{~s}\)

Since the Velocity of any electromagnetic radiation = 3 ×10⁸m.s^-1

Wavelength of that radiation \(\lambda=\frac{c}{v}=\frac{3 \times 10^{\circ}}{100 \times 10^6}=3 \mathrm{~m}\)

Since 100MHz =1006 Hz]

Wave number (v) = 1/λ = 1/3

= 0.33m^-1

Question 3. Calculate the wavelength of an electromagnetic radiation of frequency of 97.8 MHz.
Answer:

Wavelength of radiation \(\lambda=\frac{c}{v}=\frac{3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{97.8 \times 10^6 \mathrm{~s}^{-1}}\)

= 3.06 m

Question 4. How long would it take a radio wave of frequency, 6.2× 10⁸s^-1 to travel from Mars to Earth, the observation distance being 8.1 × 10⁷km?
Answer:

The velocity of any electromagnetic radiation (c) 3 × 10⁸m

Required time \(\frac{\text { Distance }}{\text { Velocity }}\)

= =\(\frac{8.1 \times 10^7 \times 10^3}{3 \times 10^8}\)

= 270s

= 4min 30s

Particle nature of electromagnetic radiation: Planck’s quantum theory

The electromagnetic wave theory could successfully explain the various phenomena of light such as interference, diffraction, polarisation, etc.

• However, it failed to explain the various phenomena involving energy transfer, namely black body radiation, photoelectric effect, etc.
• The failure of the classical electromagnetic theory of radiation led Max Planck (1900) to propose a new theory known as ‘Planck’s quantum theory’.
• Later this theory was extended by Einstein (1905). The main points of this theory are— M Radiant energy is emitted or absorbed in the form of small, discrete packets of energy called ‘quanta’ (singular quantum). In the case of light, the quantum of energy is often called ‘photon’.
• The energy of each quantum of radiation does not have a fixed value. It depends on the frequency of the emitted or absorbed radiation.
• The energy possessed by each quantum of radiation is directly proportional to its frequency, i.e.,

E ∝ v or E = hv [where E = energy of each quantum or photon, v = frequency of the radiation, h = Planck’s constant (6.626 ×10^-27erg-s or 6.626 × 10^-34J-s )].

The total amount of energy emitted or absorbed by a body must be an integral multiple of a quantum, i.e., E = nhv [where n = number of photons, absorbed or emitted by a body]. This means the amount of energy emitted or absorbed by a body can be—hv, 2hv, 3hv, etc. but never 0.5 hv or 1.4 hv.

The energy possessed by one photon is called one quantum.

The energy possessed by one-mole photons is called one Einstein.

1 einstein = NO (or 6.022 × 10²³ ) quanta.

∴ 1 einstein energy (J5) = N₀hv

Explanation of black body radiation based on quantum theory:

When a solid is heated, the atoms are set into oscillations emitting radiations of frequency v.

As heating is continued the atoms absorb more energy and emit radiations of higher frequency. Since the frequency of red light is minimal, an iron rod on heating first turns red and then yellow (higher frequency than red) and further turns into white and finally blue.

Explanation of the photoelectric effect based on quantum theory:

Einstein (1905) explained different aspects of the photoelectric effect using Planck’s quantum theory, as follows

• When a light of certain frequency strikes a metal surface, photons ofthe light collide with the electrons ofthe metal. Each photon thus transfers its entire energy (hv) to each of the colliding electrons.
• Photoelectrons are emitted only if the energy of the photon is sufficient to overcome the force of attraction between the electron and the nucleus.
• In other words, electrons are emitted from the surface of metals, when irradiated with photons of a certain minimum frequency known as threshold frequency (hv₀).
• The minimum energy of a photon that causes photoelectric emission is known as the photoelectric work function (hv₀).

If the frequency of the incident radiation (v) is greater than the threshold frequency (v₀), the excess energy of the protons Is transferred to the ejected electrons ns kinetic energy;

⇒  \(\frac{1}{2} mv-W_0\).

Hence, the kinetic energy of the ejected electrons

• Increases with an Increase In the frequency of the Incident light.
• Keeping the frequency constant, If the Intensity of the incident light Is
• Increased, the number of photons striking the metal surface increases,
• This In turn Increases the number of electrons of the metal colliding with the photons, which In turn increases the strength of the photoelectric current.
• The above observations are By experimental results, A plot of the kinetic energy of emitted photoelectrons against the frequency of the absorbed photons gives a straight line with slope, h.

Keeping the frequency constant, a plot of the kinetic energy of the entitled photoelectrons against the intensity of incident radiation gives n straight line parallel to the x-axis.

Dual nature of electromagnetic radiation:

Prom the above discussion It Is clear that the properties of radiation like Interference or diffraction can be explained if the light is considered to have a wave nature whereas, other properties of light such as black body radiation, photoelectric effect, etc.

Can be explained, if the light is considered to have a particle nature. Thus, it is concluded that light behaves both as a wave and also as a stream of particles. Therefore light is said to have a dual nature.

Numerical Examples

Question 1. A Ray of habit or frequency SOA is incident on a metal surface and thus, absorbs 10^-7J of Calculate the number of photons Incident on the mental surface
Answer:

Let the number of photons incident on the metal surface = n

∴ \(n h v=10^{-7} \text { or, } \frac{n h c}{\lambda}=10^{-7}\)

Or, \(\frac{n \times 6.626 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}}=10^{-7}\)

Or, \(n=\frac{10^{-7} \times 5000 \times 10^{-10}}{6.626 \times 10^{-34} \times 3 \times 10^8}=2.51 \times 10^{11}\)

Question 2. The threshold frequency, v0 for a metal is 7.0× 10⁴s^-1. Calculate the kinetic energy of an electron emitted, when radiation of frequency, v = 1.2 × 10¹⁵s^-1, strikes this metal.
Answer:

Kinetic energy of the photoelectrorts \(\left(\frac{1}{2} m v^2\right)=h\left(v-v_0\right)\)

= \(\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(1.2 \times 10^{15} \mathrm{~s}^{-1}-7.0 \times 10^{14} \mathrm{~s}^{-1}\right)\)

= \(3.31 \times 10^{-19} \mathrm{~J}\)

Question 3. If a light with frequency 2.0 × 10¹⁶Hz emitted photo o electrons with double the kinetic energy as emitted by the light of frequency 1.25 × 10¹⁶ Hz from the same metal surface, calculate the threshold frequency of the metal
Answer:

Kinetic energy of photoelectrons \(\left(\frac{1}{2} m v^2\right)=h\left(v-v_0\right)\)

Given, h(2.0 × 10¹⁶-v₀) = h(1.25 × 10¹⁶-v₀) ×2

or, v₀ = (2.5- 2.0) × 10¹⁶ =0.5 ×10¹⁶

= 5 × l0¹⁵Hz

Question 4. When a radiation of frequency 7.5×10¹⁴ Hz strikes a metal surface, the maximum kinetic attained by the emitted electrons is 1.6 × 10^-19J. Calculate the threshold frequency of the metal.
Answer:

Kinetic energy ofemitted electrons \(\left(\frac{1}{2} m v^2\right)=h\left(v-v_0\right)\)

Given, 1.6 × 10^-19 = h(v-v₀)

Or, \(\frac{1.6 \times 10^{-19}}{6.626 \times 10^{-34}}=7.5 \times 10^{14}-v_0\)

or, v₀ = 7.5 × 10¹⁴ –  2.41 × 10¹⁴

= 5.09 ×10¹⁴ Hz

Question 5. Calculate the energy of each quantum of electromagnetic radiation having a wavelength of 6000 Å. [h = 6.624 × 10^-27erg.s]
Answer:

Energy of each quantum, \(E=h v=\frac{h c}{\lambda}\)

∴ \(E=6.624 \times 10^{-27} \times \frac{3 \times 10^{10}}{6000 \times 10^{-8}}=3.312 \times 10^{-12} \mathrm{erg}\)

[c = 3 × 10cm.s^-1

λ = 6000Å

λ = 6000 × 10^-8cm ]

Question 6. Calculate the energy of 1 mol of photons of an electromagnetic radiation of frequency 2.6 × 10²³. [h = 6.026 × 10^-34j.s
Answer:

The energy of l mol photons = N₀hv

= 0.022 × 10²³ × 6.626 × 10^-34 × 2.5 × 10H

= 99.75 kj.mol^-`1

Question 7. How many photons of light with wavelength 400 CBM provide 1 energy? = 6.626 × 10^-34J.s
Answer:

Energy of 1 photon \(=h v=\frac{h c}{\lambda}\)

⇒ \(=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \mathrm{~J}\)

[since 1nm= 10^-9m]

∴ No. of photons required to produce 1.0J energy

= \(\frac{1.0}{\left(6.626 \times 10^{-34} \times 3 \times 10^8\right) /\left(400 \times 10^{-9}\right)} \)

= \(\frac{400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8}=2.012 \times 10^{18}\).

Question 8. Find the wave number and energy of each photon present in yellow light having wavelength 580nm. (c = 3 ×10⁸ m.s^-1 and h = 6.627 × 10^-34J.s]
Answer:

Wave number \((\bar{v})=\frac{1}{\lambda}=\frac{1}{580 \times 10^{-9}} \mathrm{~m}^{-1}\)

= 1.724 ×10^-16 m^-1

∴  1nm = 10^-9m

Energy of each photon {E) = \(h v=h \times \frac{c}{\lambda}\)

⇒ \(\left(6.627 \times 10^{-34}\right) \times \frac{3 \times 10^8}{580 \times 10^{-9}}=3.428 \times 10^{-19} \mathrm{~J}\)

Question 9. Calculate the frequency of light emitted when an electron drops from a higher to lower energy level of an atom and the difference between the two energy levels is 35.64 x 10^-13erg.[h = 6.624 × 10^-27erg-s]
Answer:

E₁ and E₂ are the two energy levels, then the difference between their energies is given by, E₂-E₁ = hv or, ΔE = hv

h = Planck’s constant and

v = Frequency of light emitted

∴ \(v=\frac{\Delta E}{h}=\frac{35.64 \times 10^{-13} \mathrm{erg}}{6.624 \times 10^{-13} \mathrm{erg} \cdot \mathrm{s}}=5.3804 \times 10^{14} \mathrm{~s}^{-1}\)

Question 10. An 80 W bulb emits a monochromatic light of wavelength 480nm. Calculate the number of photons emitted per second by the bulb.
Answer:

Power of the bulb =80 W = 80.s^-1

Energy of a photon \(=h v=\frac{h c}{\lambda}\)

= \(\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}{480 \times 10^{-9} \mathrm{~m}}\)

= \(4.14 \times 10^{-19} \mathrm{~J}\)

Number of emitted photons per second =  \(\frac{80 \mathrm{~J} \cdot \mathrm{s}^{-1}}{4.14 \times 10^{-19} \mathrm{~J}}\)

= \(1.932 \times 10^{20} \mathrm{~s}^{-1}\)

Question 11. Calculate the wavelength of a photon (in nm) having energy of 1 eV.
Answer:

Energy of 1 photon, E = 1 eV = \(1.609 \times 10^{-19} \mathrm{~J}\)

E = hv = \(=h \frac{c}{\lambda}\)

Or, λ = \(\frac{h c}{E}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right) \times\left(3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}{1.609 \times 10^{-19}}\)

= \(12.35 \times 10^{-7} \mathrm{~m}\)

= \(\frac{12.35 \times 10^{-7}}{10^{-9}}\)

Question 12. The iodine molecule absorbs radiation of wavelength 450nm to dissociate into iodine atoms. If each molecule of iodine absorbs 1 quantum of radiation, determine the kinetic energy of the iodine atom. (Bond energy of I₂ = 240kJ.mol^-1 )
Answer:

The energy required for the dissociation of 1 mol of iodine molecules = Bond energy of I₂ molecule = 240kJ

∴ The energy required for the dissociation of a single iodine molecule

= \(\frac{240 \times 10^3}{6.022 \times 10^{23}} \mathrm{~J}\)

= \(3.985 \times 10^{-19} \mathrm{~J}\)

Energy absorbed by each iodine molecule \(=h v=\frac{h c}{\lambda}\)

⇒ \(\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{450 \times 10^{-9}} \mathrm{~J}=4.417 \times 10^{-19} \mathrm{~J}\)

∴ The surplus energy after the dissociation of each molecule of iodine =(4.417-3.985) × 10^-19J =4.32 × 10^-20J

This surplus energy imparts kinetic energy to each iodine molecule.

The surplus energy that imparts kinetic energy to each iodine atom

= \(\frac{4.32 \times 10^{-20}}{2} \mathrm{~J}\)

= \(2.16 \times 10^{-20} \mathrm{~J}\)

∴ The kinetic energy of each iodine atom = 2.16 × 10^-20J.

Question 13. Calculate the energy associated with 1 mol of photon corresponding to electromagnetic radiation having a frequency of 5 × 10^-14Hz.
Answer:

Energy associated with 1 mol of a photon is given by, E=Nhv [AT = Avogadro number, h = Planck’s constant, v = frequency of radiation]

= \(6.022 \times 10^{23} \times\left(6.626 \times 10^{-34}\right) \times\left(5 \times 10^{14}\right) \mathrm{J}\)

= \(199.51 \times 10^3\) J

= 199.51 kJ

Emission And Absorption Spectra

When light or any other electromagnetic radiation is made to pass through a spectroscope, the light separates into various components of different wavelengths, producing various colored bands.

The set of colored bands formed is called a spectrum. Spectra can be of two types —

1. Emission spectrum and
2. Absorption spectrum

Emission spectrum: When the radiation emitted from any source such as any incandescent solid, liquid, or gaseous substance (i.e., by passing an electric discharge through a gas at low pressure or by heating some substance to high temperature) is directly analyzed through a spectroscope, the spectrum obtained is called an emission spectrum.

Depending on the source of radiation, the emission spectrum can be divided into two types—

Continuous spectrum:

The spectrum produced when white light emitted from any source such as the sun, a bulb, a molten metal, or any incandescent source is passed through a spectroscope is a continuous spectrum.

• In this spectrum, the colors from red to violet are arranged sequentially continuously, without any break.
• In this spectrum, red lies at one end and violet at the other, and the rest of the colors lie between these two colors.
• The colors are so continuous that each of them appears to merge into the other.

Hence, it is known as a continuous spectrum. It contains all the wavelengths of the visible range.

For example— Solar spectrum.

The spectrum emitted from an incandescent source that contains all the wavelengths of light of the visible range is known as continuous spectrum. There are no. black lines in this spectrum.

Line spectrum or atomic spectrum:

• If a gaseous element is heated or an electric discharge is passed through the gas at low pressure and the light emitted is resolved in a spectroscope, the spectrum obtained is not continuous.
• In this case, the different colored lines obtained are separated from each other by dark bands.
• The spectrum obtained is called the line spectrum.

The excited gaseous atoms are responsible for the piotirllou of such spectrum, lunco, line spectrum Is also known as the atomic spectrum.

In a lino spectrum, each line corresponds to a particular wavelength. The line spectrum of each element consists of a group of lines with certain fixed wavelengths, Every element has Its characteristic spectrum, which is different from those of other elements.

For example, (the spectrum of sodium vapor gives two yellow lines. Hence, the line spectrum of each element characterizes the atom of that element.

Therefore, the atomic spectrum of an element can be used to identify the element and Is sometimes called the fingerprint of its atoms.

Absorption spectrum:

When white light emitted from a heated incandescent substance is passed through a liquid or a gaseous substance, radiations of certain wavelengths are absorbed.

• If the emergent radiations are resolved in a spectroscope, a few dark lines are observed, in the atoms. Otherwise continuous spectrum.
• These dark lines constitute the absorption spectrum of that absorbing substance (liquid or gas).
• The dark lines in the absorption spectrum are at the same place where colored lines are obtained in the emission spectrum for the same substance.
• This indicates that the wavelengths of radiation absorbed in the absorption spectrum are the same as those emitted in the case of the emission spectrum.

Differences between emission and absorption spectra

Emission spectrum: Line spectrum of a hydrogen atom

• To study the emission spectrum of hydrogen, the gas is taken in a discharge tube at low pressure, and an electric discharge is passed through it. ill H₂ molecules dissociate into energetically excited Hatoms and emit electromagnetic radiation.
• The emitted radiation is allowed to pass through the prism in a spectroscope.
• The resulting spectrum consists of a large number of isolated sharp lines grouped into different series, named after their discoverers. These lines constitute the line spectrum (discontinuous spectrum).
• The visible region ofthe hydrogen spectrum can be viewed even with the naked eye. It was discovered by Balmer in 1885 and hence the series of isolated spectral lines involving the visible region is called the Balmer series.
• Balmer observed four prominent lines (red, bluish-green, blue and violet) in the visible region and these were designated as Ha, Hy, and respectively.

Balmer (1885) showed that the wave numbers (inverse of wavelengths) of the spectral lines in the visible region (Balmer series) can be expressed by the empirical formula,

⇒ \(\bar{v}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]\) ……[1]

Where R is the universal constant, known as Rydberg’s constant Its value is 1.09678 × 10⁵ cm^-1. n is an integer equal to or greater than 3 (i’.e., n = 3, 4, 5, 6 v (wave number) =i, where X – wavelength.

• Replacing n by 3, 4, 5, and 6 respectively the wave numbers of the spectral lines, Hy and can be calculated.
• Further investigation on the line spectrum of hydrogen (using an improved spectroscope) led to the discovery of five other series lines in the ultraviolet and infrared regions. These are indicated in the adjacent table.
• Rydberg (1890), showed that all series of lines in the hydrogen spectrum could be described by the following expression (similar to Balmer’s empirical formula)

⇒ \(\bar{v}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\) …..[2]

Where nx =1, 2, 3, 4,……, n2 = n1 + 1 , n1 + 2,…….

Series Of Lines In The Line Spectrum Of Hydrogen Atom 

Putting n1 = 1,2, 3, 4, 5, 6 and n2 = + 1, n1 + 2, n+3, … etc., die wave numbers of the spectral lines corresponding to Lyman, Balmer, Paschen, Brackett Pfund and Humphreys series can be calculated.

The cause of the formation of the line spectrum of hydrogen was not exactly known at that time.

However, it was believed that the line spectrum was obtained as a result of absorption and subsequent emission of energy discontinuously by the electron present in the H-atom. Hence, the line spectrum is also called an atomic spectrum.

Numerical Examples

Question 1. Calculate the wavelengths of Hff and in the emission spectrum of hydrogen. [R = 109678 cm^-1].
Answer:

Balmers equation is given by \(\bar{v}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]\)

n=3 for H and n = 6 for H

∴ \(\bar{v}_{\mathrm{H}_a}=109678\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=109678 \times \frac{5}{36} \mathrm{~cm}^{-1}\)

∴ \(\lambda_{\mathrm{H}_\alpha}=\frac{1}{\bar{v}_{\mathrm{H}_\alpha}}=\frac{36}{109678 \times 5}=6.565 \times 10^{-5} \mathrm{~cm}\)

Similarly \(\bar{v}_{\mathrm{H}_\delta}=109678\left(\frac{1}{2^2}-\frac{1}{6^2}\right)=109687 \times \frac{8}{36} \mathrm{~cm}^{-1}\)

∴ \(\lambda_{\mathrm{H}_8}=\frac{1}{\bar{v}_{\mathrm{H}_8}}=\frac{36}{109678 \times 8}=4.10 \times 10^{-5} \mathrm{~cm}\)

Question 2. Identify the spectral line having a wavelength of 4.863 x 10_5cm in the emission spectra of hydrogen.
Answer:

Wave no \(\bar{v}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]\)

[Given, lamba = 4.863 × 10^-5Cm]

∴ \(\bar{v}=\frac{1}{\lambda}=\frac{1}{4.863 \times 10^{-5}}=109678\left[\frac{1}{4}-\frac{1}{n^2}\right]\)

∴ \(\frac{1}{n^2}=\frac{1}{4}-\frac{1}{4.863 \times 10^{-5} \times 109678}=0.0625 \quad \text { or, } n=4\)

The spectral line with wavelength 4.863 × 10^-5cm is Hbeta.

Question 3. Determine the wavelength and frequency of the radiation having the longest wavelength in the Lyman series of hydrogen atoms.
Answer:

Rydberg’s equation \(\frac{1}{\lambda}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\): In case of lyman series. For the wavelength to be the longest, the difference in energies between the two energy levels should be minimal.

Hence, n₂= 2

∴ \(\frac{1}{\lambda}=109678\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \quad \text { or, } \frac{1}{\lambda}=109678\left(1-\frac{1}{4}\right)\)

∴ \(\lambda=\frac{4}{3 \times 109678}=1215.67 \times 10^{-8} \mathrm{~cm}=1215.67\)

So, the X radiation109678 has the longest wavelength in the Lyman series = 1215.67×10-8 cm.

∴ Frequency of the radiation

⇒ \(v=\frac{c}{\lambda}=\frac{3 \times 10^{10}}{1215.67 \times 10^{-8}} \mathrm{~s}^{-1}\)

=\(2.467 \times 10^{15} \mathrm{~s}^{-1}\)

Question 4. The wave number of a spectral line in the Lyman series of H-atom is 82260 cm-1. Show that this line has appeared in this series due to the return of the electron from the second to the first orbit.
Answer:

Wave number \(\bar{v}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\)

∴ \(\nabla=R\left[1-\frac{1}{n_2^2}\right] \text { or, } \frac{\bar{v}}{R}=1-\frac{1}{n_2^2}\)

Or, \(\frac{1}{n_2^2}=1-\frac{\bar{v}}{R}=1-\frac{82260}{109680}=\frac{27420}{109680}=\frac{1}{4}=\frac{1}{2^2}\)

∴ n2=2

This means that the electron has returned from the second to the first orbit.

Question 5. Calculate the shortest and longest wavelengths in the Lyman series of the hydrogen spectrum.
Answer:

For the Lyman series, n₁ = 1 and n₂ = 2,3,4, …..

Rydberg equation \(\vec{v}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right],\)

The difference between nx and n2 is maximum for the shortest wavelength in the Lyman series. Hence n2 = .

∴ \(\frac{1}{\lambda_{\min }}=\bar{v}_{\max }=R\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right]=R=109678 \mathrm{~cm}^{-1}\)

∴ \(\lambda_{\min }=\frac{1}{109678}=9.117 \times 10^{-6} \mathrm{~cm}=911.7\)

Similarly, the difference between n₁ and n₂ is minimal for the longest wavelength in the Lyman series. Hence, n₂ = 2

∴ \(\frac{1}{\lambda_{\max }}=\bar{v}_{\min }=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4} R\)

∴ \(\frac{1}{\lambda_{\max }}=\bar{v}_{\min }=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4} R\)

∴ \(\lambda_{\max }=\frac{4}{3 R}=\frac{4}{3 \times 109678}\)

=\(1215.7 \times 10^{-8} \mathrm{~cm}=1215.7\)

Question 6. Show that the Balmer series appears between 3647A and 6563A in the hydrogen spectrum
Answer:

For Balmer series \(\frac{1}{\lambda}=\bar{v}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right], n=3,4,5, \cdots \infty\)

The Limits Of the Balmer Series Can Be obtained when n = 3 and n = 00 respectively

⇒ \(\frac{1}{\lambda_{\min }}=\bar{v}_{\max }=R\left[\frac{1}{2^2}-\frac{1}{\infty^2}\right]=\frac{R}{4}=\frac{109678}{4} \mathrm{~cm}^{-1}\)

∴ \(\lambda_{\min }=\frac{4}{109678} \mathrm{~cm}=3647 \times 10^{-8} \mathrm{~cm}=3647 \)

Also, \(\frac{1}{\lambda_{\max }}=\bar{v}_{\min }=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5 R}{36}=\frac{5 \times 109678}{36} \mathrm{~cm}^{-1}\)

∴ \(\lambda_{\max }=\frac{36}{5 \times 109678}=6564 \times 10^{-8} \mathrm{~cm}=6564 \)

Question 7. Calculate the wavelength of the spectral line with n2 = 3 in the Lyman series of hydrogen atoms.
Answer:

In Lyman series, = 1

∴ \(\bar{v}=R\left[\frac{1}{1^2}-\frac{1}{3^2}\right]=109678 \times \frac{8}{9} \mathrm{~cm}^{-1}\)

Since n₂ = 3

⇒ \(\lambda=\frac{1}{\bar{v}}=\frac{9}{109678 \times 8}=1.026 \times 10^{-5} \mathrm{~cm}\)

Bohr’s Atomic Model

In 1913, Neils Bohr, an eminent scientist successfully explained the stability of an atom and the cause of the appearance of line spectra with the help of Planck’s quantum theory.

He rectified the defects of Rutherford’s nuclear model and put forward a new atomic model for the hydrogen atom which is known as Bohr’s atomic model or Rutherford-Bohr’s atomic model.

Postulates Of Bohr’s Model Of Atom

The theory, put forward by Bohr regarding the structure of the H-atom, is based on three revolutionary postulates—

Postulate 1:

The electron in the hydrogen atom revolves around the nucleus only in certain selected circular paths (called orbits) which are associated with definite energies. The electrons revolve only in those orbits in which the angular momentum ofthe electron is a whole number multiple of \(\frac{h}{2 \pi},\) i.e., the angular momentum of the electron, \(m v r=n \times \frac{h}{2 \pi}\), Where n= 1,2,3,4 ……….etc, m = mass of the electron, v = velocity of the electron, r = radius of the orbit, h = Planck’s constant. Thus, the angular.

Postulate 2:

When an electron revolves in any selected orbit, it neither emits nor absorbs energy. The energy of an electron in a particular orbit remains constant. These orbits are, therefore, called stationary orbits although the electrons are not stationary. Electrons revolving in a stationary orbit are said to be in the stationary state.

Explanation:

Each stationary orbit is considered as the energy level. The energy of an electron revolving in a stationary orbit is supposed to be the energy of that particular orbit although the orbit has no energy of its own. The energy of the orbit increases with an increase in its distance from the nucleus.

Depending on their distance from the nucleus, these orbits are divided into seven energy levels such as K, L, M, N… etc. and these are designated respectively by the numbers 1, 2, 3, 4 . . . etc.

The numbers signifying the energy levels are known as principal quantum numbers (re). These orbits can be arranged in the increasing order of their energy as follows:

K(n = 1) < L(n = 2) < M(n = 3) < N(n = 4)… etc.

Postulate 3:

When an electron jumps from a higher stationary energy, level to a lower stationary energy level, it emits a fixed amount of energy in the form of radiation. On the other hand, when an electron absorbs a certain amount of energy, it moves to a higher energy level.

Electrons can never occupy a position in between two successive stationary energy levels.

Mathematical expression:

If the frequency of radiation emitted or absorbed is assumed to be v and the energies of the higher and lower stationary orbits as E₂ and E₁ respectively, then the difference in energy, ΔE = E₂-E₁ =hv, [where h = Planck’s constant

[Since E₂ and E₁ have fixed values, the magnitude of v is also fixed. This accounts for the existence of some definite lines in the emission spectra of hydrogen. Energy is involved in the transition of electrons.

Concept of quantization of energy:

The concept of quantization of energy can be best understood by taking the following example: When a person moves down a staircase, his energy changes discontinuously.

He can have only certain definite values of energies corresponding to those of the various steps.

Alternatively, the energy of the person is quantized. However, if he goes down a ramp, his energy changes continuously, having any value corresponding to any point on the ramp. In short, energy is not quantized.

Calculation Of Various Quantities For Hydrogen-Like Atoms Using Bohr’s Theory

The radius of ‘faith orbit:

Suppose the total quantity of positive charge in the nucleus of a hydrogen-like atom is Ze (where Z = atomic number and e = charge of a proton or an electron).

The only electron present in that atom is revolving around the nucleus in a circular orbit of radius, r with velocity, v.

According to Coulomb’s law, the attractive force of the nucleus on the electron i.e… the centripetal force = [where Ze is the total positive charge on the nucleus and e =charge of an electron].

In case of an electron revolving with velocity v, centrifugal force = \(\frac{m v^2}{r}\) [where m = mass of electron and r = radius ofthe orbit], For the stability of an atom, the centripetal force must be equal to the centrifugal force, i.e

⇒ \(\frac{Z e^2}{r^2}=\frac{m v^2}{r} \quad \text { or, } v^2=\frac{Z e^2}{m r}\)  …………………….. (1)

According to Bohr’s theory, the angular momentum of the revolving electron in a stationary orbit is an integral multiple \(\frac{h}{2 \pi} \text { i.e., } m v r=\frac{n h}{2 \pi}\)

[where, n = 1,2,3… etc., are whole numbers]

Or, \(v=\frac{n h}{2 \pi m r}\)…………………….. (2)

Or, \(v^2=\frac{n^2 h^2}{4 \pi^2 m^2 r^2}\) …………………….. (3)

Hence, from equations [1] and [3] we have,

⇒ \(\frac{Z e^2}{m r}=\frac{n^2 h^2}{4 \pi^2 m^2 r^2} \quad \text { or, } \frac{Z e^2}{m}=\frac{n^2 h^2}{4 \pi^2 m^2 r}\)

Or, \(r=\frac{n^2 h^2}{4 \pi^2 m^2} \times \frac{m}{Z e^2} \text { or, } r=\frac{n^2 h^2}{4 \pi^2 m Z e^2}\)

i.e., radius of ‘n’ th orbit \(r_n=\frac{n^2 h^2}{4 \pi^2 m Z e^2}\)…………………….. (4)

Equation [4] indicates the radius of with orbit of the revolving electron.

The velocity of a revolving electron in ‘n’th orbit:

According to Bohr’s theory, the angular momentum of a revolving electron is given by, mvr \(=\frac{n h}{2 \pi} \quad \text { or, } r=\frac{n h}{2 \pi m v}\)

Again, the radius of ‘n’ th orbit, r \(=\frac{n^2 h^2}{4 \pi^2 m Z e^2}\)

Comparing the above two relations we may write,

⇒ \(\frac{n h}{2 \pi m v}=\frac{n^2 h^2}{4 \pi^2 m Z e^2} \quad \text { or, } \frac{1}{v}=\frac{n h}{2 \pi Z e^2} \quad \text { or, } v=\frac{2 \pi Z e^2}{n h}\)

∴ The velocity of an electron revolving in the ‘n’th Bohr orbit is given by

⇒ \(v_n=\frac{2 \pi Z e^2}{n h}\)…………………….. (5)

Substituting the values of n = 1, 2, 3, etc., in this equation, the velocity of revolving electrons in different Bohr orbits can be determined.

The velocity of resolving electron in the ‘ n ‘th orbit of the H atom.

= \(\frac{2 \times \frac{22}{7} \times 1 \times\left(4.8 \times 10^{-10}\right)^2}{n \times\left(6.626 \times 10^{-27}\right)}=\frac{2.186 \times 10^8}{n} \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

Relation between the velocities of an electron in ‘n’th and first orbit:

Velocity of an electron revolving in the’ n ‘th orbit.

⇒ \(\left(v_n\right)=\frac{2 \pi \mathrm{Z} e^2}{n h}\) and that in the first orbit, (v1) \(=\frac{2 \pi Z e^2}{1 \times h}\)

∴ \(\frac{v_n}{v_1}=\frac{2 \pi Z e^2}{n h} \times \frac{1 \times h}{2 \pi Z e^2}=\frac{1}{n}\)

∴ \(\frac{v_n}{v_1}=\frac{2 \pi Z e^2}{n h} \times \frac{1 \times h}{2 \pi Z e^2}=\frac{1}{n}\)

or, \(v_n=v_1 \times \frac{1}{n}\)

Relation between the radius of ‘n’th orbit and the first orbit.

Radius of ‘n th orbit \(\left(r_n\right)=\frac{n^2 h^2}{4 \pi^2 m Z e^2}\) and the radius of the first orbit \(\left(r_1\right)=\frac{1^2 \cdot h^2}{4 \pi^2 m Z e^2}=\frac{h^2}{4 \pi^2 m Z e^2}\)

∴ \(\frac{r_n}{r_1}=\frac{n^2 h^2}{4 \pi^2 m Z e^2} \times \frac{4 \pi^2 m Z e^2}{h^2}=n^2\)

or, rn = r₁ × n²

The total energy of an electron revolving in ‘n’th orbit:

The total energy of electron revolving in ‘ n ‘th orbit, En = Kinetic energy + Potential energy

E_n = Kinetic energy + Potential energy

= \(\frac{1}{2} m v^2+\left(-\frac{Z e^2}{r}\right)=\frac{1}{2} m \times \frac{Z e^2}{m r}-\frac{Z e^2}{r}\)

= \(\frac{1}{2} \frac{Z e^2}{r}-\frac{Z e^2}{r}=-\frac{Z e^2}{2 r}=-\frac{Z e^2}{2} \times \frac{1}{r}\)

∴ \(v^2=\frac{Z e^2}{m r}\)

Since \(v^2=\frac{Z e^2}{m r}\)

= \(-\frac{Z e^2}{2} \times \frac{4 \pi^2 m Z e^2}{n^2 h^2}\)

r = \(\frac{n^2 h^2}{4 \pi^2 m Z e^2}\)…………………….. (6)

Equation [6] Indicates the energy of the electron revolving in the orbit of a Hydrogen- like

For hydrogen atom Z =1

∴ For hydrogen atom, \(E_n=-\frac{2 \pi^2 m e^4}{n^2 h^2}\)…………………….. (7)

Substituting the values of m, e, n, and R in equation number[7] we have

⇒ \(E_n=-\frac{2 \times(22 / 7)^2 \times\left(9.108 \times 10^{-28} \mathrm{~g}\right) \times\left(4.8 \times 10^{-10} \mathrm{esu}\right)^4}{n^2 \times\left(6.626 \times 10^{-27} \mathrm{erg} \cdot \mathrm{s}\right)^2}\)

= \(-\frac{2.176 \times 10^{-11}}{n^2} \mathrm{erg} \cdot \text { atom }^{-1}\)

= \(-\frac{2.176 \times 10^{-18}}{n^2} \mathrm{~J} \cdot \mathrm{atom}^{-1}=-\frac{1312}{n^2} \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \quad \cdots[7 \mathrm{a}]\)

Energies E₁,  E₂, E₃, etc., are calculated by putting n – 1, 2, 3, etc. in equation [7a] and are represented schematically in.

It is observed from these values that successive energy levels are not equidistant. The energy gap between two successive levels decreases with an increase in distance from the nucleus.

When an electron is at an infinite distance from the nucleus, its potential energy is zero.

Work done to bring the electron from an infinite distance to a distance V concerning the nucleus gives the measure of its potential energy.

The force of attraction between the nucleus and the electron is I given by, F = (applying Coulomb’s law).

∴ The potential energy of the electron at a distance ‘r’ from the nucleus \(=\int_{r=\infty}^{r=\infty} \frac{Z e^2}{r^2} d r=-\frac{Z e^2}{r}\)

Electronic energy as negative energy:

When an electron is at an infinite distance (n = ∞) from the nucleus, it experiences no force of attraction from the nucleus. Hence, the energy of the electron at an infinite distance from / the nucleus is taken as zero (E_∞ = 0).

• When the electron moves towards the nucleus, it experiences a force of attraction from the nucleus.
• As a result, some energy is released and the energy of the electron becomes negative. Due to the release of energy, the stability of the electron occupying a fixed orbit in the atom is increased.
• As the electron comes closer to the nucleus, it experiences a greater force of attraction and hence more energy is released, thereby making the electronic energy more negative.

All these facts are per the expression of electronic energy,

⇒ \(E_n=-\frac{2 \pi^2 m Z^2 e^4}{n^2 h^2}\) As n decreases, the energy of the electron decreases (because it is associated with a -ve sign) and hence the stability of the electron in the atom increases.

Thus, the negative value ofthe electronic energy in an atom signifies that the stability of the electron increases as it occupies orbits closer to the nucleus.

The energy is minimal when it revolves around the nucleus occupying the first orbit (i.e., n = 1).

The ratio of the energies of an electron revolving in the V th orbit and the first orbit of the hydrogen-like atom:

⇒ \(\frac{E_n}{E_1}=\left(-\frac{2 \pi^2 m Z^2 e^4}{n^2 h^2}\right) \times \frac{1}{\left(-\frac{2 \pi^2 m Z^2 e^4}{1^2 \times h^2}\right)}=\frac{1}{n^2}\)

∴ \(E_n=E_1 \times \frac{1}{n^2}\) …………………….. (8)

The radius of the first orbit of the hydrogen atom:

From equation number [4], it is seen that the radius of the ‘ n ’ th orbit of a hydrogen-like atom, \(r_n=\frac{n^2 h^2}{4 \pi^2 m Z e^2}\)

Where, h = Planck’s constant = 6.626 × 10^-27erg-s , m = mass of electron = 9.1× 10^-28g, e =charge of electron = 4.8 × 10^-10esu]

Derivation of Rydberg’s equation

When an electron of a hydrogen atom jumps from ‘ n₂ ‘th orbit to‘ ‘th orbit, let the frequency of emitted radiation be v. According to Bohr’s theory,

⇒ \(h v=E_{n_2}-E_{n_1}=\left(-\frac{2 \pi^2 m e^4}{n_2^2 h^2}\right)-\left(-\frac{2 \pi^2 m e^4}{n_1^2 h^2}\right)[\text { for } \mathrm{H}, Z=1]\)

=\(\frac{2 \pi^2 m e^4}{h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

⇒ \(\frac{2 \pi^2 m e^4}{h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) …………………….. (9)

∴ \(v=\frac{2 \pi^2 m e^4}{h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) …………………….. (10)

or, \(\frac{c}{\lambda}=\frac{2 \pi^2 m e^4}{h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

Since \(h v=E_{n_2}-E_{n_1}=\left(-\frac{2 \pi^2 m e^4}{n_2^2 h^2}\right)-\left(-\frac{2 \pi^2 m e^4}{n_1^2 h^2}\right)\)

⇒ \(\frac{2 \pi^2 m e^4}{h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

∴ \(v=\frac{2 \pi^2 m e^4}{h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

Or, \(\frac{c}{\lambda}=\frac{2 \pi^2 m e^4}{h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

Since \(v=\frac{c}{\lambda}\)

Or, \(\frac{1}{\lambda}=\frac{2 \pi^2 m e^4}{c h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

Since \(\frac{1}{\lambda}=\bar{v}\) = wave Number ]

or, \(\bar{v}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) …………………….. (11)

R is a constant known as Rydberg’s constant.

∴ Rydberg’s constant,R \(=\frac{2 \pi^2 m e^4}{c h^3}\) …………………….. (12)

Note that earlier equation in article[11]no is2.4.4.comparable with equation [2]

Equation [11] is Rydberg’s equation which is related to the line spectra of the hydrogen atom:

Expression for wave number of spectral lines involving one-electron ions is given by

⇒ \(\bar{v}=Z^2 \times R\left(\frac{1}{n^2}-\frac{1}{n_2^2}\right) \text {, where } Z=\text { atomic number. }\)

⇒  \(\bar{v}=Z^2 \times R\left(\frac{1}{n^2}-\frac{1}{n_2^2}\right) \text {, where } Z=\text { atomic number. } \)

Calculation of the value of Rydberg’s constant: Substituting the values of n, m, e, c and h in equation [12] we get,

Rydberg’s constant, \(R=\frac{2 \pi^2 m e^4}{c h^3}\)

= \(\frac{2 \times(3.14)^2 \times\left(9.108 \times 10^{-28} \mathrm{~g}\right) \times\left(4.8 \times 10^{-10} \mathrm{esu}\right)^4}{\left(2.9979 \times 10^{10} \mathrm{~cm} \cdot \mathrm{s}^{-1}\right) \times\left(6.626 \times 10^{-27} \mathrm{erg} \cdot \mathrm{s}\right)^3}\)

=109737cm^-1

The experimentally measured value of Rydberg’s constant is 109677 cm-1. This value of R agrees fairly well with that ofthe calculated value.

Wavelength of radiation emitted due to electronic transition from third to second Bohr orbit: Rydberg’s equation related to the frequency ofthe spectral lines of H -atom-

\(\bar{v}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

R= 109677 cm^-1

The wavelength of radiation (A) emitted due to the transition of an electron of a hydrogen atom from the third orbit (n = 3) to the second orbit (n = 2) can be calculated with the help of this equation.

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\)

Since \(\bar{v}=\frac{1}{\lambda} \text { and } n_1=2, n_2=3\)

Or, \(\frac{1}{\lambda}=109677 \times\left(\frac{9-4}{36}\right)=\frac{109677 \times 5}{36} .\)

∴ \(\lambda=\frac{36}{109677 \times 5}=6564.7 \times 10^{-8} \mathrm{~cm}=6564.7 \)

As the wavelength of visible light lies between 4000-8000 A, the line spectrum appears in the visible region. It represents the Balmer series.

The ionization potential of the H-atom from Rydberg’s equation:

The ionization potential of hydrogen is defined as the energy required to transfer the electron from the first orbit (n = 1) of a hydrogen atom to an infinite distance to form an H+ ion.

Or, \(\frac{1}{\lambda}=\frac{2 \pi^2 m e^4}{c h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\left[ \bar{v}=\frac{1}{\lambda} \text { and } R=\frac{2 \pi^2 m e^4}{c h^3}\right]\)

Or, \(\frac{c}{\lambda}=\frac{2 \pi^2 m e^4}{h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

Or, \(v=\frac{2 \pi^2 m e^4}{h^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

Since \(v=\frac{c}{\lambda}\)

Or, \(h v=\frac{2 \pi^2 m e^4}{h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

Or, \(\Delta E=\frac{2 \pi^2 m e^4}{h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

[since E =Hv]

Substituting n₁ = 1 and n₂ = infinite in the equation, the value of the ionization potential of a hydrogen atom can be obtained.

∴ The ionization potential of H-atom (in CGS unit)

⇒ \(\frac{2 \times(3.14)^2 \times\left(9.108 \times 10^{-28}\right) \times\left(4.8 \times 10^{-10}\right)^4}{\left(6.626 \times 10^{-27}\right)^2} \times\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)\)

= 2170 × 10^-14 × (1 -0) =2170 × 10^-14erg-atom^-1

= 2170 × 10^-14× 6.24 × 10¹¹ eV-atom^-1

= 13.6 eV-atomr1 =13.6 x 1.6022 × 10^-19 atom^-1

= 13.6 x 1.6022 x 10-19 x 6.022 × 10-19-mol^-1

= 1312kJ-mol^-1

[since 1 erg = 6.24 × 10¹¹eV]

The ionization energy of other one-electron species is given by, \(I E=(I E)_{\mathrm{H}} \times \frac{Z^2}{n^2}, \text { where }(I E)_{\mathrm{H}}\) = Inonisation Energy Of H-atom, Z = Atmotic number,n = principal Quantum number from which the electron is removed.

Number of revolutions of an electron ptt second:

Velocity of an electron revolving in the ‘n’ th orbit \(v_n=\frac{2 \pi Z e^2}{n h}\)

Perimeter of the orbit = 2nrn \(=\frac{2 \pi n^2 h^2}{4 \pi^2 m Z e^2}=\frac{n^2 h^2}{2 \pi m Z e^2}\)

∴ Number of revolutions of an electron in the ‘ n ’th orbit per second = \(=\frac{\text { Velocity }}{\text { Perimeter }}\)

= \(\frac{2 \pi Z e^2}{n h} \div \frac{n^2 h^2}{2 \pi m Z e^2}\)

= \(\frac{2 \pi Z e^2}{n h} \times \frac{2 \pi m Z e^2}{n^2 h^2}=\frac{4 \pi^2 m Z^2 e^4}{n^3 h^3}\)

Explanation of hydrogen spectrum by Bohr’s theory

Bohr’s theory furnishes a logical explanation of the origin ground state) of six series of spectral lines in the atomic spectrum of hydrogen.

• Under ordinary conditions, the revolving electron ESS Merits Of Bohr’s atomic model in the hydrogen atom exists in the lowest energy state (n = 1 ), known as the ground state.
• If energy (Example: heat, light) is available for absorption, then the electron present in the kT-shell (n=1) absorbs the energy and moves to a higher energy level [L(n = 2), Af(n = 3), N(n = 4)…..etc.]. This state is known as the excited state.
• Since in a sample of hydrogen, there are a large number of atoms, the electrons in different H-atoms absorb different amounts of energies (quanta) and are promoted to different higher energy levels viz., n = 2, 3, 4,
• When the source of energy is removed, electrons jump back to any lower energy level by emitting energy (quanta) as radiations of different frequencies which give rise to different lines in the spectrum. When electrons from orbits n = 2,3,4, ••• jump back to orbit n = 1, a series of lines, known as the Lyman series is obtained.
• Similarly, if electrons from higher energy levels n = 3, 4, 5, 6, . . . jump back to energy level n = 2, the Balmer series will appear in the spectrum.

In the same way, when the electrons return to the third energy level (n = 3), Paschen series, to the fourth energy level (n = 4), Brackett series, to the fifth energy level (n = 5), Pfimd series and to the sixth energy level (n = 6), Humphreys series is obtained in the spectrum. The formation of six spectral series is illustrated below.

Thus, it is clear that a hydrogen atom containing only one electron can form several series of lines in its atomic spectrum.

Merits Of Bohr’s atomic model

Stability of an atom:

One of the main postulates of Bohr’s theory regarding the structure of an atom is that an electron while revolving in its stationary orbit does not emit any energy.

So, the electron moving continuously around the nucleus in a fixed orbit will never fall on the nucleus i.e., this postulate offers a reasonable explanation regarding the stability of an atom by eliminating the main defect of Rutherford’s nuclear model.

Line spectrum of H-atom:

Bohr’s atomic model can explain the discontinuous spectra formed in different one-electron systems such as H – atoms, He⁺ ion, Li²⁺ ion, etc.

The radius of the first orbit of H-atom:

From Bohr’s theory, it has been possible to estimate the value of the radius of the first orbit (n = 1) of a hydrogen atom.

Its value is 0.529 × 10^-8 cm or 0.529 Å. This value has been calculated later by other methods and found to be very close to that obtained based on Bohr’s theory.

Principal quantum number:

The idea of principal quantum numbers was first established through Bohr’s model.

The energy of an electron:

Based on Bohr’s theory, the energy of an electron revolving in any quantum level can be calculated and from this, it has also been possible to estimate the energy emitted or absorbed due to the transition of an electron from one energy level to another.

Thus, the relative positions of different lines in atomic spectra can be explained.

Rydberg’s constant and its value:

With the help of Bohr’s theory, mathematical expression as well as the value of Rydberg’s constant can be determined.

Limitations of Bohr’s atomic model

Spectra of atoms or ions having two or more electrons:

(i.e., multi-electron system) cannot be explained with the help of Bohr’s theory.

1. When spectroscopes with high resolving powers were used, it was found that each line in the spectrum was split into several closely spaced lines (called fine structure) which could not be explained by Bohr’s theory.
2. Bohr’s theory could not explain the splitting of spectral lines under the influence of magnetic field (Zeeman effect) or electric field (Stark effect) i.e., the formation of fine structure of atomic spectra.
3. Bohr’s atomic model is two-dimensional and hence, fails to give an idea about the actual three-dimensional electronic model of the atom.
4. According to de Broglie (1923), a tiny particle like an electron has dual character i.e., an electron has a particle as well as wave nature. Bohr treated the electron only as a particle, i.e., Bohr’s model ignored the dual character of the electron.
5. According to Bohr’s atomic model, an electron moves around the nucleus along a fixed circular path with a definite velocity.
6. However, according to Heisenberg’s uncertainty principle, it is impossible to simultaneously measure both the position and velocity (or momen- turn) of a subatomic particle-like electron with absolute accuracy at a particular instant.
7. Thus, Bohr’s theory directly contradicts Heisenberg’s uncertainty principle.

Sommerfeld’s modification Of Bohr’s Theory Idea Of Elliptical Orbits

To explain the fine structure of spectral lines in the hydrogen spectrum.

Sommerfeld (in 1915) extended Bohr’s theory and proposed that—

1. An electron moves around a positively charged nucleus in different elliptical orbits in addition to circular orbits.
2. When the electron moves in an elliptical orbit, the nucleus remains at one of the foci.
3. When the electron moves in a circular path, the angle of revolution changes while the distance from the nucleus remains the same.
4. However, in an elliptical motion, both the angle of revolution (or radial angle) and the distance (‘ r’ or radial distance) ofthe electron from the nucleus change.
5. In order to describe an elliptical path, Sommerfeld introduced the concept of a second quantum number, called the azimuthal quantum number, denoted by the letter ‘k’ in addition to the principal quantum number n. These two quantum numbers are related to each other by the equation,

⇒ \(\frac{n}{k}=\frac{\text { length of major axis }(a)}{\text { length of minor axis }(b)}\)

⇒ \(\frac{k}{n}=\frac{b}{a}=\sqrt{1-\epsilon^2}[\epsilon=\text { eccentricity }]\)

When n = k, i.e., length of major axis = length of major axis, the ellipse reduces to a circle. Thus, a circular orbit is a special case of the elliptical orbit.

For each value of n, k may have n different values. These are 1, 2, 3…, n. Thus for the Bohr orbit with n – 4, there are four Sommerfeld orbits (three elliptical orbits + one circular orbit) with k = 1, 2, 3, and 4. When k = 4, the elliptical orbit reduces to a circular orbit.

Sommerfeld thus introduced the concept of subshells in a principal quantum shell. The difference in energy between any two subshells is quite small.

The fine structure of the hydrogen spectrum can be explained by assuming several transitions between several subshells, each characterized by different values of k but the same value of n.

To explain the splitting of spectral lines under the influence of the electric or magnetic fields, Sommerfeld suggested that the electronic orbits may lie in different planes.

Discussion:

• Bohr’s idea of ‘electrons moving in circular ‘ orbits’ was extended by Sommerfeld by introducing the concept of elliptical orbits.
• According to Einstein’s theory of relativity, the equation expressing the relation between rest mass and mass of a moving particle is given by \(=m_0 /\left(1-\frac{v^2}{c^2}\right)^{1 / 2}\) between rest mass and mass of a moving particle is given by m- mass of the moving particle, v=velocity of the moving particle, c = velocity of light).
• According to Kepler’s law, the velocity of a body moving in an elliptical orbit is maximum at the perihelion (closest to the focus) and minimum at the aphelion (farthest from the focus).
• Therefore, the mass of an electron moving in an elliptical orbit is maximum at the perihelion and minimum at the aphelion.
• Consequently, the electron is compelled to deviate from its original orbit to a new and identical elliptical path that lies in the same plane. Thus, the elliptical path continues its precession slowly surrounding the nucleus.
• Due to this continuous precessional motion of the elliptical orbit, the energy of the electron undergoes a slight change depending on the position of the elliptical orbit.
• The fine structure of spectral lines observed in the atomic spectra actually originates from the slight difference in energy ofthe electron.

Merits of Sommerfeld’s modification

• Due to the introduction of azimuthal quantum numbers, the concept of multiple elliptical orbits corresponding to the same principal quantum number came into being.
• Due to multiple values of azimuthal quantum number for the same value of the principal quantum number, the transition of an electron can occur from one quantum level to another which may result in multiple lines in the spectrum. Thus the fine structure of the spectrum can be explained by Sommerfeld’s theory.
• The subshells belonging to the same principal energy levels have different eccentricities. The more is the eccentricity of a subshell, the more is the opportunity for the electrons residing in that subshell to be closer to the nucleus. Thus, the subshell wifi possesses more penetration power and the corresponding electrons belonging to the subshell will have more power to shield the nuclear charge.

Demerits of Sommerfeld’s Modification.

• Like Bohr’s atomic model, the atomic structure proposed by Sommerfeld is also two-dimensional. this model also fails to explain the three-dimensional model of atomic structure.
• Sommerfeld’s modification also fails to explain the spectra of multi-electron atoms.
• The relative intensities of the fines in the spectra cannot be explained by using this theory.
• Sommerfeld’s atomic model contradicts Heisenberg’s uncertainty principle because according to the latter, it is impossible to simultaneously determine the exact position and exact momentum of an electron.

Numerical Examples

Question 1. Calculate the energy associated with the fifth orbit of H-atom, if the energy associated with the first orbit is 2.17 x 1018J.atom-1
Answer:

⇒ \(E \propto \frac{1}{n^2} \text { or } E=\frac{K}{n^2}\)

∴ \(E_1=\frac{K}{1^2}(\text { for } n=1) \text { and } E_5=\frac{K}{5^2}(\text { for } n=5)\)

∴ \(\frac{E_5}{E_1}=\frac{1}{25} \quad \text { or, } E_5=\frac{E_1}{25}=\frac{-2.17 \times 10^{-18}}{25}\)

=-8.68 × 10^-20 J atom^-1

Question 2. The radius of the first orbit of the H-atom is 0.53A. Find the radius of the fifth orbit.
Answer:

⇒ \(r_n=r_1 \times n^2\)

∴ \(r_5=r_1 \times 5^2=0.53 \times 25=13.25\)

Question 3. Will there be regular variations in the energy associated with successive principal quantum numbers of hydrogen-like atoms?
Answer:

No, the variation will not be regular because, energy Associated with an element with an electron in n-th Orbit, En = \(=\frac{-2 \pi^2 m Z^2 e^4}{n^2 h^2}\)

For a hydrogen-like atom, \(\frac{-2 \pi^2 m Z^2 e^4}{n^2 h^2}\) = constant

∴ \(E_n=\frac{K}{n^2}\)

∴ \(E_1=\frac{K}{1^2}, E_2=\frac{K}{2^2}, E_3=\frac{K}{3^2}, E_4=\frac{K}{4^2}, \cdots \text { etc. }\)

From the values of the energy associated with the electrons, it is clear that the variation of energy in successive principal quantum numbers of hydrogen-like atoms is not regular. the spacing between the energy levels decreases, as we move outwards from the nucleus.

Question 4. Energy associated with the n-th orldt of 11 -atom Is I ‘l (i given by the expression, \(E_n=-\frac{13.6}{n^2}\). Show that \(E_{(n+1)}-E_n=\frac{13.6 \times 2}{n^3} \mathrm{eV} \text {, when ‘ } n \text { ‘ is very large. }\)
Answer:

⇒ \(E_{(n+1)}-E_n=\left[-\frac{13.6}{(n+1)^2}-\left(-\frac{13.6}{n^2}\right)\right]\)

= \(\left[\frac{13.6}{n^2}-\frac{13.6}{(n+1)^2}\right] \mathrm{eV}=\frac{13.6(2 n+1)}{n^2(n+1)^2} \mathrm{eV}\)

If the value of n is very large, then (2n + l)=2n and(n+l)=n

∴ \(E_{(n+1)}-E_n=\frac{13.6 \times 2 n}{n^2 \times n^2}=\frac{13.6 \times 2}{n^3} \mathrm{eV}\)

Question 5. If an electron Is promoted from the first orbit to the third orbit of a hydrogen atom, then by how many times will the radius of the orbit be increased?
Answer:

The radius ofthe n -th orbit of H -atom \(r_n=\frac{n^2 h^2}{4 \pi^2 m e^2}\)

Radius ofthe first orbit \(\left(r_1\right)=\frac{1^2 \times h^2}{4 \pi^2 m e^2}\)

Radius of the third orbit (r3) \(=\frac{3^2 \times h^2}{4 \pi^2 m e^2}\)

∴ \(\frac{r_3}{r_1}=\frac{3^2 \times h^2}{4 \pi^2 m e^2} \times \frac{4 \pi^2 m e^2}{h^2}=9 \text { i.e., } r_3=9 \times r_1 \text {. }\)

∴ The radius ofthe orbit will be increased by 9 times

Question 6. If an electron drops from the third orbit (n = 3) to the first orbit (n = 1) of the H-atom, what will be the frequency and wavelength of the radiation emitted? What would have happened if the electron jumped from the first orbit to the third orbit?
Answer: Let the frequency of the emitted radiation be v.

Then \(v=R \times c \times\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \mathrm{cps}\)

=\(109677 \times 3 \times 10^{10} \times\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\)

=\(329031 \times 10^{10} \times\left(1-\frac{1}{9}\right)\)

Since R = 109677 cm-1 and c = 3 × 10¹⁰ em s^-1 ]

= 3.29 × 10¹⁵ × 0.889 = 2.925 ×10¹⁵  cps.

The wavelength of the emitted radiation,

⇒ \(\lambda=\frac{c}{v}=\frac{3 \times 10^{10}}{2.925 \times 10^{15}}=1.025 \times 10^{-5} \mathrm{~cm}=1025 \mathrm{~A}\)

When an electron jumps from the 1st orbit to the 3rd orbit then energy is absorbed.

The frequency and wavelength of absorbed radiation are the same i.e., the frequency and wavelength of absorbed radiation will be 2.925 x 1015 cps and 1025A respectively.

Quantities 7. Find the wavelength (in angstrom) of the photon| emitted when an electron jumps from the second Bohr orbit to the first Bohr orbit of the hydrogen atom. The ionization potential of the hydrogen atom in its ground energy state =2.17 x 10-11 erg-atom-1
Answer:

The energy of the electron in the electron in first orbit (n = 1 ) of hydrogen atom \(E_1=-\frac{2 \pi^2 m e^4}{h^2}\)

The ionization potential of hydrogen atom le., the energy required to move the electron from n = 1 energy level to an infinite distance

=2.17 × 10¹¹ erg.

∴ Energy of electron in1st (n = 1) orbit =-2.17 × 10¹¹ erg

\(\text { i.e., }-\frac{2 \pi^2 m e^4}{h^2}=-2.17 \times 10^{-11}\)

=1221 × 10^-8 cm =1221Å

Since 1Å = 10^-8 cm]

Question 8. Determine Balmer the series wavelengths of H and H Lines in the Balmer Series [ R= 109670 cm^-1 ]
Answer:

For Balmer series:

⇒ \(\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\)

For Balmer Series n₁ =2 And For The Line Hα, n₂ = 3

∴ \(\frac{1}{\lambda}=109670\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=109670\left(\frac{1}{4}-\frac{1}{9}\right)\)

∴ \(\lambda=\frac{36}{109670 \times 5}=6.565 \times 10^{-5} \mathrm{~cm}\)

For the line H_β, n₂ = 4

⇒ \(\frac{1}{\lambda}=109670\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=109670\left(\frac{1}{4}-\frac{1}{16}\right)\)

∴ \(\lambda=\frac{16}{109670 \times 3}=4.863 \times 10^{-5} \mathrm{~cm}\)

Question 9. Find the velocity of the electron revolving in the third orbit of the hydrogen atom. Also, determine the number of revolutions of the electron per second around the nucleus.
Answer:

The velocity of the electron revolving in the 3rd orbit around the nucleus of the hydrogen atom,

\(v_3=\frac{2 \pi e^2}{3 \times h}=\frac{2 \times 3.14 \times\left(4.8 \times 10^{-10}\right)^2}{3 \times\left(6.627 \times 10^{-27}\right)}=7.278 \times 10^7 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

The radius of the n ’ th orbit of H-atom, rn \(=\frac{n^2 h^2}{4 \pi^2 m e^2}\)

∴ Radius ofthe third orbit (r³) \(=\frac{3^2 h^2}{4 \pi^2 m e^2}\)

= \(\frac{9 \times\left(6.627 \times 10^{-27}\right)^2}{4 \times(3.14)^2 \times\left(9.108 \times 10^{-28}\right) \times\left(4.8 \times 10^{-10}\right)^2}\)

= 4.77 × 10^-8 cm [m = 9.108 × 10^-28 g]

Circumference ofthe third orbit

= 27πr³ = 2 × 3.14 × 4.77 × 10^-8 =2.99 × 10^-7cm

∴ Number of revolutions of electron per second around the nucleus \(=\frac{\text { Velocity }}{\text { Circumference of the orbit }}\)

= \(\frac{7.278 \times 10^7}{2.99 \times 10^{-7}}=2.43 \times 10^{14}\)

Question 10. Prove that the velocity of an electron revolving in the 1st orbit of H-atom is nearly 10^-2 times that of light.
Answer:

According to Bohr’s theory, mvr \(=\frac{n h}{2 \pi}\)

The radius of‘ n ‘th orbit, r \(=\frac{n^2 h^2}{4 \pi^2 m Z e^2}\)

From equations [1] and [2], we have \(\frac{n h}{2 \pi m \nu}=\frac{n^2 h^2}{4 \pi^2 m Z e^2}\)

⇒ \(\text { or, } \frac{1}{v}=\frac{n h}{2 \pi \mathrm{Z} e^2} \text { or, } v=\frac{2 \pi \mathrm{Z} e^2}{n h}\)

For hydrogen atom, Z = 1 ; Hence, \(\nu=\frac{2 \pi e^2}{n h}\)

For the first orbit, n = 1 , e = 4.8 x 10^-1 esu,

h = 6.627 × 10^-27 erg-s

∴ The velocity of moving electron in the first orbit

⇒ \(v=\frac{2 \times 3.14 \times\left(4.8 \times 10^{-10}\right)^2}{1 \times 6.627 \times 10^{-27}}\)

= 2.183 × 10⁸ =(2.183 × 10^-10) × 10^-2 cm-s^-1

Again, the velocity of light =3 × 10¹⁰ cm-s^-1

Hence, the velocity of an electron revolving in the first orbit of the H-atom is almost 10-2 times ofthe velocity of light.

Question 11. If the energy of the first Bohr orbit is – 13.58 eV, then what will be the energy of the third Bohr orbit?
Answer:

Energy of n -th Bohr-orbit of H-atom, En \(=-\frac{2 \pi^2 m e^4}{n^2 h^2}\)

∴ Energy first Bohr orbit, E1 \(=-\frac{2 \pi^2 m e^4}{1^2 \times h^2}\)

∴ Energy of third Bohr orbit, E3 \(=-\frac{2 \pi^2 m e^4}{3^2 \times h^2}\)

Hence \(\frac{E_3}{E_1}=\frac{2 \pi^2 m e^4}{9 \times h^2} \times \frac{h^2}{2 \pi^2 m e^4}=\frac{1}{9}\)

∴ \(E_3=\frac{1}{9} \times E_1=\frac{1}{9} \times(-13.58) \mathrm{eV}=-1.509 \mathrm{eV}\)

Question 12. The velocity of an electron revolving in a certain orbit of the H-atom is \(\frac{1}{275}\) times the velocity of light. Find the orbit in which the electron is revolving.
Answer: Let the electron is revolving in the n-th orbit.

The velocity ofthe electron in the n -th orbit, \(V_n=\frac{2 \pi Z e^2}{n h}\)

[since Z=1

e= 4.8 × 10¹⁰ esu

h = 6.626 × 10^-27erg.s]

= \(\frac{2 \times 3.14 \times\left(4.8 \times 10^{-10}\right)^2}{n \times 6.626 \times 10^{-27}}\)

Since Z = 1

e = 4.8 × 10^-10esu

h= 6.626 × 10^-27 erg.s

= \(\frac{2.18 \times 10^8}{n} \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

Given, the velocity of revolution of the electron

= \(\frac{1}{275} \times\left(3 \times 10^{10}\right) \mathrm{cm} \cdot \mathrm{s}^{-1}=1.09 \times 10^8 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

∴ \(\frac{2.18 \times 10^8}{n}=1.09 \times 10^8\)

or, \(n=\frac{2.18 \times 10^8}{1.09 \times 10^8}=2\)

∴ The electron is revolving in the 2nd orbit of H-atom.

Question 13. According to Bohr’s theory, the energy of an electron in n-th, \(E_n=-\frac{21.76 \times 10^{-19}}{n^2} \mathrm{~J}\) J. Find the longest Wavelength of radiation required to remove one electron from the 3rd orbit of He+ ion.
Answer:

The energy of the electron in n -th orbit of H-atom,

⇒ \(E_n=-\frac{2 \pi^2 m e^4}{n^2 h^2}\)

[Since Z=1]

∴ \(-\frac{2 \pi^2 m e^4}{n^2 h^2}=-\frac{21.76 \times 10^{-19}}{n^2}=21.76 \times 10^{-19}\)

Now the energy ofthe electron in’ n ‘th orbit of He+ ion.

⇒ \(E_n^{\prime}=-\frac{2 \pi^2 Z^2 m e^4}{n^2 h^2}=-\frac{2 \pi^2 \times 2^2 \times m e^4}{n^2 h^2}\)

Since Z=2.

Therefore, E’n \(=-\frac{2 \pi^2 m e^4}{n^2 h^2} \times 4\)

∴ \(E_3=-\frac{2 \pi^2 m e^4}{3^2 \times h^2} \times 4\)

Removal of an electron from the 3rd orbit means a transfer of electron from the 3rd orbit to n = infinite.

∴ Energy ofthe electron at an infinite distance

\(E_{\infty}=-\frac{2 \pi^2 m e^4}{\infty^2 \times h^2} \times 4=0\)

Therefore the energy required to remove the electron,

⇒ \(\Delta E=E_{\infty}-E_3=0-\left(-\frac{2 \pi^2 m e^4}{3^2 \times h^2} \times 4\right)=\frac{2 \pi^2 m e^4}{h^2} \times \frac{4}{9}\)

∴ \(\Delta E=21.76 \times 10^{-19} \times \frac{4}{9} \mathrm{~J}\)

Since \(\frac{2 \pi^2 m e^4}{h^2}=21.76 \times 10^{-19}\)

We know, AE = \(h \times \frac{c}{\lambda}\)

⇒ \(\text { or, } \quad \frac{\left(6.627 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\lambda}=21.76 \times 10^{-19} \times \frac{4}{9}\)

[Since h = 6.627 × 10^-34j-s and c = 3 × 10⁸m-s^-1 ]

Hence \(\lambda=\frac{9 \times\left(6.627 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{21.76 \times 10^{-19} \times 4}\)

= 2056 × 10^-19m = 2056 Å

1 Å = 10^-10 m = 20556 Å

[Since 1 Å = 10-10m]

Question 14. Calculate the energy emitted when electrons of 1.0g. An og atom of hydrogen undergoes transition emitting the spectra) line of lowest energy in the visible region of Its atomic spectrum [R_H = 1.1 ×  10⁷m^-1].
Answer:

The given spectral line in the visible region corresponds to the Balmer series. For the Balmer series,

⇒ \(\vec{v}=R_M\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\)

For Lowest Energy, n₁ =2 and n₂= 3.

∴ \(\frac{1}{\lambda}=R_H\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=1.1 \times 10^7 \times \frac{5}{36}\)

The amount of energy emitted during the transition of an electron, E = hv \(=h \times \frac{c}{\lambda}=h \times c \times \frac{1}{\lambda}\)

= \(6.626 \times 10^{-34} \times 3 \times 10^8 \times\left(1.1 \times 10^7 \times \frac{5}{36}\right)=3.037 \times 10^{-19} \mathrm{~J}\)

∴ The energy emitted when electrons of gram-atom of hydrogen undergo transition =N0 x hv

= 6.022  ×  10²³× (3.037 × 10^-19) J = 18.28 ×  10⁴ J = 182.8 kj

Question 15. The atomic spectrum of hydrogen contains a series of four lines having wavelengths 656.5, 486.3, 434.2, and 410.3 nm. Determine the wavelength of the next line in the same series [R_H = 109678 cm^-1].
Answer:

As the given wavelengths lie in the visible region, they should belong to the Balmer series. For Balmer series, nx = 2. The value of n1=2 the shortest wavelength (410.3nm) can be determined using the equation,

⇒\(\bar{v}=\frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{n_2^2}\right)\)

Or, \(\frac{1}{410.3 \times 10^{-7} \mathrm{~cm}}=109678 \mathrm{~cm}^{-1}\left(\frac{1}{4}-\frac{1}{n_2^2}\right)\)

Or, \(\frac{1}{n_2^2}=\frac{1}{4}-\frac{1}{410.3 \times 10^{-7} \times 109678} \quad \text { or, } \frac{1}{n_2^2}=0.25-0.22\)

Or, \(n_2^2=\frac{1}{0.03}=33.33\)

∴ N₂ =6

Thus, the next line is obtained as a result of the transition of an electron from n₂ = 7 to = 2.

∴ \(\frac{1}{\lambda}=109678\left(\frac{1}{2^2}-\frac{1}{7^2}\right) \mathrm{cm}^{-1}=25181 \mathrm{~cm}^{-1}\)

Or, \(\lambda=\frac{1}{25181}=3.971 \times 10^{-5} \mathrm{~cm}=397.1 \mathrm{~nm}\)

Question 16. The angular momentum of an electron in a Bohr’s orbit of a hydrogen atom is 3.1655× 10³⁴kgm²s^-1. Calculate the wavelength of the spectral line emitted when an electron falls from this level to the next lower level.
Answer:

Angular momentum of an electron in ‘n-th’ Bohr orbit of H-atom, mvr \(=\frac{n h}{2 \pi}\)

Or, \(3.1655 \times 10^{-34}=\frac{n \times 6.626 \times 10^{-34}}{2 \times 3.14}\)

⇒ \(\bar{v}=\frac{1}{\lambda}=109678\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \mathrm{cm}^{-1}\)

= \(109678\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\)

= \(\lambda=6.564 \times 10^{-5} \mathrm{~cm}\)

Since n₁ = 2, n₂ = 3

Question 17. Calculate the distance of separation between the second and third orbits of the hydrogen atom.
Answer:

The radius of the n-th orbit of the H-atom is given by \(r_n=0.529 \times n^2\)

∴ r₃– r₂ =0.529(32-22)Å

= 0.529 × 5

= 2.645 Å

Question 18. After absorbing an energy of 2.044¹∪10-19J, the electron of the H-atom will jump to which orbit?
Answer:

The energy ofthe electron in the n-th orbit of the H-atom is,

⇒ \(E_n=\frac{-2.18 \times 10^{-19}}{n^2} \mathrm{~J}\)

The energy of an electron in the ground state (n = 1) is

Ex = \(E_1=-21.8 \times 10^{-19} \mathrm{~J}\)

If tire electron absorbs an energy of 2.044 × 10—19 J, the total energy

= \(\left(-21.8 \times 10^{-19}+20.44 \times 10^{-19}\right) \mathrm{J}=-1.36 \times 10^{-19} \mathrm{~J}\)

∴ \(\frac{-21.8 \times 10^{-19}}{n^2}=-1.36 \times 10^{-19} \mathrm{~J}\)

∴ \(n=\sqrt{\frac{-21.8 \times 10^{-19}}{-1.36 \times 10^{-19}}}=4\)

Thus, the electron will jump to the fourth orbit.

Towards The Development Of Quantum Mechanical Model Of Atom

The limitations of Bohr’s model of the atom led to the development of a new improved model for atoms. Two significant concepts that led to the formulation of such a model were:

Dual nature of matter particles, Heisenberg’s uncertainty principle.

Dual Nature Of Matter Particles: De Broglie Theory

Both the scientists, Niels Bohr and Sommerfeld regarded electrons as negatively charged particles. However, the particle nature of the electron failed to explain some of its characteristics.

In 1905, Einstein suggested that light has a dual nature i.e., both wave nature as well as particle nature.

In 1924, French physicist, Louis de Broglie concluded that just like radiations, moving electrons or other such microscopic particles are associated with wave nature i.e., they exhibit wave nature as well as particle nature.

The wave associated with a particle is called a matter wave or de Broglie wave.

The wavelength associated with a moving particle is given by the de Broglie equation \(\lambda=\frac{h}{m v}=\frac{h}{p}\) [A = wavelength of the moving particle, m = mass of the particle, v = velocity of the particle, p = momentum of tire particle, h = Planck’s constant.] All the properties of electrons and other matter particles can be explained by its wave-particle duality.

From de Broglie equation, \(\lambda \propto \frac{1}{m v}\)

Since h=Constant i.e., wavelength \(\propto \frac{1}{\text { momentum }}\)

Thus, the wavelength of the wave associated with a fast-moving particle is inversely proportional to its momentum.

Derivation of de Broglie Equation:

The de Broglie equation was derived by using Planck’s quantum theory and Einstein’s equation of grass-energy equivalency. According to Planck’s quantum theory,

\(E=h v=h \times \frac{c}{\lambda}\)

where, c = velocity of light, E = energy of a photon, v = frequency and A = wavelength.

According to Einstein’s equation, E = mc²

Where, m = mass ofa photon, c = velocity of a photon.

From [1] and [2] we have \(\frac{h c}{\lambda}=m c^2 \text { or, } \lambda=\frac{h}{m c} \cdots[3]\) de Broglie pointed out that equation (3) can be applied to fast-moving matter particles such as electrons.

∴ \(\lambda=\frac{h}{m v}\) [m = mass & v = velocity ofthe particle]

This is de Broglie equation.

The wave associated with fast-moving matter particles is called matter wave or de Broglie wave.

This type of wave is distinctively different from electromagnetic waves.

de Broglie equation has no significance for moving particles having large mass because in such cases the wavelength of the associated wave is too small for ordinary observation.

For example, the wavelength associated with a cricket ball of mass 200 g (0.2 kg) moving with a velocity of 1 m-s-1 is—

⇒ \(\lambda=\frac{h}{m v}=\frac{6.62 \times 10^{-34}}{0.2 \times 1} \mathrm{~m}=3.31 \times 10^{-23} \mathrm{pm}\)

Such a small value of A cannot be measured.

Application of wave nature of electrons:

The concept of the wave nature of electrons is used in electron microscopes to get images of particles as tiny as 10A and for studying the surface structure of solid substances.

Angular momentum of Bohr electron from do Broglie equation:

According to de Broglie, a tiny particle like an electron, revolving in a circular orbit must have a wave character associated with it. Thus, for the wave (associated with the moving electron) to be completely in phase, the circumference of the orbit should be an integral multiple of the wavelength, λ.

∴ 2πλ = nλ or, λ \(=\frac{2 \pi r}{n}\)

[where, r = radius of the orbit and n = an integer] From de Broglie equation \(\lambda=\frac{h}{m v}\)

[where, m = mass of electron, v = velocity of electron.]

∴ \(\frac{2 \pi r}{n}=\frac{h}{m v} \text { or, } m v r \text { (angular momentum) }=\frac{n h}{2 \pi}\)

[where, r = radius of the orbit and n = an integer] From de Broglie Equation

⇒ \(\lambda=\frac{h}{m v}\)

This is the same relation as predicted by Bohr.

Circumference of the electronic orbit and the wavelength associated with the electronic motion:

From Bohr’s theory, angular momentum of an electron revolving in a circular orbit is an integral multiple of \(\frac{h}{2 \pi} \text { i.e., } m v r=\frac{n h}{2 \pi}\) orbit is an integral multiple of \(\frac{h}{2 \pi} \text { i.e., } m v r=\frac{n h}{2 \pi}\)

or, \(m v=\frac{n h}{2 \pi r}\)

r = radius ofthe orbit, v = velocity ofelectron. From de Broglie equation, \(\lambda=\frac{h}{m v} \quad \text { or, } m v=\frac{h}{\lambda}\)

[where λ = wavelength associated with moving electron.]
From [1] and [2] we have, \(\frac{n h}{2 \pi r}=\frac{h}{\lambda} \quad \text { or, } 2 \pi r=n \lambda\) or, 2πr = nλ

So, the circumference of the electronic orbit is an integral multiple of the wavelengths associated with the motion of electrons.

The kinetic energy of a moving particle and the de Broglie wavelength associated with it:

Kinetic energy(E) of a particle (mass = m) moving with velocity v is given by,

\(E=\frac{1}{2} m v^2 \quad \text { or, } m E=\frac{1}{2} m^2 v^2 \quad \text { or, } m v=\sqrt{2 m E}\) …………………..(1)

From de Broglie equation \(\lambda=\frac{h}{m \nu} \quad \text { or, } m v=\frac{h}{\lambda}\)…………………..(2)

[λ = wavelength associated with the moving particle] From 1 and 2 we have

⇒ \(\frac{h}{\lambda}=\sqrt{2 m E} \text { or, } \lambda=\frac{h}{\sqrt{2 m E}}\) …………………..(3)

Calculation of de Broglie wavelength of the electron from the potential applied:

If an accelerating potential V is applied to an electron beam, the energy acquired by each electron

= e (charge in coulomb)×  V(potential in volt)

= eV electron – volt

This energy becomes the kinetic energy of the electron.

∴ \(\frac{1}{2} m v^2=e V \quad \text { or, } \quad v=\sqrt{\frac{2 e V}{m}}\)

Substituting the values of h, m, and e in equation [1], we get \(\lambda=\frac{1.226 \times 10^{-9}}{\sqrt{V}} \mathrm{~m}\)

In place of the electron, if any other charged particle carrying Q coulomb is accelerated under a potential difference of V volt, then kinetic energy = QV. Hence \(\lambda=\frac{h}{\sqrt{2 m Q V}}\)

Numerical Examples

Question 1. Calculate the wavelength of the de Broglie wave associated with an electron moving with a velocity of 2.05 x 107m-s-1
Answer:

From de Broglie equation:

⇒ \(\lambda=\frac{h}{m v}\)

Where λ = wavelength , m = mass ofthe electron, v = velocity of the electron,

⇒ \(\lambda=\frac{6.626 \times 10^{-34}}{\left(9.11 \times 10^{-31}\right) \times\left(2.05 \times 10^7\right)}\)

= \(3.548 \times 10^{-11} \mathrm{~m}\)

Question 2. Calculate the velocity of an electron having de Broglie wavelength of 200A \(\left[m=9.11 \times 10^{-31} \mathrm{~kg}, h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right]\)
Answer:

According to de Broglie equation, \(\lambda=\frac{h}{m v}\)

∴ \(v=\frac{h}{m \lambda}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right) \times\left(200 \times 10^{-10} \mathrm{~m}\right)}\)

= 3.64 ×  × 10 ^–34 m.s

[Since 1Å =10¹⁰m]

Question 3. Calculate the ratio of velocities of a moving electron to that of a proton associated with the same de Broglie wavelength. [mg = 9.11 × 10^-31 kg, mp = 1.67 × 10^-27 kg, h = 6.626 × 10^-34 J.s]
Answer:

As given in the question, Ae = Ap

∴ \(\frac{h}{m_e \nu_e}=\frac{h}{m_p \nu_p}\)

∴ \(\frac{v_e}{v_p}=\frac{m_p}{m_e}=\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}=1833\)

Question 4. Calculate the momentum of the particle which has a de Broglie wavelength of O.lA
Answer:

⇒ \(\lambda=\frac{h}{m v}\)

∴ \(m v=\frac{h}{\lambda}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{0.1 \times 10^{-10} \mathrm{~m}}\)

= 6.626 ×10  ^–23kg-m-s-1.

Since 1A= 10-10m]

Question 5. Calculate the de Broglie wavelength of a proton that is moving with a kinetic energy of 5 x 10-23J.
Answer:

Mass of a proton = 1.67 x 10-27kg

Kinetic energy of a proton =5 x 10 23J

∴ \(\frac{1}{2} m v^2=5 \times 10^{-23}\)

Or, \(\frac{1}{2} \times 1.67 \times 10^{-27} \times v^2=5 \times 10^{-23}\)

∴ \(v=\left(\frac{10 \times 10^{-23}}{1.67 \times 10^{-27}}\right)^{1 / 2}=244.7 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ \(\lambda=\frac{h}{m \nu}=\frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 244.7}=1.62 \times 10^{-9} \mathrm{~m}\)

∴ \(\lambda=\frac{h}{m \nu}=\frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 244.7}=1.62 \times 10^{-9} \mathrm{~m}\)

Question 6. Calculate the de Broglie wavelength of an electron moving With A speed that is 1% of the speed of light.
Answer:

⇒ \(\lambda=\frac{h}{m v}\)

=\(\frac{6.626 \times 10^{-34}}{\left(9.11 \times 10^{-31}\right) \times\left(3 \times 10^8 \times \frac{1}{100}\right)} \)

= \(2.42 \times 10^{-10} \mathrm{~m}\)

Question 7. Calculate the kinetic energy of an a- -particle that has a de Broglie wavelength of 8 pm.
Answer:

Mass of an a -particle \(=\frac{4 \times 10^{-3}}{6.022 \times 10^{23}} \mathrm{~kg}=6.64 \times 10^{-27} \mathrm{~kg}\)

⇒ \(\lambda=\frac{h}{m v} \text { or, } v=\frac{h}{m \lambda}=\frac{6.626 \times 10^{-34}}{\left(6.64 \times 10^{-27}\right) \times\left(8 \times 10^{-12}\right)}\)

= 1.247× 10¹⁴m.s^-1

[since 1 pm = 10^-121m]

The kinetic energy of the α -particle = 1/2 mv²

=\(\frac{1}{2} \times\left(6.64 \times 10^{-27}\right) \times\left(1.247 \times 10^4\right)^2 \mathrm{~J}=5.16 \times 10^{-19} \mathrm{~J}\)

Question 8. Calculate the de Broglie wavelength of an electron accelerating in a particle accelerator through a potential difference of 110 million volt
Answer:

The kinetic energy of an electron under the potential difference of of110 million volts = llOMeV =110x 106eV

∴ \(\frac{1}{2} m v^2=110 \times 10^6 \mathrm{eV}=110 \times 10^6 \times 1.602 \times 10^{-19} \mathrm{~J}\)

Or, \(\frac{1}{2} \times 9.11 \times 10^{-31} \times v^2=110 \times 10^6 \times 1.602 \times 10^{-19}\)

∴ \(v=\left(\frac{2 \times 110 \times 10^6 \times 1.602 \times 10^{-19}}{9.11 \times 10^{-31}}\right)^{1 / 2}\)

= 6.22 × 10⁹m.s^-1

∴ \(\frac{h}{m v}=\frac{6.626 \times 10^{-34}}{\left(9.11 \times 10^{-31}\right) \times\left(6.22 \times 10^9\right)} \)

= \(1.17 \times 10^{-13} \mathrm{~m}\)

Question 9. Find de Broglie wavelength associated with a tennis ball of mass 60 g moving with a velocity of 10 m.s^-1
Answer:

de Broglie wavelength, \(\lambda=\frac{h}{m v}\)

Here, h = 6.626 × 10^-34 J.s^-1

m = 60g = 60 ×10^-3 kg = 6 × 10^-2 kg, v = 10 m.s^-1

∴ \(\lambda=\frac{6.626 \times 10^{-34}}{6 \times 10^{-2} \times 10}=1.105 \times 10^{-33} \mathrm{~m}\)

Question 10. Calculate the wavelength (in nm) associated with a beam of protons moving with a velocity of 10³m-s^-1. [Mass of proton 1.67 ×10^-27kg, h = 6.63 × 10^-34 J.s^-1]
Answer:

de Broglie wavelength \(\lambda=\frac{h}{m v}\)

= \(\frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 10^3} \mathrm{~m}=0.40 \times 10^{-9} \mathrm{~m}\)

= 0.40 nm

Question 11. Calculate the wavelength of an a -particle having an energy of 6.8 × 10^-18J.
Answer:

⇒ \(\lambda=\frac{h}{\sqrt{2 m E}}=\frac{6.626 \times 10^{-34}}{\sqrt{2 \times 6.8 \times 10^{-18} \times\left(4 \times 1.67 \times 10^{-27}\right)}}\)

= 2.198 × 10^-12m

Question 12. Calculate the wavelength of the wave associated with an electron beam, if the beam is accelerated by a potential difference of 5000 volts.
Answer:

Kinetic energy ofthe electron = 5000 eV

= 5000 × 1.602 ×10^-19J

The velocity of an electron due to the applied potential difference is ms^-1. Hence, kinetic energy \(=\frac{1}{2} m v^2\)

∴ \(\frac{1}{2} m v^2=5000 \times 1.602 \times 10^{-19} \mathrm{~J}\)

= 5000 × 1.602 ×10^-19  Kg.m².s^-1

∴ \(v=\sqrt{\frac{2 \times 5000 \times 1.602 \times 10^{-19}}{9.11 \times 10^{-31}}}=4.193 \times 10^7 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

[since mass of electron = 9.11 × 10^-31kg]

From de Brogile Equation , we get, \(\lambda=\frac{h}{m v}\)

⇒ \(\frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 4.193 \times 10^7}\)

= \(=0.1736 \times 10^{-10} \mathrm{~m}\)

=0.1736 Å

Question 13. The electron of 2 -atom in the ground state absorbs energy equal to 1.5 times the minimum energy, required to remove the electron from the hydrogen atom. Calculate the wavelength of the electron emitted, [mass of electron =9.11 X 10-31kg]
Answer:

The energy required to remove an electron from the ground state of H-atom = 13.6eV. Therefore, energy absorbed by the electron of H-atom = 1.5 × 13.6eV = 20.4eV

∴ Residual energy after removal ofthe electron

= (20.4- 13.6)eV = 6.8eV = 6.8 × 1.602 ×10^-19

This residual energy is converted into kinetic energy.

∴ \(\frac{1}{2} m v^2=6.8 \times 1.602 \times 10^{-19}\)

∴ \(v=\sqrt{\frac{6.8 \times 1.602 \times 10^{-19}}{9.11 \times 10^{-31}}}\)

= \(1.546 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

λ = \(\frac{h}{m v}\)

=\(\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right) \times\left(1.546 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)}\)

= \(4.70 \times 10^{-10} \mathrm{~m}\)

Question 14. Find the velocity of an electron so that its momentum is equal to that of a photon Of Wavelength 650nm.
Answer:

⇒ \(\lambda=\frac{h}{p}\) [from de Broglie Equation]

∴ If the momentum of electron = momentum of a photon, then A of electron =A of photon = 650nm

Applying de Broglie equation to electron, \(\lambda=\frac{h}{p}=\frac{h}{m v}\)

∴v =  \(\frac{h}{m \lambda}=\frac{6.63 \times 10^{-34}}{\left(9.11 \times 10^{-31}\right)\left(650 \times 10^{-9}\right)}\)

= 118.97 m.s^-1

Question 15. The kinetic energy of a subatomic particle is 3.60 x 10_24J. Calculate the frequency of the corresponding particle wave.
Answer:

Kinetic energy \(=\frac{1}{2} m v^2=3.60 \times 10^{-24} \mathrm{~J}\)

∴ mv² = 2 × 3.60 × 10^-24J

=7.2× 10^-24J For any given wave

∴ \(=\frac{v}{v}\)

∴ \(\frac{h}{m v}=\frac{v}{v} \text { or, } v=\frac{m v^2}{h}\) \(=\frac{7.2 \times 10^{-24}}{6.626 \times 10^{-34}}=1.086 \times 10^{10} \mathrm{~s}^{-1}\)

Question 16. Calculate the mass of a photon with wavelength 3.6A.
Answer:

A = 3.2Å =3.2× 10^-10m Velocity of photons (v) = velocity of light

= 3.2× 10^-8m.s^-1

From de Broglie equation, \(\lambda=\frac{h}{m v}\)

Or, \(m=\frac{h}{\lambda v}=\frac{6.626 \times 10^{-34}}{\left(3.2 \times 10^{-10}\right) \times\left(3 \times 10^8\right)}=6.9 \times 10^{-33} \mathrm{~kg}\)

Heisenberg’s Uncertainty Principle Numerical Examples

Question 1. Calculate the uncertainty in velocity (m.s^-1 ) of a moving object of mass 25 g, if the uncertainty in its position is 10-5m. [h = 6.6 × 10^-34 J.s]
Answer:

We, know \(\Delta x \cdot \Delta p=\frac{h}{4 \pi} \text { or, } \Delta x \cdot m \Delta v=\frac{h}{4 \pi}\)

∴ \(\Delta v=\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 10^{-5} \times 0.025}\)

= \(2.099 \times 10^{-28} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= \(\Delta x=10^{-5} \mathrm{~m}, m=25 \mathrm{~g}\)

= 0.025kg

Question 2. An electron has a velocity of 600m-s_1 [accuracy: 0.005%]. With what accuracy can we locate the position of this electron?[mass of an electron = 9.1 × 10^-31kg, h = 6.6 ×  10^-34 J.s ]
Answer:

According to uncertainty principle, \(\Delta x \cdot m \Delta v=\frac{h}{4 \pi}\)

Or, \(\frac{6.6 \times 10^{-34}}{4 \times 3.14 \times\left(9.1 \times 10^{-31}\right) \times\left(600 \times \frac{0.005}{100}\right)}\)

⇒ \(\frac{6.6 \times 10^{-34}}{4\times 3.14 \times 9.1 \times 10^{-31} \times 0.03}\)

⇒\(1.92 \times 10^{-3} \mathrm{~m}\)

Question 3. The uncertainties in position and velocity of a particle arc 10-1° in and 5.27× 10^-24 m-s^-1 respectively. Calculate the mass of the particle.
Answer:

According to uncertainty principle \(\Delta x \cdot \Delta p=\frac{h}{4 \pi}\)

or, \((\Delta p)^2=\frac{h}{4 \pi}\)

[since \(\Delta x=\Delta p\)

Therefore \(\Delta p=\sqrt{\frac{h}{4 \pi}} \quad \text { or, } m \Delta v=\sqrt{\frac{h}{4 \pi}} \text { or, } \Delta v=\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

Question 4. If uncertainties in the position and momentum of a
moving object are the same, find uncertainty in velocity

Answer:

According to uncertainly principle, \(\Delta x \cdot \Delta p=\frac{h}{4 \pi}\)

Or, \((\Delta p)^2=\frac{h}{4 \pi}\)

∴ Δx = Δp

∴ Δp =   \(\sqrt{\frac{h}{4 \pi}}\)

Or, mΔv =  \(\sqrt{\frac{h}{4 \pi}} \text { or, } \Delta v=\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

Question 5. Calculate the uncertainty in the velocity of an electron if
the uncertainty in its position is of the order ±12pm
Answer:

According to uncertainly principle, Δx.mΔv = h/4π

Δv = \(\)

Mass of electron =9.11 ×  10^-31kg

And 1pm=10^-12 m

= \(4.82 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Question 6. Calculate the minimum uncertainty in the position of a bullet of mass 2.5g having a probable velocity between 60,000,000 and 60,000,001 m.s^-1
Answer:

The maximum uncertainty in velocity

Av =60,000,001-60,000,000 =l m.s^-1

According to the uncertainty principle, Ax – mAv = h/mv

∴ \(\Delta x=\frac{h}{4 \pi m \times \Delta v}\)

= \(\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times\left(2.5 \times 10^{-3}\right) \times 1}\)

= \(2.12 \times 10^{-32} \mathrm{~m}\)

Question 7. The uncertainty in the determination of the velocity of a dust particle (of mass O.lmg) is 4.5 X 10-20 m-s Calculate the least uncertainty in its position.
Answer:

⇒  \(\Delta x \times \Delta p \frac{h}{4 \pi} \text { or, } \Delta x \times m \Delta v \frac{h}{4 \pi} \text { or, } \Delta x \frac{h}{4 \pi m \times \Delta v}\)

Least uncertainty in determining its position,

Δx  = \(\frac{h}{4 \pi m \times \Delta v}=\frac{6.626 \times 10^{-5}}{4 \times 3.14 \times\left(0.1 \times 10^{-6}\right)\times\left(4.5 \times 10^{-20}\right)}\)

= \(1.172 \times 10^{-8} \mathrm{~m}\)

Question 8. If uncertainties in the measurement of the position and momentum of an electron are found to be equal in magnitude, then what is the uncertainty in the measurement of velocity? Comment on the result.
Answer:

According to uncertainty principle, \(\Delta x \times \Delta p=\frac{h}{2 \pi}\)

Given, Δx =Δp

∴ \((\Delta p)^2=\frac{h}{4 \pi} \text { or, } \Delta p=\sqrt{\frac{h}{4 \pi}}\)

Or, \(m \Delta v=\sqrt{\frac{h}{4 \pi}}, \text { or } \Delta v=\frac{1}{m} \sqrt{\frac{h}{4 \pi}}\)

= \(\frac{1}{9.11 \times 10^{-31}} \sqrt{\frac{6.626 \times 10^{-34}}{4 \times 3.14}}=7.97 \times 10^{12} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Question 9. Calculate the product of uncertainties in the position and velocity of an electron of mass 9.1 X 10-31kg, according to Heisenberg’s uncertainty principle.
Answer:

According to uncertainty principle, \(\Delta x \cdot \Delta p=\frac{h}{4 \pi}\)

Or, \(\Delta x \cdot m \Delta v=\frac{h}{4 \pi} \quad \text { or, } \Delta x \cdot \Delta v=\frac{h}{4 \pi m}\)

=  \(\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{4 \times 3.14 \times 9.11 \times 10^{-31} \mathrm{~kg}}\)

= \(5.79 \times 10^{-5} \mathrm{~m}^2 \cdot \mathrm{s}^{-1}\)

Shapes Of Orbitals From Wave Function

It has been stated earlier that the three-dimensional space around the nucleus in which the probability of finding an electron is maximum is called an orbital.

To obtain a clear idea about the shapes of orbitals, we will first discuss the variation of—

1. The radial part of the wave function,
2. Square of the radial wave function, and
3. Radial distribution function with an increase in distance from the nucleus.

Variation Of Radial Part Of Wave Function With Distance From The Nucleus

Schrodinger wave equation for the electron in a one-electron atom (H-atom) can be solved to get different expressions for wave function \((\psi)\) for different orbitals.

The orbital wave function for an electron in an atom has no physical meaning. It is simply a mathematical function of the coordinates of the electron.

However, for different orbitals the plots of the radial part of the corresponding wave functions as a function of r (distance from the nucleus) are different. depicts such plots for Is, 2s, 2p and 3s orbitals.

For is -orbital, the radial part of the wave function [ψ(r) or R] decreases sharply with increasing distance, r, from the dying nucleus.

For 2s -orbital ψ (r) or R, decreases sharply in the beginning, becomes zero at a particular distance, and then becomes negative as r increases.

FM 3s-orbilal \(\psi\). decreases sharply in the Beginning with an Increase In r, becomes zero At A Particcular Distance, And Then Becomes negative. On Further Increases In \(r, \psi(r)\) again becomes zero and finally becomes positive.

For 2p -orbital if \(\psi(r)\) rises from zero to a maximum, then decreases with increasing distance (r) from the nucleus. On further increase in distance, ifr(r) approaches almost zero.

For 3p -orbital,\(\psi(r)\) rises from zero and attains a maximum value. On further increase in ψ(r) begins to decrease and becomes zero at a particular distance. Then it becomes negative with a further increase in r.

Characteristic features observed in the plots of r vs Ψ(r):

1. The radial part of the wave functions for 2s, 3s, 3p, etc. orbitals can be positive or negative depending upon the distance (r) from the nucleus. These are not related to the positive and negative charges.
2. For each orbital, the radial part of the wave function Ψ(r) approaches zero as r→∞.
3. For 2s, 3s, and 3p -orbitals, one common feature for the variation of wave function Ψ(r)) with distance is that Ψ(r) becomes zero at a finite distance from the nucleus. However, for different orbitals if Ψ(r) becomes zero at different distances Ψ(r).
4. The distance Ψ(r) at which becomes zero is called a nodal point radial node or simply node. At the nodal point, the radial wave function if Ψ(r) changes its sign from positive to negative or vice versa.
5. For different orbitals, the number of radial nodes =(n-1-1).
6. This indicates that the number of radial nodes is determined by the values of the principal quantum number ( n) and azimuthal quantum number (Z) of the orbital under consideration.

There is no relation between the positive and negative values of the wave function with the positive and negative charges.

Radial probability density [Ψ²(r) Or R²] graphs variation Of the square Of Radical Wave Function With Distance From The nucleus (r)

The square of the radial wave function, Ψ²(r) or R2 for an orbital gives the radial density.

According to the German physicist, Max Bom, the radial density, Ψ²(r) at a point gives the probability density of the electron at that point along a particular radial line.

The variation of Ψ²(r) as a function of r for different orbitals is given in the figure. The nature of these curves is different for different orbitals.

For Is -orbital, probability density is maximum near the nucleus (r≈0) and decreases sharply as we move from it.

For 2s -orbital the probability density is maximum near the nucleus (r≈0).

With increasing distance, Ψ2(r) first decreases sharply to zero and starts increasing again. After reaching a small maxima it decreases again and approaches zero as the value of r increases further.

The intermediate region (a spherical shell) where this probability density reduces to zero is called the nodal surface or simply node.

In general ns -orbital has (n- 1) nodes. Thus, the number of nodes for 2s -orbital is one, two for 3s and so on, i.e., the number of nodes increases with an increase of principal quantum number n.

The probability density variation for Is and 2s orbitals can be visualized in terms of charge cloud diagrams. In these diagrams, the density of the dots in a region represents the electron probability density in that region.

For 2p- Orbital Probability Density Is zero at r = 0. However, with increasing distance, it begins to increase and reaches a maximum and then decreases gradually as the distance (r) from the nucleus increases and ultimately approaches zero.

From similar plots of various orbitals, it has been found that all orbitals except s, have zero electron density at r = 0.

Radial probability distribution curve: Variation of radial distribution function (RDF) with distance from the nucleus (r)

The plot of Ψ²(r) versus r gives the probability density for the electron around the nucleus. However, in order to determine the total probability in an infinitesimally small region, we have to multiply probability density if Ψ2(r) by the volume of the region i.e., probability = Ψ²(r) x dv [where dv = volume of the region].

Since the atoms have spherical symmetry, it is more useful to discuss the probability of finding the electron in a spherical shell between the spheres of radii r and (r + dr).

The volume of such a shell of extremely small thickness, dr, is 4nr2dr. So we have, Probability = R2 x 4rrr2dr = 47tr2Ψ²(r)(r)dr [since R = Ψ²(r))].

This gives the probability of finding the electron at a particular distance (r) from the nucleus. This is called radial distribution function (RDF).

Radial distribution function (RDF) = 4πr²ψ²(r)dr

Important information obtained from the plots of RDF vs r:

1. For all orbitals, the probability is zero at the nucleus.
2. If the point r = 0 is neglected, then it can be seen that,
3. The number of radial nodes for any orbital -n-l- 1,
4. The number of maxima (peak) for any orbital =(n-l- 1) +1 = (n-/). The peak in any curve gives the distance from the nucleus to that point where the probability of finding the electron is maximum. This is called the radius of maximum probability.
5. All the s -orbitals, except the first one (Is), have a shell-like structure, rather like an onion, or a hailstone, consisting of concentric layers of electron density. Similarly, all but the first p -p-orbital (2p) and the first dorbital (3d) have shell-like structures.
6. The first s -s-orbital (Is), first p -p-p-orbital (2p) and first orbital (3d) have two important characteristics—
7. they do not contain radial nodes and contain only one maxima.
8. Examination of the plots for Is, 2s, and 3s -orbitals shows that the most probable distance of maximum probability density increases markedly as the principal quantum number increases.
9. Furthermore, by comparing the plots for 2s and 2p, or 3s, 3p, and 3d -orbitals it is seen that the most probable radius decreases slightly as the azimuthal quantum number increases.

Shapes Of Orbitals

The angular part of the wave function, A; m(6, 0), depends on the azimuthal (/) and magnetic (m) quantum numbers but is independent ofthe principal quantum number (n).

On the other hand, the radial part of the wave function, Rn t(r), depends on the principal quantum number (n).

Thus the principal quantum number ( n) determines the size, while the azimuthal (/)and magnetic (m) quantum numbers determine the shape of an orbital.

Shape of s-orbital:

From the solutions of the Schrodinger equation for s -s-orbital of H-atom, it has been known that the value of wave function ψ, or the probability of finding the electron ψ2 in space around the nucleus depends only on the distance from the nucleus but not on the direction.

• In other words, there is an equal probability of finding the electron at a given distance in all directions Around the nucleus, i.e. ψ=f(r)
• Thus all s -orbitals are spherical. For Is -orbital, RDF (the probability of finding the electron) increases as the distance from the nucleus increases and reaches a maximum value at a particular distance.
• (This distance is 0.529A for the electron in the ground state of H-atom). Then this probability begins to decrease and becomes negligible at large distances.
• Thus for Is -orbital, the probability of finding the electron is zero at r = 0 and also at r =∞

Like Is -orbital, 2s -orbital is also spherical.

However, 2s -orbital differs from Is -orbital in the fact that the probability of finding the electron is zero not only at r = 0 and r = ∞ but also at a particular distance between r = 0 and r – ∞.

• In fact, for 2s -orbital, RDF (the probability of finding the electron) increases as the distance from the nucleus increases and reaches a maximum value.
• Then it begins to decrease and becomes zero at a particular distance.
• The spherical shell of zero electron density is called a nodal surface or simply a node.
• After crossing the nodal surface the probability of finding the electron begins to increase again and reaches a second maxima.

This second maximum represents the region of the highest electron density in a 2s orbital. This is known as an antinode.

• After crossing this region of highest electron density, die probability of finding the electron again begins to decrease and approaches zero as the value of r increases further.
• Thus 2s -orbital has a shell-like structure consisting of concentric layers of electron density.
• 3s -orbital is also spherical. Its structure is similar to that of the 2s -orbital, but it differs from the 2s orbital as it has two nodal surfaces and three regions of maximum electron density.
• The probability of finding the electron at the 3rd maxima is the highest and is called the antinode in a 3s -orbital.
• Thus there are two nodes before the arrival of the highest probability region. These are pictorially represented in
• Angular wave function \(\left[\mathrm{A}_{L, m}(\theta, \phi)\right]\) for s -orbital does not depend on θ and Φ.
• It has non-zero values (with equal magnitude) in all possible directions. So s -orbital has no nodal plane.

Points to remember: Is, 2s, and 3s -orbitals are all spherical, but they differ from each other in the following respects—

Size and energy of ns -orbital increases as the magnitude of n increases. Thus, we have Is < 2s < 3s—.

Depending on the value of n, there are different numbers of nodes in ns -orbital. Thus, there are 0, 1, and 2 nodes for Is, 2s, and 3s -orbitals respectively.

Shapes of ρ-orbitals:

On solving the Schrodinger equation for 2p orbital of H-atom, it has been known that wave function depends on—

• The distance (r) from the nucleus and also the orientation of the orbital in three-dimensional space (x, y, z).
• Probability density calculation shows that each p -p-orbital consists of two sections, which are the regions of maximum electron density.
• These two sections are called lobes, which are on either side ofthe plane passing through the nucleus.
• The probability density of the electron is zero on this plane. It is called the nodal plane.
• Again, the probability density of the electron is equal in both lobes, but the wave function has opposite signs in the two lobes.

• Now, for p-orbitals, f= 1 and hence m = -1,0, +1. Thus there are three p-orbitals in any quantum level (except n = 1 ). The size, shape, and energy of the three p -orbitals are identical.
• They differ, however, in the way the lobes are oriented. Since the lobes may be considered to lie along the three axes x, y, and z, they are assigned the designations px, py, and pz respectively.
• It should be noted, however, that there is no simple relation between the values of m(-l; 0, +1) and the x, y, and z directions.

The wave functions for the three p -p-orbitals are—

⇒ \(\psi_x=f(r) \cdot f(x) ; \psi_y=f(r) \cdot f(y) ; \psi_z=f(r) \cdot f(z)\)

The three p-orbitals corresponding to a particular quantum level are said to be degenerate because they have the same energy.

• Thus there are three degenerate p-orbitals in each of the second n = 2, third n = 3, fourth n = 4, etc. quantum levels.
• Like s -s-orbitals, p -p-orbitals increase in size and energy with an increase in the principal quantum number (n). Hence the order of increasing size and energy of various p -p-orbitals is 2p< 3p < 4p < …..
• Now the number of radial nodes for 2p -orbital -n-1-1 = 2-1-1 = 0. Thus 2p -orbitals [i.e., 2px, 2py, 2pz) have no radial nodes.

The number of radial nodes for p -p-orbitals of some of the higher quantum levels is given in the following table.

For px -orbital probability density is zero in the yz -plane. So yz -plane is the nodal plane of px -orbital. Similarly zx and xy -planes are the nodal planes for py and pz orbitals respectively.

• It should be remembered that the number of nodal planes for any orbital is equal to the value of azimuthal quantum number l corresponding to that orbital.
• The number of nodal planes is, however independent of the principal quantum number ‘ n ‘.

The no. of nodal planes for s, p, d, and f-orbitals (belonging to any principal quantum level) are 0, 1, 2, and 3 respectively.

Shapes of d-orbitals:

From solutions of the Schrodinger wave equation for 3d -orbitals of H-atom it has been known that the wave function depends on the distance from the nucleus (r) and also on two different directions in space,

For example f(f)  = f(r)-f(x)-f(y) Now, for 3d -orbital, 1 = 2. Hence m = -2, -1, 0, +1, +2.

Thus there are five d -d-orbitals in the 3rd quantum level.

These d -d-orbitals have the same energy and hence they are degenerate. These orbitals are designated as dxydyzdzx>dx2-y2 and dz2- Probability density calculation shows that the orientations of these orbitals in space are different.

The shapes of the first four d -d-orbitals are similar to each other (which has four lobes) whereas that ofthe fifth dz2 is different (which has only two lobes).

The d -orbitals for which n is greater than three {viz., 4d, 5d, etc.) also have shapes similar to 3d -orbitals, but they differ in energy and size.

dxy -orbital has four lobes, lying in the xy -plane. The hatch lobe makes an angle of 45° with x and y-axes.

The wave function for d -orbital has the same sign (either + or – ) in each pair of opposite lobes, but opposite signs (+ and – ) I in adjacent lobes, relating to the symmetry of the angular wave function.

The wave function if =0 along the x, y, and z axes indicates that the xz – and yz – planes represent the nodal planes of the d -orbital.

These planes are at right angles with each other. Similarly, dyz -the orbital has four lobes lying in the yz -plane, each lobe makes an angle of 45° with the y and z axes.

The wave function dxy =0 along the three axes (x, y, and z) indicates that the xy and xz planes constitute the nodal planes of the dyz -orbital. The planes are at right angles with each other.

Similarly, dxz -the orbital has four lobes lying in the xz-plane, each lobe makes an angle of 45° with x and z-axes. The wave function ifrd =0 along the three axes (x, y, and z), indicates that xy and yz -planes constitute the nodal planes of dxz -orbital. These planes are at right angles with each other.

dx2-y2 -orbital has four lobes which lie along the x and y axes in the xy -plane. The wave function xy = ψ ax2-y2 along the z-axis indicates that this orbital has two nodal planes that pass through the origin and make angles of 45° with xz and yz -planes. These two planes are at right angles with each other.

dz² -orbital has two lobes (having the same sign which lie along the z-axis and contain an annular portion surrounding the nucleus in the xy-plane. This annular portion of the orbital is called the doughnut or belly band.

A conical surface passing through the origin constitutes the nodal plane of the dz2 -orbital.

F-Orbitals Schrodinger wave equation gives a set of solutions when the azimuthal quantum number f has the value 3. These are called f-orbitals. For l = 3, m has the values -3, -2, -1, 0, +1, +2, +3. Thus there are seven f-orbitals.

For the existence of f-orbitals, the minimum value of the principal quantum number, n, has to be 4, as the value of l cannot be greater than (n-l).

These seven orbitals have the same energy (degenerate) but they differ in their orientations in space. The seven f-orbitals are designated.

⇒ \(\text { as }-f_{x\left(x^2-y^2\right)}, f_{y\left(x^2-y^2\right)}, f_{z\left(x^2-y^2\right)}, f_{x y z}, f_{z^3}, f_{y z^2}, f_{x z^2} \text {. }\)

1. No. spherical/radial nodes in any orbital = n-1- 1
2. No. of planar or angular nodes in any orbital = Z
3. Total no. of nodes in any orbital = n-1

Rules For Filling Up Of Electrons In Different Orbitals

The correct ground state electronic configuration of an atom is obtained based on the following principles—Pauli’s exclusion principle, Hund’s rule, and the Aufbau principle.

Pauli’s exclusion principle

Principle:

The knowledge of four quantum numbers is important in assigning the exact location of the electron within an atom.

• After meticulous study of the line spectra of atoms, Wolfgang Pauli in 1925 proposed his exclusion principle which is widely known as Pauli’s exclusion principle.
• According to this principle, no two electrons in an atom will have the same values for all four quantum numbers (n, l, m, and s).
• If three of the quantum numbers of any two electrons are the same then they must differ in their fourth quantum number.
• If the quantum numbers n, l, and m of two electrons have identical values, then the value of s should be different (+i for one and for the other).
• Therefore, the corollary of this principle may be stated as—each orbital can accommodate a maximum of two electrons having an opposite spin.

With the help of Pauli’s exclusion principle, the maximum number of electrons a subshell can accommodate can be calculated. For example—

s -subshell:

In the case of s -subshell, 1 = 0. Therefore m = 0. Number of orbitals in s -subshell = 1.

According to Pauli’s exclusion principle, each orbital can hold a maximum number of two electrons. So, s -subshell can accommodate a maximum of 2 electrons.

p -subshell:

For p -subshell, 1=1 and m = —1,0, +1. The number of orbitals in the -subshell is three (px, py, and pz ).

According to Pauli’s exclusion principle, since each orbital can hold a maximum of 2 electrons, the maximum accommodating capacity of p -subshell {i.e., three p orbitals)

=3 × 2 = 6 electrons.

d -subshell:

In the case of d -subshell, 1 = 2, m = -2, -1, 0 +1, +2. Thus, m has 5 values indicating the presence of 5 orbitals. As the maximum number of electrons that each orbital can hold is 2, the maximum number of electrons that a d -d-subshell can accommodate is 5× 2 = 10.

f-subshell:

For f-subshell, l = 3, m = -3, -2, -1, 0, +1, +2, +3. Seven values of m indicate the presence of seven orbitals. Hence the maximum number of electrons that may be present in f-subshell is 7 × 2 = 14.

Pauli’s exclusion principle also permits the determination of the maximum number of electrons that can be present in a certain orbit or shell.

Example:

For L -shell (n = 2), l has two values, i.e., 1 = 0 [ssubshell] and l = 1 [p -subshell].

The s -subshell can hold 2 electrons and p -subshell can accommodate 6 electrons. Therefore, the maximum accommodating capacity for L shell =(2 + 6) = 8 electrons.

Similarly, it can be shown that, the maximum number of electrons that can be accommodated in M-shell (n = 3) = 18 and the maximum number of electrons that may be present in IVshell (n = 4) =32.

Electron accommodating capacity of K, L, M, and V-shell

Thus, it is seen that the maximum number of electrons accommodated in any electronic orbit with the principal quantum number’ n’ is 2n2.

Number of orbitals and electron accommodating capacity of different shells.

Hund’s multiplicity rule:

This rule helps decide the mode of filling of the orbitals ofthe same energy level with electrons.

Rule:

The pairing of electrons in the orbitals within the same subshell does not take place until the orbitals are singly filled up with electrons having parallel spin.

Discussion:

The rule implies that orbitals with the same energy are filled up first with one electron and then the additional electron occupies the singly filled orbital orbital to form paired electrons (with opposite spin).

The energy order of the orbitals, the Aufbau principle, and the electronic configuration of atoms

The German word ‘Aufbau’ means ‘to build one by one! The Aufbau principle gives the sequence of gradual filling up of the different subshells of multi-electron atoms.

Aufbau principle:

The Aufbau principle states that electrons are added progressively to the various orbitals in the order of increasing energy, starting with the orbital with the lowest energy.

Electrons never occupy the die orbital of higher energy leaving the orbital of lower energy vacant.

A study of the results of spectral analysis has led to the arrangement of the shells and subshell in the increasing order of their energies in the following sequence:

Is < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f< Sd < 6p < 7s < 5f< 6d ..

Electronic configuration always conforms to Pauli’s Exclusion Principle.

According to Hund’s rule, pairing electrons in the orbitals within the same subshell (degenerate orbitals hating the same n ) cannot occur until the orbitals are singly filled up.

The energy of the subshell increases with an increase in the value of {n + l). In a multi-electron atom, the energy of a subshell, cannot be determined only by principal quantum number (n ), in exclusion of azimuthal quantum number (Z).

The correct order of energies of various subshells is determined by the (n + 1) rule or Bohr-Bury rule.

The implication of the rule can be better understood with the help ofthe following example.

In case of 3d -subshell, (n + Z) = (3 + 2) = 5, but for 4s -subshell, (n + Z) = (4 + 0) = 4 .

From this, it is clear that the energy of the 4s -subshell is less than that of the 3d -subshell. Hence, the electron goes to the 4s subshell first, in preference to the 3d -subshell.

If Two subshells have the same value for{n + 1), then the electron enters that subshell which has a lower value of n.

For example, for 3d -subshell, (n + 1) = (3 + 2) = 5 and for 4p -subshell, (n + 1) = (4 + 1) = 5 In this case, the electron first enters the 3d -subshell which has a.lower value of n.

The sequence in which the subshells are filled with electrons.

The figure depicts the sequence of filling up of the subshells with electrons. The electronic configuration of any atom can be easily predicted from this diagram.

Exceptions to (n+1) rule:

Exceptions to the {n + Z) rule are found to occur in the case of filling up of electrons in Lanthanum (La) and Actinium (Ac).

The values of (n + 1) in the case of both the subshells 4/ and 5d (4 + 3 = 7 = 5 + 2) are found to be the same.

Similarly the values of (n +1) in the case of both the subshells 5/ and 6d (5 + 3 = 8 = 6 + 2) are equal. So, the order of energies of these subshells is 4f< 5d and 5f< 6d.

According to the (n + Z) rule, the expected electronic configuration of La (57) and Ac (89) should be [Xe]4/15d06s2 and [Rn]5/16d°7s2 respectively.

However, the electronic configuration of La and Ac are actually [Xe]4/ and [Rn]5/°6d17s2 respectively. In other words, lanthanum and actinium are exceptions to the (n + 1) rule.

Method of writing electronic configuration of an atom 1) To express the electronic configuration of an atom, the principal quantum number (n = 1, 2, 3… etc.) is written first.

The symbol ofthe subsheU(s, p, d, f, etc.) is written to the right ofthe principal quantum number. For example, s -subshell of the first shell is expressed as Is; sand subshells of the second shell are expressed as 2s and 2p respectively.

The total number of electrons present in any subshell is then written as the right superscript of the subshell symbol.

For example, the electronic configuration, ls²2s²2p⁵ suggests that the s -subshell of the first shell contains 2 electrons, and the s, and p -subshells of the second shell contain 2 electrons and 5 electrons respectively. Thus, the total number of electrons present is equal to 9.

Examples: Electronic configuration of ₁₇CL atom:

The atomic number of chlorine is 17. Number of electrons present in chlorine atom is 17.

Out of these 17 electrons, 2 electrons are present in the s -subshell of the first shell (K-shell), 2 electrons and 6 electrons in the s – and p -subshell of the second shell (L -shell) respectively, and 2 and 5 electrons are present in the s – and p -subshell of the third shell (Mshell) respectively.

Thus, the electronic configuration of the chlorine atom is ls²2s²2p63s²3p5.

Electronic configuration of ₂₆Fe atom:

The atomic number of iron is 26. Number of electrons present in an atom of iron is 26. These 26 electrons are distributed in K, L, M, and N-shells in such a way that their electronic configuration becomes ls²2s²2p⁶3s²3pe³de4s².

Here the symbol signifies an orbital and the arrow sign (↑) means an odd electron and the paired arrow sign (↓↑) stands for a pair of electrons with opposite spins.

Stability of half-filled or filled subshells The electronic configurations of some atoms have certain characteristic features.

It is seen that half-filled and filled subshells are more stable compared to nearly half-filled or nearly-filled subshells.

Hence, if the (n-1)d -subshell of any atom contains 4 or 9 electrons and the ns -subshell contains 2 electrons, then one electron from the ns -subshell gets shifted to the (n-1) d subshell, thereby making a total number of either 5 or 10 electrons in it. As a result, ns -subshell is left with 1 electron instead of 2.

The extra stability of half-filled and filled subshells can be explained in terms of the symmetrical distribution of electrons and exchange energy.

Symmetrical distribution of electrons:

The subshells with half-filled or filled electrons are found to have a more symmetrical distribution of electrons.

Consequently, they have lower energy which ultimately results in greater stability of the electronic configuration.

Electrons present in the same subshell have equal energy but their spatial distribution is different. As a result, the magnitude of the shielding effect of another is quite small and so, the electrons are more strongly attracted by the nucleus.

Interelectronic repulsion:

Two types of interactions are possible between electrons of the same subshell due to interelectronic repulsive force.

Interaction due to electronic charge:

The magnitude of the repulsive force acting between two electrons situated at n distance r from each other is inversely proportional to the square of the distance between them.

Consequently, the stability of two-electron or multi-electron ions or atoms increases with an increase in distance r. Thus, die two electrons present in the d -d-subshell prefer to be in two separate d -orbitals instead of one leading to the increased stability ofthe atom or ion.

Interaction due to rotation of electrons:

Two electrons tend to remain close to each other if they have opposite spins. On the other hand, if both the electrons have parallel spin, then they prefer to remain far from each other.

The electrons occupying degenerate orbitals (orbitals of the same energy) can exchange their positions with other electrons with the same spin. In this process, exchange energy is released.

The greater the probability of exchange, the more stable the configuration. The probability of exchange is greater in the case of a half-filled or filled subshell.

Thus, the magnitude of exchange energy is greatest for half-filled or filled subshells leading to their exceptionally high stability.

This exchange energy forms the basis of Hund’s multiplicity rule. The relative magnitude of exchange energy can be calculated by the formula,

No. of exchanges \(=\frac{n !}{2 \times(n-2) !}\)

(n = number of degenerate electrons with parallel spin.)

Number of interactions in case of d4 electronic configuration

Total number of exchanges for d4 electronic configuration

=3+2+1=6

Number of interactions in case of d5 electronic configuration

Electronic configuration of ions

When an additional electron is added to an orbital of an atom, a negatively charged ion called an anion is formed while the removal of an electron from the orbital of an atom produces a positively charged ion called cation.

1. Electronic configuration of anions:

The total number of electrons present in an anionic species is = (Z + n) where Z = atomic number and n = number of electrons gained. The electronic configuration ofthe anion is written based on the total number of electrons present in it.

Examples: 

1. Fluoride ion (F^–): Total number of electrons present in F^– ion

= (9 + 1) = 10

Electronic configuration of F^– ion =  ls²2s²2p^6-

2. Nitride ion (N³¯ ): Total number of electrons present in N³¯ ion

= (7 + 3) = 10

Electronic configuration of N³¯ ion = ls²2s²2p⁶

3. Oxide ion (O²¯): Total number of electrons present in  O²¯ ion

= (8 + 2) = 10.

Electronic configuration of O²¯ ion = ls²2s²2p⁶

4. Sulphide Ion (S²¯): Total number ofelectrons present in S²¯ ion

=(1.6 + 2) = 18

Electronic configuration of S²¯ ion = ls²2s²2p⁶3s²3p⁶

2. Electronic Configuration of cations:

• A total number of electrons present in a cationic species = (Z-n) where Z = atomic number and n = number of electrons lost.
• For writing the electronic configuration of the cation, the electronic configuration of the neutral atom is written first.
• Then requisite no. of electrons is removed from the outermost shell. Electrons from the ns -subshell should be removed before removing any electron from the (n- 1)d -subshell.
• The total number of electrons present in a cationic species = (Z-n) where Z = atomic number and n = number of electrons lost.
• For writing the electronic configuration of the cation, the electronic configuration of the neutral atom is written first.
• Then requisite no. of electrons is removed from the outermost shell. Electrons from the ns -subshell should be removed before removing any electron from the (n- 1)d -subshell.

Examples:

1. Sodium ion (Na⁺):

Electronic configuration of = \({ }_{11} \mathrm{Na}: 1 s^2 2 s^2 2 p^6 3 s^1\)

⇒ \(\mathrm{Na}^{+} \text {lon: } 1 s^2 2 s^2 2 p^6\)

2. Chromium Ion (Cr³⁺):

Electronic Configuration of =   \({ }_{24} \mathrm{Cr}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^1\)

⇒ \(\mathbf{C r}^{3+} \text { ion: } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^3\)

3. Manganese ion (Mn²⁺):

Electronic Configuration of = \(25^{\mathrm{Mn}}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^2\)

⇒ \(\mathbf{M n}^{2+} \text { ion: } 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5\)

4. Ferrous (Fe²⁺) and Ferric (Fe³⁺) ion:

Electronic Configuration of = \({ }_{26} \mathrm{Fe}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6 4 s^2\)

⇒ Ferrous ion (Fe²⁺): =\(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^6\)

⇒ Similarly, ferric ion (Fe³⁺):

Electronic Configuration of =\(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5\)

5. Cuprous (Cu⁺) and Cupric (Cu²⁺) ion:

⇒  Cu⁺+ ion:

Electronic Configuration of = ls²2s²2p⁶3s²3p⁶63d¹⁰

⇒ Similarly, cupric ion (Cu²⁺):

Electronic Configuration of =  ls²2s²2p⁶3s²3p⁶63d⁹

Numerical Examples

Question 1. A sample of gaseous oxygen contains only 180 isotopes. How many neutrons are present in 11.2 L of the gas at STP?
Answer:

No. of neutrons present in an atom of 180 isotope

=(18-8) = 10

∴ No. of neutrons present in 11.2L of  the gas

= \(\frac{2 \times 10 \times 6.022 \times 10^{23} \times 11.2}{22.4}=6.022 \times 10^{24}\)

Question 2. Calculate the energy required for the promotion of electrons from the 1st to 5th Bohr orbit of all the atoms present in 1 mole of H-atoms.
Answer:

Electronic energy in the n-th orbit ofH-atom

⇒ \(E_n=-\frac{1312}{n^2} \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

∴ Total energy required

= \(\left(E_5-E_1\right)=-\frac{1312}{5^2}-\left(-\frac{1312}{1^2}\right)=1259.52 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 3. Calculate the velocity (cm-s-1) and frequency of revolution of electron present in the 3rd orbit of H-atom.
Answer:

Velocity of revolving electron present in the 3rd orbit of H atom

= \(\frac{2 \pi z e^2}{n h}=\frac{2 \times \pi \times 1 \times\left(4.8 \times 10^{-10}\right)^2}{3 \times 6.626 \times 10^{-27}}\)

Frequency of revolution ofthe electron

= \(\frac{v}{2 \pi r}=\frac{v}{2 \pi\left(\frac{n^2 h^2}{4 \pi^2 m z e^2}\right)}=\frac{2 \pi m v z e^2}{n^2 h^2}\)

= \(\frac{2 \times 3.14 \times\left(9.11 \times 10^{-28}\right) \times\left(7.27 \times 10^7\right) \times 1 \times\left(4.8 \times 10^{-10}\right)^2}{(3)^2 \times\left(6.626 \times 10^{-27}\right)^2}\)

= 2.4242 ×  10¹⁴

Question 4. Calculate the wavelength and frequency associated with the spectral line having the longest wavelength in the fund series of hydrogen spectra.
Answer:

In the case of the Pfund series, the spectral line with the longest wavelength is obtained when the electronic transition occurs from n2 = 6 to nl = 5. Thus

⇒ \(\bar{v}=\frac{1}{\lambda}=109678\left(\frac{1}{5^2}-\frac{1}{6^2}\right)=1340.5\)

⇒ \(\lambda=7.4 \times 10^{-4} \mathrm{~cm}\)

= \(\frac{c}{\lambda}=\frac{3 \times 10^{10}}{7.4 \times 10^{-4}}\)

v = \(\frac{c}{\lambda}=\frac{3 \times 10^{10}}{7.4 \times 10^{-4}}=4.05 \times 10^{13} \mathrm{~s}^{-1}\)

= \(4.05 \times 10^{13} \mathrm{~Hz}\)

Question 5. Calculate the energy of 1 mol of photons associated with a frequency of 5 ×10¹⁰s^-1
Answer:

E = N₀hv=  \(6.022 \times 10^{23}\left(6.626 \times 10^{-34}\right)\left(5 \times 10^{10}\right)\)

=19.95J

Question 6. The wavelength associated with a moving particle of mass 0.1 mg is 3.3× 10^-29m. Find its velocity, [h = 6.6×  10^-34 kg.m².s^-1]
Answer:

λ =  \(\frac{h}{m v}\)

v = \(\frac{h}{m \lambda}=\frac{6.6 \times 10^{-34}}{\left(0.1 \times 10^{-6}\right) \times\left(3.3 \times 10^{-29}\right)}\)

= \(200 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Question 7. Calculate the kinetic energy of a moving electron associated with a wavelength of 4.8 pm.
Answer:

v = \(\frac{h}{m \lambda}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right) \times\left(4.8 \times 10^{-12} \mathrm{~m}\right)}\)

= \(1.51 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Kinetic energy

⇒  \(=\frac{h}{m \lambda}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1}}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right) \times\left(4.8 \times 10^{-12} \mathrm{~m}\right)}\)

=\(1.038 \times 10^{-14} \mathrm{~J}\)

Question 8. Calculate the frequency and wavelength of the energy emitted when the electron jumps from the 4th orbit to the 1st orbit of the H-atom.
Answer:

⇒  \(\bar{v}=\frac{1}{\lambda}=1.09678 \times 10^7 \times\left(\frac{1}{1^2}-\frac{1}{4^2}\right)=10.28 \times 10^6 \mathrm{~m}^{-1}\)

⇒  \(\text { Frequency, } v=\frac{c}{\lambda}=\frac{3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}}{9.73 \times 10^{-8} \mathrm{~m}}=3.1 \times 10^5 \mathrm{~s}^{-1}\)

Question 9. The wavelength of the first line in the Balmer series of H-atom is 15200 cm^-1. Calculate the wavelength of the first line in the same series of Li²⁺ ions. 10. The ionization potential of sodium is 4.946 × 102kJ.mol^-1 Calculate the wavelength of the radiation required to ionize a sodium atom

⇒  \(\bar{v}_{\mathrm{H}}=R_H\left(\frac{1}{2^2}-\frac{1}{n^2}\right)=15200 \mathrm{~cm}^{-1}\)

= \(\bar{v}_{\mathrm{Li} \mathrm{i}^{3+}}=\bar{v}_{\mathrm{H}} \times z^2=15200 \times 3^2=136800 \mathrm{~cm}^{-1}\)

Question 11. The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz) (Inm = 10⁹m)
Answer:

For violet light, \(v_1=\frac{c}{\lambda_1}=\frac{3 \times 10^8}{400 \times 10^{-9}}=7.5 \times 10^{14} \mathrm{~Hz}\)

For red light, \(v_2=\frac{c}{\lambda_2}=\frac{3 \times 10^8}{750 \times 10^{-9}}=4.0 \times 10^{14} \mathrm{~Hz}\)

Thus the frequency range of visible light extends from 11.0 × 10¹⁴HZ to 7.5 × 10¹⁴ Hz.