NEET Physics Class 12 Chapter 3 Modern Physics Notes

Modern Physics Photoelectric Effect Hertz’s observations

The phenomenon of photoelectric emission was discovered in 1887 by Heinrich Hertz (1857–1894), during his electromagnetic wave experiments. In his experimental investigation on the production of electromagnetic waves through a spark discharge, Hertz observed that high voltage sparks across the detector loop were enhanced when the emitter plate was illuminated by ultraviolet light from an arc lamp.

Light shining on the metal surface somehow facilitated the escape of free, charged particles which we now know as electrons. When light falls on a metal surface, some electrons near the surface absorb enough energy from the incident radiation to overcome the attraction of the positive ions in the surface material.

After gaining sufficient energy from the incident light, the electrons escape from the surface of the metal into the surrounding space.

Hallwach’s And Lenard’s Observations

Wilhelm Hallwachs and Phillipp Lenard investigated the phenomenon of photoelectric emission in detail during 1886–1902.

Lenard (1862–1947) observed that when ultraviolet radiations were allowed to fall on the emitter plate of an evacuated glass tube enclosing two electrodes (metal plates), current flows in the circuit figure.

As soon as the ultraviolet radiations were stopped, the current flow also stopped. These observations indicate that when ultraviolet radiations fall on the emitter plate C, electrons are ejected from it which are attracted towards the positive, collector plate A by the electric field. The electrons flow through the evacuated glass tube, resulting in the current flow.

Thus, light falling on the surface of the emitter causes current in the external circuit. Hallwachs and Lenard studied how this photocurrent varied with collector plate potential and with the frequency and intensity of incident light.

Hallwachs, in 1888, undertook the study further and connected a negatively charged zinc plate to an electroscope. He observed that the zinc plate lost its charge when it was illuminated by ultraviolet light.

Further, the uncharged zinc plate became positively charged when it was irradiated by ultraviolet light. The positive charge on a positively charged zinc plate was found to be further enhanced when it was illuminated by ultraviolet light.

From these observations, he concluded that negatively charged particles were emitted from the zinc plate under the action of ultraviolet light. After the discovery of the electron in 1897, it became evident that the incident light caused electrons to be emitted from the emitter plate.

Due to the negative charge, the emitted electrons are pushed toward the collector plate by the electric field. Hallwachs and Lenard also observed that when ultraviolet light fell on the emitter plate, no electrons were emitted at all when the frequency of the incident light was smaller than a certain minimum value, called the threshold frequency. This minimum frequency depends on the nature of the material of the emitter plate.

It was found that certain metals like zinc, cadmium, magnesium, etc. responded only to ultraviolet light, having a short wavelength, to cause electron emission from the surface. However, some alkali metals such as lithium, sodium, potassium, cesium, and rubidium were sensitive even to visible light.

All these photosensitive substances emit electrons when they are illuminated by light. After the discovery of electrons, these electrons were termed photoelectrons. The phenomenon is called the photoelectric effect.

Properties Of Photons

• A photon is a packet of energy emitted from a source of radiation. Photons are carrier particles of electromagnetic interaction.
• Photons travel in straight lines with speed of light c = 3 × 108 m/s.
• The energy of photons is given as \(\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}=\mathrm{mc}^2\) where v is frequency, lamba is wavelength, h is Planck’s constant.
• The effective or motional mass of photon is given as \(m=\frac{E}{c^2}=\frac{h \nu}{c^2}=\frac{h}{\lambda c}\)
• The momentum of a photon is given as \(\mathrm{p}=\mathrm{mc}=\frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{h} v}{\mathrm{c}}=\frac{\mathrm{h}}{\lambda}\)
• The photon is a chargeless particle of zero rest mass
• Photons are electrically neutral. They are not deflected by electric and magnetic fields.
• If E is the energy of source in joule then number of photons emitted is \(\mathrm{n}=\frac{\text { total energy radiated }}{\text { energy of each photon }}=\frac{E}{h \nu}=\frac{E \lambda}{h c}\)
• The intensity of photons is defined as the amount of energy carried per unit area per unit time. or power carried per unit area Intesity \(\left(I_p\right)=\frac{\text { Energy }}{\text { area } \times \text { time }}=\frac{\text { Power }}{\text { area }}, I_p=n h \nu=\frac{N}{4 \pi r^2} P\)

where n = number of photons per unit area per unit time N = number of photons, P = power of source

For a point source \(\mathrm{I}_{\mathrm{p}}=\mathrm{nh} v=\frac{\mathrm{N}}{4 \pi \mathrm{r}^2} \mathrm{P}\)

For a line source \(\mathrm{I}=\mathrm{nh} v=\frac{\mathrm{N}}{2 \pi \mathrm{r} \ell} \mathrm{P}\)

Solved Examples

Example 1. Find the number of photons in 6.62 joules of radiation energy of frequency 10¹²Hz.
Solution: Number Of Photons \(n=\frac{E}{h \nu}=\frac{6.62}{6.62 \times 10^{-34} \times 10^{12}}=10^{22}\)

Example 2. Calculate the energy and momentum of a photon of wavelength 6600Å.
Solution: energy of photon \(\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{6600 \times 10^{-10}}=3 \times 10^{-19} \mathrm{~J}\)

Momentum Of Photon \(\mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{6.6 \times 10^{-34}}{6600 \times 10^{-10}}=10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{sec}\)

Important Terms Related To The Photoelectric Effect

When electromagnetic radiations of suitable wavelength are incident on a metallic surface then electrons are emitted, this phenomenon is called the photoelectric effect.

Photoelectron: The electron emitted in the photoelectric effect is called a photoelectron.

Photoelectric current: If current passes through the circuit in a photoelectric effect then the current is called photoelectric current.

Work function: The minimum energy required to make an electron free from the metal is called work function. It is constant for a metal and denoted by  or W. It is the minimum for Cesium. It is relatively less for alkali metals.

Work functions of some photosensitive metals

To produce a photoelectric effect only metal and light are necessary but for observing it the circuit is completed. Figure shows an arrangement used to study the photoelectric effect.

Here the plate (1) is called emitter or cathode and the other plate (2) is called collector or anode.

Saturation current: When all the photoelectrons emitted by the cathode reach the anode then the current flowing in the circuit at that instant is known as saturated current, this is the maximum value of photoelectric current.

Stopping potential: The minimum magnitude of the negative potential of the anode to the cathode for which the current is zero is called stopping potential. This is also known as cutoff voltage. This voltage is independent of intensity.

Retarding potential: Negative potential of the anode to the cathode which is less than the stopping potential is called retarding potential.

Observations: (Made By Einstein)

A graph between the intensity of light and photoelectric current is found to be a straight line as shown in the figure. Photoelectric current is directly proportional to the intensity of incident radiation. In this experiment, the frequency and retarding potential are kept constant.

A graph between photoelectric current and a potential difference between cathode and anode is found as shown in the figure.

In the case of saturation current, rate of emission of photoelectrons = rate of flow of photoelectrons, here, vs→ stopping potential and it is a positive quantity Electrons emitted from the surface of metal have different energies. Maximum kinetic energy of photoelectron on the cathode = eVs KEmax = eV s

Whenever the photoelectric effect takes place, electrons are ejected out with kinetic energies ranging from 0 to K.Emax

i.e. 0 ≤ KEc ≤ eVs

The energy distribution of photoelectrons is shown in the figure.

Stopping potential (Vs) eVs = hv – W (Work function W = hv0)

⇒ \(V_s=\frac{h\left(v-v_0\right)}{e}\)

If the intensity is increased (keeping the frequency constant) then the saturation current is increased by the same factor by which intensity increases. The stopping potential is the same, so the maximum value of kinetic energy is not affected.

If light of different frequencies is used then the obtained plots.

It is clear from the graph, that as v increases, stopping potential increases, which means the maximum value of kinetic energy increases.

Graphs between the maximum kinetic energy of electrons ejected from different metals and the frequency of light used are found to be straight lines of the same slope.

Graph between Kmax and v m₁, m₂, m₃: Three different metals.

It is clear from the graph that there is a minimum frequency of electromagnetic radiation that can produce a photoelectric effect, which is called threshold frequency.

vth = Threshold frequency

For photoelectric effect v>vth

For no photoelectric effect v < vth Minimum frequency for the photoelectric effect.

Threshold wavelength \(\begin{aligned}
& v_{\min }=v_{t h} \\
& \left(\lambda_{t h}\right) \rightarrow
\end{aligned}\) The maximum wavelength of radiation that can produce a photoelectric effect. \(\lambda \leq \lambda_{\text {th }}\) for photo electric effect Maximum wavelength for photoelectric effect \(\lambda_{\max }=\lambda_{\mathbb{E n}} .\) Now write the equation of a straight line from the graph.

We have \(\mathrm{K}_{\max }=\mathrm{Av}+\mathrm{B}\)

When \(v=v_{\text {th }}, K_{\max }=0\)

And B = – Avth

Hence [Kmax = A(v – vth)]

and A = tan 0 = 6.63 × 10–34 J-s (from experimental data) later on ‘A’ was found to be ‘h’.

It is also observed that the photoelectric effect is an instantaneous process. When light falls on the surface electrons start ejecting without taking any time.

Three Major Features Of The Photoelectric Effect Cannot Be Explained In Terms Of The Classical Wave Theory Of Light.

Intensity: The energy crossing per unit area per unit time perpendicular to the direction of propagation is called the intensity of a wave.

Consider a cylindrical volume with an area of cross-section A and length c Δt along the X-axis. The energy contained in this cylinder crosses the area A in time Δt as the wave propagates at speed c. The energy contained.

The intensity is \(\begin{aligned}
& \mathrm{U}=\mathrm{U}_{\mathrm{av}}(\mathrm{c} . \Delta \mathrm{t}) \mathrm{A} \\
& \mathrm{I}=\frac{\mathrm{U}}{\mathrm{A} \Delta \mathrm{t}}=\mathrm{U}_{\mathrm{av}} \mathrm{c} .
\end{aligned}\)

In terms of maximum electric field, \(\mathrm{I}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{C}.\)

If we consider light as a wave then the intensity depends upon the electric field. If we take work function W = Δ . A . t, then \(t=\frac{W}{I A}\)

so for the photoelectric effect, there should be a time lag because the metal has a work function. However, it is observed that the photoelectric effect is an instantaneous process. Hence, light is not of a wave nature.

The intensity problem: Wave theory requires that the oscillating electric field vector E of the light wave increases in amplitude as the intensity of the light beam is increased. Since the force applied to the electron is eE, this suggests that the kinetic energy of the photoelectrons should also be increased as the light beam is made more intense. However, observation shows that maximum kinetic energy is independent of the light intensity.

The frequency problem: According to the wave theory, the photoelectric effect should occur for any frequency of the light, provided only that the light is intense enough to supply the energy needed to eject the photoelectrons. However, observations show that there exists for each surface a characteristic cutoff frequency vth, for frequencies less than that, the photoelectric effect does not occur, no matter how intense is light beam.

The time delay problem: If the energy acquired by a photoelectron is absorbed directly from the wave incident on the metal plate, the “effective target area” for an electron in the metal is limited and probably not much more than that of a circle of diameter roughly equal to that of an atom. In classical theory, the light energy is uniformly distributed over the wavefront.

Thus, if the light is feeble enough, there should be a measurable time lag, between the impinging of the light on the surface and the ejection of the photoelectron. During this interval, the electron should be absorbing energy from the beam until it has accumulated enough to escape. However, no detectable time lag has ever been measured. Now, quantum theory solves these problems by providing the correct interpretation of the photoelectric effect.

Planck’s Quantum Theory:

The light energy from any source is always an integral multiple of a smaller energy value called the quantum of light. Hence energy Q = NE,

where E = hv and N (number of photons) = 1,2,3,…

Here energy is quantized. hv is the quantum of energy, it is a packet of energy called a photon.

⇒ \(E=h \nu=\frac{h c}{\lambda} \quad \text { and } \quad h c=12400 \mathrm{eVA}\)

Einstein’s Photon Theory

In 1905 Einste made a remarkable assumption about the nature of light; namely, that, under some circumstances, it behaves as if its energy is concentrated into localized bundles, later called photons. The energy E of a single photon is given by E = hv, If we apply Einstein’s photon concept to the photoelectric effect, we can write hv = W + K max, (energy conservation)

The equation says that a single photon carries an energy h into the surface where it is absorbed by a single electron. Part of this energy W (called the work function of the emitting surface) is used in causing the electron to escape from the metal surface.

The excess energy (hv – W) becomes the electron kinetic energy; if the electron does not lose energy by internal collisions as it escapes from the metal, it will still have this much kinetic energy after it emerges. Thus Kmax represents the maximum kinetic energy that the photoelectron can have outside the surface. There is complete agreement of the photon theory with the experiment.

Now IA = Nhv \(\mathrm{N}=\frac{\mathrm{IA}}{\mathrm{h} \nu}\) no. of photons incident per unit time on an area ‘A’ when the light of intensity ‘I’ is incident normally.

If we double the light intensity, we double the number of photons and thus double the photoelectric current; we do not change the energy of the individual photons or the nature of the individual photoelectric processes.

The second objection (the frequency problem) is met if Kmax equals zero, we have

⇒ \(\mathrm{h} \mathrm{v}_{\mathrm{th}}=\mathrm{W} \text {, }\)

Which asserts that the photon has just enough energy to eject the photoelectrons and none extra to appear as kinetic energy. If vth is reduced below th, h will be smaller than W and the individual photons, no matter how many of them there are (that is, no matter how intense the illumination), will not have enough energy to eject photoelectrons.

The third objection (the time delay problem) follows from the photon theory because the required energy is supplied in a concentrated bundle. It is not spread uniformly over the beam cross-section as in the wave theory.

Hence Einstein’s equation for the photoelectric effect is given by

⇒ \(\mathrm{h} v=\mathrm{h} v_{\mathrm{th}}+\mathrm{K}_{\max } \quad \mathrm{K}_{\max }=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{\mathrm{th}}}\)

Solved Examples

Example 3. In an experiment on photoelectric emission, the following observations were made;

1. Wavelength of the incident light = 1.98 × 10–7 m;
2. Stopping potential = 2.5 volt.

Find:

1. The kinetic energy of photoelectrons with maximum speed.
2. Work function and
3. Threshold frequency;

Solution:

Since vs = 2.5 V, Kmax = eVs

so, K max = 2.5 eV

Energy of incident photon 12400 eV E = 1980 = 6.26 eV W = E – K max = 3.76 eV

⇒ \(\mathrm{E}=\frac{12400}{1980} \mathrm{eV}=6.26 \mathrm{eV} \quad \mathrm{W}=\mathrm{E}-\mathrm{K}_{\max }=3.76 \mathrm{eV}\)

⇒ \(\begin{aligned}
& \mathrm{h} v_{\text {th }}=\mathrm{W}=3.76 \times 1.6 \times 10^{-19} \mathrm{~J} \\
& v_{\text {th }}=\frac{3.76 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} \approx 9.1 \times 10^{14} \mathrm{~Hz}
\end{aligned}\)

Example 4. A beam of light consists of four wavelengths 4000 Å, 4800 Å, 6000 Å, and 7000 Å, each of intensity 1.5 × 10–3 Wm–2. The beam falls normally on an area of 10–4 m2 of a clean metallic surface of work function 1.9 eV. Assuming no loss of light energy (i.e. each capable photon emits one electron) calculate the number of photoelectrons liberated per second.
Solution:

⇒ \(\begin{aligned}
& E_1=\frac{12400}{4000}=3.1 \mathrm{eV}, \quad E_2=\frac{12400}{4800}=2.58 \mathrm{eV} \quad E_3=\frac{12400}{6000}=2.06 \mathrm{eV} \\
& \text { and } \quad E_4=\frac{12400}{7000}=1.77 \mathrm{eV}
\end{aligned}\)

Therefore, light of wavelengths 4000 Å, 4800 Å, and 6000 Å can only emit photoelectrons.

Number of photoelectrons emitted per second = No. of photons incident per second)

⇒ \(=\frac{I_1 A_1}{E_1}+\frac{I_2 A_2}{E_2}+\frac{I_3 A_3}{E_3} \quad=I A\left(\frac{1}{E_1}+\frac{1}{E_2}+\frac{1}{E_3}\right)=\frac{\left(1.5 \times 10^{-3}\right)\left(10^{-4}\right)}{1.6 \times 10^{-19}}\left(\frac{1}{3.1}+\frac{1}{2.58}+\frac{1}{2.06}\right)\)

Example 5. A small potassium foil is placed (perpendicular to the direction of incidence of light) at a distance r (= 0.5 m) from a point light source whose output power P0 is 1.0W.

Assuming the wave nature of light how long would it take for the foil to soak up enough energy (= 1.8 eV) from the beam to eject an electron? Assume that the ejected photoelectron collected its energy from a circular area of the foil whose radius equals the radius of a potassium atom (1.3 × 10–10 m.

Solution: If the source radiates uniformly in all directions, the intensity  of the light at a distance r is given by \(\mathrm{I}=\frac{P_0}{4 \pi \mathrm{r}^2}=\frac{1.0 \mathrm{~W}}{4 \pi(0.5 \mathrm{~m})^2}=0.32 \mathrm{~W} / \mathrm{m}^2 \text {. }\)

The target area A is π(1.3 × 10–10 m)2 or 5.3 × 10–20 m2, so the rate at which energy falls on the target is given by P = πA = (0.32 W/m2) (5.3 × 10–20 m2) = 1.7 × 10–20 J/s.

If all this incoming energy is absorbed, the time required to accumulate enough energy for the electron to escape is \(\mathrm{t}=\left(\frac{1.8 \mathrm{eV}}{1.7 \times 10^{-20} \mathrm{~J} / \mathrm{s}}\right)\left(\frac{1.6 \times 10^{-19} \mathrm{~J}}{1 \mathrm{eV}}\right)=17 \mathrm{~s}.\)

Our selection of a radius for the effective target area was somewhat arbitrary, but no matter what reasonable area we choose, we should still calculate a “soak-up time” within the range of easy measurement. However, no time delay has ever been observed under any circumstances, the early experiments setting an upper limit of about 10–9 s for such delays.

Example 6. A metallic surface is irradiated with monochromatic light of variable wavelength. Above a wavelength of 5000 Å, no photoelectrons are emitted from the surface. With an unknown wavelength, the stopping potential is 3 V. Find the unknown wavelength.
Solution: using equation of photoelectric effect \(\mathrm{K}_{\max }=\mathrm{E}-\mathrm{W} \quad\left(\mathrm{K}_{\max }=\mathrm{eV}_{\mathrm{s}}\right)\)

Therefore \(3 \mathrm{eV}=\frac{12400}{\lambda}-\frac{12400}{5000}=\frac{12400}{\lambda}-2.48 \mathrm{eV} \text { or } \lambda=2262 \mathrm{~A}\)

Example 7. Illuminating the surface of a certain metal alternately with light of wavelengths λ1 = 0.35 λm and λ2 = 0.54 μm, it was found that the corresponding maximum velocities of photoelectrons have a ratio μ = 2. Find the work function of that metal.
Solution: Using the equation for two wavelengths

⇒ \(\begin{aligned}
&\frac{1}{2} m v_1^2=\frac{h c}{\lambda_1}-W\\
&\frac{1}{2} m v_2^2=\frac{h c}{\lambda_2}-W
\end{aligned}\)

Dividing Eq. 1 with Eq. 2 with v1 = 2v2, we have \(4=\frac{\frac{h c}{\lambda_1}-W}{\frac{h c}{\lambda_2}-W}\)

⇒ \(3 \mathrm{~W}=4\left(\frac{\mathrm{hc}}{\lambda_2}\right)-\left(\frac{\mathrm{hc}}{\lambda_1}\right)=\frac{4 \times 12400}{5400}-\frac{12400}{3500}=5.64 \mathrm{eV}\)

Example 8. Light described at a place by the equation E = (100 V/m) [sin (5 × 10 15 s–1) t + sin (8 × 1015 s–1)t] falls on a metal surface having work function 2.0 eV. Calculate the maximum kinetic energy of the photoelectrons.
Solution: The light contains two different frequencies. The one with a larger frequency will cause photoelectrons with the largest kinetic energy. This larger frequency is

⇒ \(v=\frac{\omega}{2 \pi}=\frac{8 \times 10^{15} \mathrm{~s}^{-1}}{2 \pi}\)

The maximum kinetic energy of the photoelectrons is

⇒ \(\begin{aligned}
\mathrm{K}_{\max } & =\mathrm{hu}-\mathrm{W} \\
& =\left(4.14 \times 10^{-15} \mathrm{eV}-\mathrm{s}\right) \times\left(\frac{8 \times 10^{15}}{2 \pi} \mathrm{s}^{-1}\right)-2.0 \mathrm{eV}=5.27 \mathrm{eV}-2.0 \mathrm{eV}=3.27 \mathrm{eV} .
\end{aligned}\)

Compton Effect

The scattering of a photon by an electron in which the wavelength of the scattered photon is greater than the wavelength of the incident photon is called the Compton effect. Conservation of energy gives h_v + m₀ c² = h_v’ + mc²

Conservation of momentum along y-axis gives \(0=\frac{h v^{\prime}}{c} \sin \theta-p \sin \theta\)

Conservation of momentum along x-axis gives \(\frac{\mathrm{h} v}{\mathrm{c}}=\frac{\mathrm{h} v^{\prime}}{\mathrm{c}} \cos \theta+\mathrm{p} \cos \phi\)

Compton shif \(\Delta \lambda=\lambda^{\prime}-\lambda=\frac{h}{m_0 c}(1-\cos \theta)\)

The quantity \(\lambda_{\mathrm{c}}=\frac{\mathrm{h}}{\mathrm{m}_0 \mathrm{c}}=2.42 \times 10^{12} \mathrm{~m}\) s called Compton wavelength of electron.

For maximum shift \(\theta=\pi \quad \text { so } \quad(\Delta \lambda)_{\max }=\frac{2 \mathrm{~h}}{\mathrm{~m}_0 \mathrm{c}}=4.48 \times 10^{-12} \mathrm{~m}\)

The Compton effect shows the quantum concept and particle nature of photons.

De-Broglie Wavelength Of Matter Wave

A photon of frequency v and wavelength λ has energy

⇒ \(E=h \nu=\frac{h c}{\lambda}\)

By Einstein’s energy-mass relation, E = mc2 the equivalent mass m of the photon is given by,

⇒ \(\begin{gathered}
m=\frac{E}{c^2}=\frac{h \nu}{c^2}=\frac{h}{\lambda c} \\
\lambda=\frac{h}{m c} \quad \text { or } \quad \lambda=\frac{h}{p}
\end{gathered}\)

Here p is the momentum of the photon. By analogy de-Broglie suggested that a particle of mass m moving with speed v behaves in some ways like waves of wavelength (called de-Broglie wavelength and the wave is called matter wave) given by,

⇒ \(\lambda=\frac{h}{m v}=\frac{h}{p}\)

where p is the momentum of the particle. Momentum is related to the kinetic energy by the equation, \(p=\sqrt{2 K m}\) and a charge q when accelerated by a potential difference V gains a kinetic energy K = qV. Combining all these relations Eq. (3), can be written as, \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{Km}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{qVm}}} \text { (de-Broglie wavelength) }\) de-Broglie wavelength for an electron If an electron (charge = e) is accelerated by a potential of V volts, it acquires a kinetic energy, K = eV Substituting the values of h, m, and q in Eq., we get a simple formula for calculating deBroglie wavelength of an electron.

⇒ \(\lambda(\text { in } \quad A)=\sqrt{\frac{150}{V(\text { in volts })}} \frac{12.27 }{\sqrt{V}}\)

The de-Broglie wavelength of a proton:

Velocity \(V_p=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}_{\mathrm{p}}}}\)
Momentum ,\(p_p=\sqrt{2 m_p e V}\)

So Wavelength is \(\lambda_p=\frac{h}{\sqrt{2 m_p e V}}=\frac{0.286}{\sqrt{V}} A\)

de-Broglie wavelength of a gas molecule: Let us consider a gas molecule at absolute temperature T. The Kinetic energy of a gas molecule is given by

⇒ \(K. E .=\frac{3}{2} k T \text {; }\) ,therefore \(\lambda_{\text {gas molecule }}=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}\)

Force Due To Radiation (Photon)

Each photon has a definite energy and a definite linear momentum. All photons of light of a particular wavelength λ have the same energy E = hc/λ and the same momentum p = h/λ. When light of intensity λ falls on a surface, it exerts force on that surface.

Assume the absorption and reflection coefficient of the surface to be ‘a’ and ‘r’ and assume no transmission. Assume the light beam falls on the surface of surface area ‘A’ perpendicularly as shown in the figure.

For calculating the force exerted by the beam on the surface, we consider the following cases

Case: 1

a = 1, r = 0

Initial momentum of the photon \(=\frac{\mathrm{h}}{\lambda}\)

Inal momentum of photon = 0

Change in momentum of photon = \(\frac{\mathrm{h}}{\lambda}\) \(\Delta \mathrm{P}=\frac{\mathrm{h}}{\lambda}\) energy incident per unit time = IA A no. of photons incident per unit time \(=\frac{I A}{h v}=\frac{I A \lambda}{h c}\)

Therefore total change in momentum per unit time \(=\mathrm{n} \Delta \mathrm{P}=\frac{\mathrm{IA} \lambda}{\mathrm{hc}} \times \frac{\mathrm{h}}{\lambda}=\frac{\mathrm{IA}}{\mathrm{c}}\)

Force on photons = total change in momentum per unit time \(=\frac{I A}{c}\)

force on plate due to photons(F) \(=\frac{\mathrm{IA}}{\mathrm{c}}\) pressure \(=\frac{F}{A}=\frac{I A}{c A}=\frac{I}{c}\)

Case : (2)

when r = 1, a = 0

Initial momentum of the photon \(=\frac{\mathrm{h}}{\lambda}\) (downward)

Final momentum of photon \(=\frac{\mathrm{h}}{\lambda}\) (upward)

Change in momentum \(=\frac{\mathrm{h}}{\lambda}+\frac{\mathrm{h}}{\lambda}=\frac{2 \mathrm{~h}}{\lambda}\)

∴ Energy incident per unit time = IA

Number of photons incidient per unit = \(\frac{\mathrm{IA} \lambda}{\mathrm{hc}}\)

∴ Total change in momentum per unit time \(=n \cdot \Delta P=\frac{I A \lambda}{h c} \cdot \frac{2 h}{\lambda}=\frac{2 I A}{C}\)

Force = total change in momentum per unit time

⇒ \(F=\frac{2 I A}{c}\) (upward on photons and downward on the plate)

Pressure \(P=\frac{F}{A}=\frac{2 I A}{c A}=\frac{2 I}{c}\)

Case 3

When o < r < 1 a + r = 1

change in momentum of a photon when it is reflected \(=\frac{2 h}{\lambda}\) upward

change in momentum of a photon when it is absorbed \(=\frac{\mathrm{h}}{\lambda}\) (upward)

No. of photons incident per unit time \(=\frac{\mathrm{IA} \lambda}{\mathrm{hc}}\)

No. of photons reflected per unit time \(=\frac{\mathrm{IA} \lambda}{\mathrm{hc}} \cdot \mathrm{r}\)

No. of photon absorbed per unit time \(=\frac{I A \lambda}{h c}(1-r)\)

force due to absorbed photon (Fa) \(=\frac{I A \lambda}{h c}(1-r) \cdot \frac{h}{\lambda}=\frac{I A}{c}(1-r) \quad \text { (downward) }\)

Force due to reflected photon (Fr) \(=\frac{I A \lambda}{h c} \cdot r \frac{2 h}{\lambda}=\frac{2 I A \lambda}{c}\)

Total force = Fa+F \((\text { downward })=\frac{1 A}{c}(1-r)+\frac{2 \mid A r}{c}=\frac{1 A}{c}(1+r)\)

Now Pressure \(P=\frac{I A}{c}(1+r) \times \frac{1}{A}=\frac{I}{c}(1+r)\)

Example 9. An electron is accelerated by a potential difference of 50 volts. Find the de-Broglie wavelength associated with it.
Solution: For an electron, the de-Broglie wavelength is given by,

\(\lambda=\sqrt{\frac{150}{\mathrm{~V}}}=\sqrt{\frac{150}{50}}=\sqrt{3}=1.73 \mathrm{~A}\)

Example 10. Find the ratio of De-Broglie wavelength of molecules of hydrogen and helium which are at temperatures 27ºC and 127ºC respectively.
Solution: de-Broglie wavelength is given by

Therefore \(\frac{\lambda_{\mathrm{H}_2}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_2} \mathrm{~T}_{\mathrm{H}_2}}}=\sqrt{\frac{4}{2} \cdot \frac{(127+273)}{(27+273)}}=\sqrt{\frac{8}{3}}\)

Example 11. A plate of mass 10 gm is in equilibrium in the air due to the force exerted by a light beam on a plate. Calculate the power of the beam. Assume the plate is perfectly absorbing.

Solution: Since the plate is in the air, so gravitational force will act on this

⇒ \(\begin{aligned}
\mathrm{F}_{\text {gravitational }} & =\mathrm{mg} \\
& =10 \times 10^{-3} \times 10 \\
& =10^{-1} \mathrm{~N}
\end{aligned}\)

for equilibrium force exerted by light beam should be equal to \(\text { gravitational }\)

⇒ \(\mathrm{F}_{\text {photon }}=\mathrm{F}_{\text {gravitational }}\)

Let the power of the light beam be P

Therefore \(\begin{aligned}
& F_{\text {photon }}=\frac{P}{C} \\
& P=3.0 \times 10^8 \times 10^{-1} \\
& P=3 \times 10^7 \mathrm{~W}
\end{aligned}\)

⇒ \(\frac{P}{c}=10^{-1}\)

Davisson And Germer Experiment

• The experiment demonstrates the diffraction of electron beam by crystal surfaces
• The experiment provides the first experimental evidence for the wave nature of the material particles

• The electrons are diffracted like X-rays. The Bragg’s law of diffraction are
  D sin Φ = nλ and 2d sinθ = nλ
  • where D = interatomic distance and d = interplanar distance
  • θ = angle between the scattering plane and the incident beam
  • Φ = scattering angle 2θ + Φ = 180º
• The electrons are produced and accelerated into a beam by an electron gun.
  • The energy of electrons is given as \(\mathrm{E}=\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}\)

The accelerated electron beam is made to fall on a Ni crystal. The scattered electrons are detected by a detector

The experimental results are shown in the form of polar graphs plotted between scattering angle Φ and intensity of scattered electron beam at different accelerating voltages. The distance of the curve from point O is proportional to the intensity of the scattered electron beam.

Important Results

• The intensity of scattered electrons depends on the scattering angle Φ
• The kink at Φ = 50° is observed at all accelerating voltage.
• The size of the link becomes maximum at 54 volts. \(\text { For } \phi=50^{\circ} \quad \text { For } \theta=\frac{180-\phi}{2}=65^{\circ}\)
• D = sinλ = nλ 2 d sinθ = nθ
• D = 2.15Å & n = 1 (for Ni)
• n = 1 and d = 0.91Å (for Ni)
  • ∴ λ = 2.15 sin 50 = 1.65Å
• λ = 2 × 0.91 sin65 = 1.65Å
• For V = 54 volt

de-Broglie wavelength

This value of λ is in close agreement with the experimental value. Thus this experiment verifies de Broglie’s hypothesis.

Solved Example

Example 12. An electron beam of energy 10 KeV is incident on metallic foil. If the interatomic distance is 0.55Å. Find the angle of diffraction.
Solution: \(\lambda=D \sin \phi \quad \text { and } \quad \lambda=\frac{12.27}{\sqrt{V}} A \text { so } \frac{12.27}{\sqrt{V}}=D \sin \phi\)

⇒ \(\frac{12.27 \times 10^{-10}}{\sqrt{10 \times 10^3}}=0.55 \times 10^{-10} \sin \phi\) \(\sin \phi=\frac{12.27}{0.53 \times 100}=0.2231\)

⇒ \(\text { or } \phi=\sin ^{-1}(0.2231) \approx 12.89^{\circ}\)

Thomson’s Atomic Model:

J.J. Thomson suggested that atoms are just positively charged lumps of matter with electrons embedded in them like raisins in a fruit cake. The total charge of the atom is zero and the atom is electrically neutral. Thomson’s model called the ‘plum pudding’ model is illustrated in the figure.

Thomson played an important role in discovering the electron, through a gas discharge tube by discovering cathode rays. His idea was taken seriously. But the real atom turned out to be quite different.

Rutherford’s Nuclear Atom:

Rutherford suggested that; “ All the positive charge and nearly all the mass were concentrated in a very small volume of the nucleus at the center of the atom. The electrons were supposed to move in circular orbits around the nucleus (like planets around the sun). The electron static attraction between the two opposite charges is the required centripetal force for such motion.

Hence \(\frac{m v^2}{r}=\frac{k Z e^2}{r^2}\) and total energy = potential energy + kineitc energy \(=\frac{-k Z e^2}{2 r}\)

Rutherford’s model of the atom, although strongly supported by evidence for the nucleus, is inconsistent with classical physics. This model suffers from two defects.

Regarding the stability of an atom: An electron moving in a circular orbit around a nucleus is accelerating and according to electromagnetic theory it should, therefore, emit radiation continuously and thereby lose energy.

If total energy decreases then radius increases as given by the above formula. If this happened the radius of the orbit would decrease and the electron would spiral into the nucleus in a fraction of a second. But atoms do not collapse. In 1913 an effort was made by Neils Bohr to overcome this paradox.

Regarding the explanation of line spectrum: In Rutherford’s model, due to continuously changing radii of the circular orbits of electrons, the frequency of revolution of the electrons must be changing. As a result, electrons will radiate electromagnetic waves of all frequencies, i.e., the spectrum of these waves will be ‘continuous’ in nature. But experimentally the atomic spectra are not continuous. Instead, they are line spectra.

The Bohr’s Atomic Model

In 1913, Prof. Niel Bohr removed the difficulties of Rutherford’s atomic model by the application of Planck’s quantum theory. For this, he proposed the following postulates

An electron moves only in certain circular orbits, called stationary orbits. In stationary orbits, electrons do not emit radiation, contrary to the predictions of classical electromagnetic theory.

According to Bohr, there is a definite energy associated with each stable orbit, and an atom radiates energy only when it makes a transition from one of these orbits to another. If the energy of an electron in the higher orbit is E2 and that in the lower orbit is E1, then the frequency of the radiated waves is given by

hv=E2-E1 or \(v=\frac{E_2-E_1}{h}\)

Bohr found that the magnitude of the electron’s angular momentum is quantized, and this magnitude for the electron must be an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\) The magnitude of the angular momentum is L = mvr for a particle with mass m moving with speed v in a circle of radius r. So, according to Bohr’s postulate, \(m v r=\frac{n h}{2 \pi} \quad(n=1,2,3 \ldots .)\)

Each value of n corresponds to a permitted value of the orbit radius, which we will denote by rn. The value of n for each orbit is called the principal quantum number for the orbit. Thus, \(m v_n r_n=m v r=\frac{n h}{2 \pi}\)

According to Newton’s second law, a radially inward centripetal force of magnitude \(F=\frac{m v^2}{r_n}\) is needed by the electron which is provided by the electrical attraction between the positive proton and the negative electron.

Thus \(\frac{m v_n^2}{r_n}=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r_n^2}\)

Solving Eqs. (2) and (3), we get 2

⇒ \(\begin{aligned}
& r_n=\frac{\varepsilon_0 n^2 h^2}{\pi m e^2} \\
& v_n=\frac{e^2}{2 \varepsilon_0 n h}
\end{aligned}\)

The smallest orbit radius corresponds to n = 1. We’ll denote this minimum radius, called the Bohr radius as a 0. Thus, \(\mathrm{a}_0=\frac{\varepsilon_0 \mathrm{~h}^2}{\pi m \mathrm{e}^2}\)

Substituting values of e0, h, p, m, and e, we get a 0 = 0.529 × 10–10 m = 0.529 Å

Eq. 4, in terms of a0 can be written as, r n = n2 a 0 or rn  n2

Similarly, substituting values of e, 0, and h with n = 1 in Eq. (v), we get

v1 = 2.19 × 106 m/s

This is the greatest possible speed of the electron in the hydrogen atom. Which is
approximately equal to c/137 where c is the speed of light in vacuum.
Eq. (v), in terms of v1, can be written as,

⇒ \(v_n=\frac{v_1}{n} \quad \text { or } \quad v_n \propto \frac{1}{n}\)

Energy levels: Kinetic and potential energies Kn and U n in the nth orbit are given by

⇒ \(\mathrm{K}_{\mathrm{n}}=\frac{1}{2} \mathrm{mv}_{\mathrm{n}}{ }^2=\frac{\mathrm{me}^4}{8 \varepsilon_0^2 \mathrm{n}^2 \mathrm{~h}^2} \quad \text { and } \quad \mathrm{U}_{\mathrm{n}}=-\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{e}^2}{\mathrm{r}_{\mathrm{n}}} \quad=-\frac{m e^4}{4 \varepsilon_0^2 \mathrm{n}^2 \mathrm{~h}^2}\) (assuming infinity as a zero potential energy level)

The total energy En is the sum of the kinetic and potential energies.

So, \(E_n=K_n+U_n=-\frac{m e^4}{8 \varepsilon_0{ }^2 n^2 h^2}\)

Substituting values of m, e, e0, and h with n = 1, we get the least energy of the atom in the first orbit, which is –13.6 eV. Hence, E1=-13.6ev and \(E_n=\frac{E_1}{n^2}=-\frac{13.6}{n^2} \mathrm{eV}\)

Substituting n = 2, 3, 4, …., etc., we get the energies of atoms in different orbits.

E2 = – 3.40 eV, E3 = – 1.51 eV, …. E∞ = 0

Solved Examples

Example 13. An -particle with kinetic energy 10 MeV is heading towards a stationary point-nucleus of atomic number 50. Calculate the distance of the closest approach.
Solution: Therefore \(T E_{\mathrm{A}}=\mathrm{TE}_{\mathrm{B}}\) \(10 \times 10^6 e=\frac{K \times(2 e)(50 e)}{r_0}\)

r0 = 1.44 × 10–14 m

r0 = 1.44 × 10–4 Å

Example 14. A beam of α-particles of velocity 2.1 × 107 m/s is scattered by a gold (z = 79) foil. Find out the distance of the closest approach of the α-particle to the gold nucleus. The value of charge/mass for α-particle is 4.8 × 107 C/kg.
Solution:

⇒ \(\frac{1}{2} m_\alpha V_\alpha{ }^2=\frac{K(2 e)(Z e)}{r_0}\)

⇒\(r_0=\frac{2 \mathrm{~K}\left(\frac{2 \mathrm{e}}{\mathrm{m}_\alpha}\right)(79 \mathrm{e})}{\mathrm{V}_\alpha^2}=\frac{2 \times\left(9 \times 10^9\right)\left(4.8 \times 10^7\right)\left(79 \times 1.6 \times 10^{-19}\right)}{\left(2.1 \times 10^7\right)^2} ; r_0=2.5 \times 10^{-14} \mathrm{~m}\)

Hydrogen-Like Atoms The Bohr model of hydrogen can be extended to hydrogen-like atoms, i.e., one electron atom, the nuclear charge is +ze, where z is the atomic number, equal to the number of protons in the nucleus.

The effect in the previous analysis is to replace e2 everywhere with ze2. Thus, the equations for, rn, vn, and E n are altered as under:

⇒ \(r_n=\frac{\varepsilon_0 n^2 h^2}{n m z e^2}=\frac{n^2}{z} a_0 \quad \text { or } \quad r_n \propto \frac{n^2}{z}\)

where a 0 = 0.529 Å (radius of the first orbit of H)

⇒ \(v_n=\frac{z e^2}{2 \varepsilon_0 n h}=\frac{z}{n} v_1 \quad \text { or } \quad v_n \propto \frac{z}{n}\)

where v 1= 2.19 × 106 m/s (speed of an electron in the first orbit of H)

⇒ \(E_n=-\frac{m z^2 e^4}{8 \varepsilon_0^2 n^2 h^2}=\frac{z^2}{n^2} E_1 \text { or } \quad E_n \propto \frac{z^2}{n^2}\)

Where E1 = –13.60 eV (energy of the atom in the first orbit of H)

Destinations valid for single electron system

Ground state: The lowest energy state of any atom or ion is called the ground state of the atom. Ground state energy of H atom = –13.6 eV Ground state energy of He+ Ion = –54.4 eV

Ground state energy of Li++ Ion = –122.4 eV

Excited State: The state of an atom other than the ground state is called its excited state.

n = 2 first excited state

n = 3-second excited state

n = 4 third excited state

n = n

0 + 1 n0th excited state

Ionization energy (IE.): The minimum energy required to move an electron from the ground state to n = ∞ is called the ionization energy of the atom or ion.

• The ionization energy of H atom = 13.6 eV
• Ionisation energy of He+ Ion = 54.4 eV
• Ionisation energy of Li++ Ion = 122.4 eV

Ionization potential (I.P.): The potential difference through which a free electron must be accelerated from rest such that its kinetic energy becomes equal to the ionization energy of the atom is called the ionization potential of the atom.

• I.P of H atom = 13.6 V
• I.P. of He+ Ion = 54.4 V

Excitation energy: Energy required to move an electron from the ground state of the atom to any other exited state of the atom is called excitation energy of that state.

Energy in the ground state of H atom = –13.6 eV

Energy in the first excited state of H-atom = –3.4 eV

1st excitation energy = 10.2 eV.

Excitation Potential: The potential difference through which an electron must be accelerated from rest so that its kinetic energy becomes equal to the excitation energy of any state is called the excitation potential of that state.

1st excitation energy = 10.2 eV.

1st excitation potential = 10.2 V.

Binding energy or Separation energy: Energy required to move an electron from any state to n = is called binding energy of that state. or energy released during the formation of an H-like atom/ion from n = ∞ to some particular n is called binding energy of that state. The binding energy of the ground state of H-atom = 13.6 eV.

Solved Examples

Example 15. First excitation potential of a hypothetical hydrogen-like atom is 15 volts. Find the third excitation potential of the atom.
Solution: Let the energy of the ground state = E0

⇒ \(E_0=-13.6 Z^2 e V \quad \text { and } \quad E_n=\frac{E_0}{n^2} \quad \Rightarrow \quad n=2, E_2=\frac{E_0}{4}\) Given \(\frac{E_0}{4}-E_0=15-\frac{3 E_0}{4}=15 \quad \text { for } \quad n=4, \quad E_4=\frac{E_0}{16}\) third excitation energy

⇒ \(=\frac{E_0}{16}-E_0=-\frac{15}{16} E_0=-\frac{15}{16} \cdot\left(\frac{-4 \times 15}{3}\right)=\frac{75}{4} \mathrm{eV}\)

Therefore Excitation potential is \(\frac{75}{4} V\)

Third excitation potential is \(\frac{75}{4} V\)

Emission spectrum of hydrogen atom:

Under normal conditions, the single electron in the hydrogen atom stays in the ground state (n = 1). It is excited to some higher energy state when it acquires some energy from external sources. But it hardly stays there for more than 10–8 seconds.

A photon corresponding to a particular spectrum line is emitted when an atom makes a transition from a state in an excited level to a state in a lower excited level or the ground level.

Let ni be the initial and nf the final energy state, then depending on the final energy state following series are observed in the emission spectrum of the hydrogen atom.

On Screen:

A photograph of spectral lines of the Lyman, Balmer, Paschen series of atomic hydrogen.

1, 2, 3….. represents the 1, 2 & 3 line of Lyman, Balmer, Paschen series.

The hydrogen spectrum (some selected lines)

Series limit: The line of any group having maximum energy of the photon and minimum wavelength of that group is called the series limit.

For the Lyman series nf = 1, for the Balmer series n f = 2, and so on.

Wavelength of Photon Emitted in De-excitation

According to Bohr when an atom makes a transition from a higher energy level to a lower level it emits a photon with energy equal to the energy difference between the initial and final levels. If E i is the initial energy of the atom before such a transition, Ef is its final energy after the transition, and the photon’s energy is \(h v=\frac{h c}{\lambda}\) then conservation of energy gives, \(h \nu=\frac{h c}{\lambda}=E_i-E_f\) (energy of emitted photon)

By 1913, the spectrum of hydrogen had been studied intensively. The visible line with longest wavelength, or lowest frequency is in the red and is called Hα, the next line, in the blue-green, is called Hb, and so on. In 1885, Johann Balmer, a Swiss teacher found a formula that gives the wavelengths of these lines. This is now called the Balmer series. The Balmer’s formula is,

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

Here, n = 3, 4, 5 …., etc. R = Rydberg constant = 1.097 × 107 m–1 andα is the wavelength of light/photon emitted during the transition, For n = 3, we obtain the wavelength of Hα line. Similarly, for n = 4, we obtain the wavelength of Hα line. For n = , the smallest wavelength (=3646 Å) of this series is obtained. Using the relation \(\mathrm{E}=\frac{\mathrm{hc}}{\lambda}\) we can find the photon energies corresponding to the wavelength of the Balmar series.

This formula suggests that, \(E_n=-\frac{R h c}{n^2}, n=1,2,3 \ldots . .\)

The wavelengths corresponding to other spectral series (Lyman, Paschen, (etc.) can be
represented by a formula similar to Balmer’s formula.

The wavelengths corresponding to other spectral series (Lyman, Paschen, (etc.) can be
represented by a formula similar to Balmer’s formula.

Lymen Series: \(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right), n=2,3,4 \ldots .\)

Paschen Serie: \(\frac{1}{\lambda}=R\left(\frac{1}{3^2}-\frac{1}{n^2}\right), n=4,5,6 \ldots . .\)

Brackett Series: \(\frac{1}{\lambda}=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right), n=5,6,7 \ldots . .\)

Pfund Series: \(\frac{1}{\lambda}=R\left(\frac{1}{5^2}-\frac{1}{n^2}\right), n=6,7,8\)

The Lyman series is in the ultraviolet and the Paschen. Brackett and Pfund’s series are in the infrared region

Solved Examples

Example 16. Calculate the wavelength and the frequency of the Hb line of the Balmer series for hydrogen.
Solution: The hb line of the Balmer series corresponds to the transition from n = 4 to n = 2 level. The corresponding wavelength for the Hβ line is, \(\frac{1}{\lambda}=\left(1.097 \times 10^7\right)\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\)

= 0.2056 × 107 = 4.9 × 10–7 m

⇒ \(v=\frac{c}{\lambda}=\frac{3.0 \times 10^8}{4.9 \times 10^{-7}}\)

⇒ \(v=\frac{c}{\lambda}=\frac{3.0 \times 10^8}{4.9 \times 10^{-7}}\)

= 6.12 × 1014 Hz

Example 17. Find the largest and shortest wavelengths in the Lyman series for hydrogen. In what region of the electromagnetic spectrum does each series lie?
Solution: The transition equation for the Lyman series is given by

⇒ \(\frac{1}{\lambda}=R\left[\frac{1}{(1)^2}-\frac{1}{n^2}\right] \quad n=2,3, \ldots . .\) for the largest wavelength, n = 2

⇒ \(\frac{1}{\lambda_{\max }}=1.097 \times 10^7\left(\frac{1}{1}-\frac{1}{4}\right)=0.823 \times 10^7\)

The shortest wavelength corresponds to n = infinity

Example 18. How many different wavelengths can be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number n?
Solution: From the nth state, the atom may go to (n – 1)th state, …., 2nd state, or 1st state. So there are (n – 1) possible transitions starting from the nth state. The atoms reaching (n – 1)th state may make (n – 2) different transitions. Similarly for other lower states. The total number of possible transitions is \((n-1)+(n-2)+(n-3)+\ldots \ldots \ldots \ldots 2+1=\frac{n(n-1)}{2}\)

Example 19. Find the kinetic energy potential energy and total energy in the first and second orbits of the hydrogen atom if the potential energy in the first orbit is taken to be zero.
Solution: \(\begin{array}{llll}
E_1=-13.60 \mathrm{eV} & \mathrm{K}_1=-\mathrm{E}_1=13.60 \mathrm{eV} & \mathrm{U}_1=2 \mathrm{E}_1=-27.20 \mathrm{eV} \\
\mathrm{E}_2=\frac{E_1}{(2)^2}=-3.40 \mathrm{eV} & \mathrm{K}_2=3.40 \mathrm{eV} & \text { and } & \mathrm{U}_2=-6.80 \mathrm{eV}
\end{array}\)

Now U 1 = 0, i.e., potential energy has been increased by 27.20 eV while kinetic energy will remain unchanged. So values of kinetic energy, potential energy, and total energy in the first orbit are 13.60 eV, 0, and 13.60 respectively and for the second orbit, these values are 3.40 eV, 20.40 eV, and 23.80 eV.

Example 20. A small particle of mass m moves in such a way that the potential energy U = ar2 where a is a constant and r is the distance of the particle from the origin. Assuming Bohr’s model of quantization of angular momentum and circular orbits, find the radius of n nth allowed orbit.
Solution: The force at a distance r is, \(\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dr}}=-2 \mathrm{ar}\)

Suppose r is the radius of the nth orbit. The necessary centripetal force is provided by the above force. Thus, \(\frac{m v^2}{r}=2 a r\)

Further, the quantization of angular momentum gives, \(m v r=\frac{n h}{2 \pi}\) Solving Eqs. 1 and 2 for r, we get \(r=\left(\frac{n^2 h^2}{8 a m \pi^2}\right)^{1 / 4}\)

Example 21. An electron is orbiting in a circular orbit of radius r under the influence of a constant magnetic field of strength B. Assuming that Bohr’s postulate regarding the quantization of angular momentum holds good for this electron, find

• The Allowed Values Of The Radius ‘R’ Of The Orbit.
• The Kinetic Energy Of The Electron In Orbit
• The potential energy of interaction between the magnetic moment of the orbital current
  due to the electron moving in its orbit and the magnetic field B.
• The total energy of the allowed energy levels.

Solution: Radius of circular path

⇒ \(\begin{aligned}
&r=\frac{m v}{B e}\\
&\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}
\end{aligned}\)

Solving these two equations, we get

⇒ \(r=\sqrt{\frac{n h}{2 \pi B e}} \quad \text { and } v=\sqrt{\frac{n h B e}{2 \pi m^2}}\)

⇒ \(K=\frac{1}{2} m v^2=\frac{n h B e}{4 \pi m}\)

⇒ \(M=i A=\left(\frac{e}{T}\right)\left(\pi r^2\right)=\frac{e v r}{2}=\frac{e}{2} \sqrt{\frac{n h}{2 \pi B e}} \sqrt{\frac{n h B e}{2 \pi m^2}}=\frac{\text { nhe }}{4 \pi m}\)

Now Potential energy U=-M.B

⇒ \(=\frac{\text { nheB }}{4 \pi m}\)

⇒ \(E=U+K=\frac{n h e B}{2 \pi m}\)

Calculation of recoil speed of atom on the emission of a photon momentum of photon \(=\mathrm{mc}=\frac{\mathrm{h}}{\lambda}\)

⇒ \(\mathrm{mv}=\frac{\mathrm{h}}{\lambda^{\prime}}\)

According to energy conservation

⇒ \(\frac{1}{2} m v^2+\frac{h c}{\lambda^{\prime}}=10.2 \mathrm{eV}\)

Since the mass of the atom is much larger than the photon Hence \(\frac{1}{2} m v^2\)

⇒ \(\begin{aligned}
& \frac{\mathrm{hc}}{\lambda^{\prime}}=10.2 \mathrm{eV} \Rightarrow \frac{\mathrm{h}}{\lambda^{\prime}}=\frac{10.2}{\mathrm{c}} \mathrm{eV} \\
& \mathrm{m} v=\frac{10.2}{\mathrm{c}} \mathrm{eV} \Rightarrow \quad v=\frac{10.2}{\mathrm{~cm}} \\
&
\end{aligned}\)

Recoil speed of atom \(=\frac{10.2}{\mathrm{~cm}}\)

X-RAYS

• X-rays were discovered by Wilhelm Roentgen in 1895. They are also called Roentgen rays.
• X-rays are produced by bombarding high-speed electrons on a target of high atomic weight and high melting point.
• The wavelength of X-rays lies between -rays and UV rays.
• The wavelength range for X-rays is 0.1 Å to 10Å.
• The frequency range for X-rays is 1016 Hz to 1018 Hz.
• The energy range for X-rays is 100 to 10000 eV.
• Hard X-rays: High-frequency X-rays are called hard X-rays.
• Hard X-rays have high penetration power
• The wavelength range is from 0.1A to 10A.
• They have a high frequency of 1018 Hz and a high energy of ~104 eV.
• Soft X-rays: Low-frequency X-rays are called soft X-rays.
• These have low penetrating power
• They have higher wavelengths (10Å to 100Å)
• They have a low frequency of 1016 Hz and a low energy of 102 eV.
• It was discovered by ROENTGEN. The wavelength of X-rays is found between 0.1 Å to 10 Å. These rays are invisible to the eye. They are electromagnetic waves and have speed c = 3 × 10 8 m/s in a vacuum. Its photons have energy around 1000 times more than the visible light.

When fast-moving electrons having the energy of the order of several KeV strike the metallic target then x-rays are produced.

Production of x-rays by coolidge tube:

The melting point, specific heat capacity, and atomic number of the target should be high. When voltage is applied across the filament then the filament on being heated emits electrons from it. Now for giving the beam shape of electrons, the collimator is used. Now when an electron strikes the target then x-rays are produced.

When electrons strike the target, some part of the energy is lost and converted into heat. Since the target should not melt or absorb heat at the melting point, the specific heat of the target should be high.

Here copper rod is attached so that the heat produced can go behind and can absorb heat and the target does not get heated very high. For more energetic electrons, accelerating voltage is increased. For more no. of photons, the voltage across the filament is increased. The X-rays were analyzed by mostly taking their spectrum

Continuous x-ray: When high energy electrons (accelerated by coolidge tube potential) strike the target element they are defeated by coulomb attraction of the nucleus & due to numerous glancing collisions with the atoms of the target, they lose energy which appears in the form of electromagnetic waves (bremsstrahlung or braking radiation) & the remaining part increases the kinetic energy of the colliding particles of the target. The energy received by the colliding particles goes into heating the target. The electron makes another collision with its remaining energy.

• When highly energetic electrons enter into target material, they are decelerated. In this process emission of energy takes place. The spectrum of this energy is continuous. This is also called bremsstrahlung.
• Continuous spectrum (v or lamba) depends upon the potential difference between filament and target.
• It does not depend upon the nature of the target material.
• If V is the potential difference & v is the frequency of emitted x-ray photon then.

Variation of frequency (v) and wavelength (λ) of x-rays with a potential difference is plotted as shown in the figure:

⇒ \(\mathrm{eV}=\frac{1}{2} \mathrm{~m} v^2=h v_{\max }=\frac{h c}{\lambda_{\text {min }}}\)

Variation of Intensity of x-rays with λ is plotted as shown in the figure:

The minimum wavelength corresponds to the maximum energy of the x-rays which in turn is
equal to the maximum kinetic energy eV of the striking electrons thus

⇒ \(\mathrm{eV}=\frac{1}{2} \mathrm{mv} v^2=\mathrm{h} v_{\max }=\frac{\mathrm{hc}}{\lambda_{\min }} \Rightarrow \lambda_{\min }=\frac{\mathrm{hc}}{\mathrm{eV}}=\frac{12400}{\mathrm{~V} \text { (involts) }} \mathrm{A} .\)

We see that cutoff wavelength λmin depends only on the accelerating voltage applied between the target and filament. It does not depend upon the material of the target, it is the same for two different metals (Z and Z’)

Solved Examples

Example 22. An X-ray tube operates at 20 kV. A particular electron loses 5% of its kinetic energy to emit an X-ray photon at the first collision. Find the wavelength corresponding to this photon.
Solution: Kinetic energy acquired by the electron is K = eV = 20 × 103 eV.

The energy of the photon = 0.05 × 20 = 103 eV = 103 eV.

⇒ \(\text { Thus, } \quad \frac{\mathrm{h} \nu}{\lambda}=10^3 \mathrm{eV}=\frac{\left(4.14 \times 10^{-15} \mathrm{eV}-\mathrm{s}\right) \times\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{10^3 \mathrm{eV}}=\frac{1242 \mathrm{eV}-\mathrm{nm}}{10^3 \mathrm{eV}}=1.24 \mathrm{~nm}\)

Characteristic X-rays

The sharp peaks obtained in graph are known as characteristic X-rays because they are characteristic of the target material. λ1, λ2, λ3, λ4, …….. = characteristic wavelength of material having atomic number Z is called characteristic x-rays, and the spectrum obtained is called a characteristic spectrum. If the target of atomic number Z’ is used then peaks are shifted.

Characteristic x-ray emission occurs when an energetic electron collides with a target and removes an inner shell electron from an atom, the vacancy created in the shell is filled when an electron from a higher level drops into it.

Suppose the vacancy created in the innermost K-shell is filled by an electron dropping from the next higher level L-shell then Kα characteristic x-ray is obtained. If a vacancy in the K-shell is filled by an electron from the M-shell, the Kβ line is produced, and so on.

similarly, Lα, Lβ,…..Mα, Mβ lines are produced.

Solved Examples

Example 23. Find which is the Kx and Kb

Solution: \(\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}, \quad \lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}\)

since the energy difference of Kx is less than Kb

⇒ \(\begin{aligned}
& \Delta \mathrm{E}_{\mathrm{k} \alpha}<\Delta \mathrm{E}_{\mathrm{k} \beta} \\
& \lambda_{\mathrm{k} \beta}<\lambda_{k \alpha}
\end{aligned}\)

Example 24. Find which is Kα and Lα
Solution:

1 is Kα and 2 is Lα

Properties Of X-Rays

• X-rays are electromagnetic waves of short wavelengths that travel in straight lines with the speed of light.
• They are chargeless and are not deflected in electric and magnetic fields.
• They cause fluorescence in many substances like zinc sulfide, cadmium tungstate, and barium platino-cyanide.
• They produce photochemical reactions and affect a photographic plate more severely than light.
• Like light, they show reflection, refraction, interference, diffraction, and polarization.
• They ionize the gases through which they pass.
• When they fall on matter they produce photoelectric effect and Compton effect.
• They are highly penetrating and can pass through many solids. Example: They pass through 1 mm thick.
• Aluminum sheet while being absorbed by a sheet of lead of the same thickness.
• The penetration power depends on the applied potential difference and atomic number of the cathode. It destroys tissues of animal bodies and white blood cells.

Absorption Of X-Rays

The intensity of the X-ray beam is defined as the amount of energy carried per unit area per sec perpendicular to the direction of the flow of energy.

• When a beam of X-rays with incident intensity l 0 passes through material then the intensity of emergent X-rays (I) is I = I 0 e–ex
• where u is absorption coefficient and x is thickness of medium
• The intensity of transmitted X-rays reduces exponentially with the thickness of the material.
• The absorption coefficient of the material is defined as the reciprocal of thickness after which the intensity of X-rays falls to \(\frac{1}{\mathrm{e}}\) times the original intensity.
• At \(\mu=\frac{1}{x} \quad \Rightarrow \quad \mathrm{I}=\mathrm{I}_0 / \mathrm{e}\)
• The absorption coefficient depends on the wavelength of X-rays (lamba), the atomic number (Z) of the material, and the density (p) of the material.
• Absorption coefficient \(\mu=C Z^4 \lambda^3 \rho\)
  1. \(\mu \propto \lambda^3\)
  2. \(\mu \propto \frac{1}{v^3}\)
  3. \(\mu \propto Z^4\)
  4. \(\mu \propto \rho\)
• The best absorber of X-rays is lead while the lowest absorption takes place in air.
• Half Thickness (X1/2): The thickness of a given sheet which reduces the intensity of incident X-rays to half of its initial value is called half thickness \(x=x_{1 / 2} \text { so } \quad x_{1 / 2}=\frac{0.693}{\mu}\left(\frac{I}{I_0}\right)=\left(\frac{1}{2}\right)^{x / x_{1 / 2}}\)
• For photographing human body parts BaSO4 is used.
• If the number of electrons striking the target is increased, the intensity of X-rays produced also increases.
• The patients are asked to drink BaSO4. the solution before X-ray examination because it is a good absorber of X-rays.

Solved Examples

Example 25. The absorption coefficient of AI for soft X-rays is 1.73 per cm. Find the percentage of transmitted X-rays from a sheet of thickness 0.578 cm.
Solution: \(\begin{aligned}
& I=I_0 e^{-1 \dot{x}} \\
& \text { so } \quad \frac{I}{I_0}=e^{-\mu \alpha}=e^{-1.73 \times 0.578} \quad \text { or } \quad \frac{I}{I_0}=e^{-1}=\frac{1}{e}=\frac{1}{2.718}=37 \% \\
&
\end{aligned}\)

Example 26. When X-rays of wavelength 0.5Å pass through a 10 mm thick AI sheet then their intensity is reduced to one-sixth. Find the absorption coefficient for Aluminium.
Solution: \(\mu=\frac{2.303}{\mathrm{x}} \log \left(\frac{\mathrm{I}_0}{\mathrm{I}}\right)=\frac{2.303}{10} \log _{10} 6=\frac{2.303 \times 0.7781}{10}=0.198 / \mathrm{mm}\)

Diffraction Of R- Rays

• The diffraction of X-rays by a crystal was discovered in 1912 by Von Laue.
• The diffraction of X-rays is possible because interatomic spacings in a crystal is of the order of wavelength of X-rays.
• The diffraction of X-rays takes place according to Bragg’s law 2d sin= n.
• This helps to determine the crystal structure and wavelength of X-rays.

X-Rays Dose:

• The dose of X-ray is measured in terms of produced ions of free energy via ionization.
• These are measured in Roentgen
• Roentgen does not measure energy but it measures ionization power.
• The safe dose for the human body per week is one Roentgen.
• One Roentgen is the number of x-rays that emit (2.5 × 104 J) free energy through the ionization of 1 gram of air at NTP.

Uses Of X-Rays

• Surgery X-rays pass through flesh but are stopped by bones. So they are used to detect fractures, foreign bodies, and diseased organs. The photograph obtained is called a radiograph.
• Radiotherapy The X-rays can kill the diseased tissues of the body. They are used to cure skin
  disease, malignant tumors, etc.
• Industry To detect defects in motor tires, golf and tennis balls, wood, and wireless valves. Used to test the uniformity of insulating material and for detecting the presence of pearls in oysters.
• Scientific research is Used to study the structure of crystals, the structure and properties of atoms, arrangement of atoms and molecules in matter.
• Detective departments were Used at customs ports to detect goods like explosives, opium concealed in parcels without opening, and detection of precious metals like gold and silver in the bodies of smugglers.

Moseley’s Law:

Moseley measured the frequencies of characteristic x-rays for a large number of elements and plotted the square root of frequency against position number in the periodic table. He discovered that the plot is very close to a straight line not passing through the origin.

Moseley’s observations can be mathematically expressed as \(\sqrt{v}=a(Z-b)\) a and b are positive constants for one type of x-rays and all elements (independent of Z). Moseley’s Law can be derived based on Bohr’s theory of the atom, frequency of x-rays is given by

⇒ \(\sqrt{v}=\sqrt{C R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)} \cdot(Z-b)\) by using the formula \(\frac{1}{\lambda}=R z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) with modification for multi-electron system. b → known as screening constant or shielding effect, and (Z – b) is the effective nuclear charge. for Kx line n1=1,n2=2 \(\sqrt{v}=\sqrt{\frac{3 R C}{4}}(Z-b) \quad \Rightarrow \quad \sqrt{v}=a(Z-b)\)

Here \(a=\sqrt{\frac{3 R C}{4}} \text {, }\) b=1 for lines

Solved Examples

Example 27. Find in Z1 and Z2 which one is greater.

Solution: since \(\sqrt{v} \equiv \sqrt{c R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)} \cdot(Z-b)\)

If Z is greater then v will be greater, and lamba will be less.

therefore \(\lambda_1<\lambda_2\)

Dividing yields \(\sqrt{\frac{\lambda_{c_0}}{\lambda_x}}=\frac{Z_x-1}{Z_{c_0}-1}\)

⇒ \(\sqrt{\frac{178.9 \mathrm{pm}}{143.5 \mathrm{pm}}}=\frac{Z_{\mathrm{x}}-1}{27-1} .\)

Solving for the unknown, we find Zx = 30.0; the impurity is zinc

Example 29. Find the constants a and b in Moseley’s equation from the following data.

Element Z Wavelength of Kα X-ray
Mo 42 71 pm
Co 27 178.5 pm

Solution: Moseley’s equation is

⇒ \(\sqrt{v}=a(Z-b)\)

Thus, \(\sqrt{\frac{c}{\lambda_1}}=a\left(Z_1-b\right)\)

and \(\sqrt{\frac{c}{\lambda_2}}=a\left(Z_2-b\right)\)

From (1) and (2) ,\(\sqrt{c}\left(\frac{1}{\sqrt{\lambda_1}}-\frac{1}{\sqrt{\lambda_2}}\right)=a \quad\left(Z_1-Z_2\right)\)

Or, \(a=\frac{\sqrt{c}}{\left(Z_1-Z_2\right)}\left(\frac{1}{\sqrt{\lambda_1}}-\frac{1}{\sqrt{\lambda_2}}\right)\)

⇒ \(=\frac{\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)^{1 / 2}}{42-27}\left[\frac{1}{\left(71 \times 10^{-12} \mathrm{~m}\right)^{1 / 2}}-\frac{1}{\left(178.5 \times 10^{-12} \mathrm{~m}\right)^{1 / 2}}\right]=5.0 \times 10^7(\mathrm{~Hz})^{1 / 2}\)

Dividing 1 by 2

⇒ \(\sqrt{\frac{\lambda_2}{\lambda_1}}=\frac{Z_1-b}{Z_2-b} \text { or, } \quad \sqrt{\frac{178.5}{71}}=\frac{42-b}{27-b} \quad \text { or, } \quad b=1.37\)

Solved Miscellaneous Problems

Problem 1. Find the momentum of a 12.0 MeV photon.
Solution: \(p=\frac{E}{c}=12 \mathrm{MeV} / \mathrm{c}\)

Problem 2. Monochromatic light of wavelength 3000 Å is incident normally on a surface of area 4 cm 2. If the intensity of the light is 15 × 10–2 W/m2, determine the rate at which photons strike the surface.
Solution: Rate at which photons strike the surface

⇒ \(=\frac{\mathrm{IA}}{\mathrm{hc} / \lambda}=\frac{6 \times 10^{-5} \mathrm{~J} / \mathrm{s}}{6.63 \times 10^{-19} \mathrm{~J} / \text { photon }}=9.05 \times 10^{13} \text { photon } / \mathrm{s} \text {. }\)

Problem 3. The kinetic energies of photoelectrons range from zero to 4.0 × 10 –19 J when light of wavelength 3000 Å falls on a surface. What is the stopping potential for this light?
Solution: \(\mathrm{K}_{\max }=4.0 \times 10^{-19} \mathrm{~J} \times \frac{1 \mathrm{eV}}{1.6 \times 10^{-19} \mathrm{~J}}=2.5 \mathrm{eV}\)

Then, from eV s = K max , V s = 2.5 V

Problem 4. Find the de Broglie wavelength of a 0.01 kg pellet having a velocity of 10 m/s.
Solution: \(\lambda=\mathrm{h} / \mathrm{p}=\frac{6.63 \times 10^{-34} \mathrm{~J} s}{0.01 \mathrm{~kg} \times 10 \mathrm{~m} / \mathrm{s}}=6.63 \times 10^{-23} \mathrm{~A} \text {. }\)

Problem 5. Determine the accelerating potential necessary to give an electron a de Broglie wavelength of 1 Å, which is the size of the interatomic spacing of atoms in a crystal.
Solution: \(\mathrm{V}=\frac{\mathrm{h}^2}{2 \mathrm{~m}_0 \mathrm{e} \lambda^2}=151 \mathrm{~V}\)

Problem 6. Determine the wavelength of the second line of the Paschen series for hydrogen.
Solution: \(\frac{1}{\lambda}=\left(1.097 \times 10^{-3} A^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{5^2}\right) \quad \text { or } \quad \lambda=12,820 A \text {. }\)

Problem 7. How many different photons can be emitted by hydrogen atoms that undergo transitions to the ground state from the n = 5 state?
Solution: No possible transition from n = 5 are = 70. 10 Photons

Problem 8. An electron rotates in a circle around a nucleus with a positive charge Ze. How is the electrons’ velocity related to the radius of its orbit?
Solution: The force on the electron due to the nuclear provides the required centripetal force

⇒ \(\frac{1}{4 \pi \varepsilon_0} \frac{Z e . ~}{r^2}=\frac{\mathrm{mv}^2}{\mathrm{r}} \Rightarrow v=\sqrt{\frac{\mathrm{Ze}^2}{4 \pi \varepsilon_0 \cdot \mathrm{rm}}}\)

⇒ \(v=\sqrt{\frac{Z e^2}{4 \pi \varepsilon_0 \cdot \mathrm{rm}}} .\)

Problem 9. A H-atom in the ground state is moving with initial kinetic energy K. It collides head-on with a He + ion in the ground state kept at rest but free to move. Find the minimum value of K so that both the particles can excite to their first excited state.
Solution: Energy available for excitation \(=\frac{4 K}{5}\)

Total energy required for excitation = 10.2 ev + 40.8 eV = 51.0 ev

⇒ \(\frac{4 K}{5}=51 \quad \Rightarrow \quad k=63.75 \mathrm{eV}\)

Problem 10. A TV tube operates with a 20 kV accelerating potential. What are the maximum–energy X–rays from the TV set?
Solution: The electrons in the TV tube have an energy of 20 keV, and if these electrons are brought to rest by a collision in which one X-ray photon is emitted, the photon energy is 20 keV.v a(Z b)=

Problem 11. In the Moseley relation, which will have the greater value for the constant a for Kx or Kb transition?
Solution: A is larger for the K transitions than for the K transitions.

Problem 12. A He+ ion is at rest and is in the ground state. A neutron with initial kinetic energy K collides head-on with the He+ ion. Find the minimum value of K so that there can be an inelastic collision between these two particles.
Solution: Here the loss during the collision can only be used to excite the atoms or electrons. So according to quantum mechanics loss = {0, 40.8eV, 48.3eV, ……, 54.4eV}

⇒ \(E_n=-13.6 \frac{Z^2}{n^2} e V\)

Now according to Newtonion mechanics

Minimum loss = 0

the maximum loss will be for perfectly inelastic collision.

let v

0 is the initial speed of the neutron and vf is the final common speed.

so by momentum conservation mv0 = mvf + 4mv \(v_f=\frac{v_0}{5}\)

where m = mass of Neutron

mass of He+ ion = 4m

so the final kinetic energy of the system

⇒ \(\text { K.E. }=\frac{1}{2} m v_f^2+\frac{1}{2} 4 m v_f^2=\frac{1}{2} \cdot(5 m) \cdot \frac{v_0^2}{25}=\frac{1}{5} \cdot\left(\frac{1}{2} m v_0^2\right)=\frac{K}{5}\) so loss will be \(\left[0, \frac{4 K}{5}\right]\)

For inelastic collision, there should be at least one common value other than zero in set (1) and

since \(\frac{4 \mathrm{~K}}{5}>40.8 \mathrm{eV}\)

K > 51 eV minimum value of K = 51 eV.

Problem 13. A moving hydrogen atom makes a head-on collision with a stationary hydrogen atom. Before the collision, both atoms are in the ground state, and after the collision, they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation states?
Solution: Let K be the kinetic energy of the moving hydrogen atom and K’, the kinetic energy of combined mass after collision. From conservation of linear momentum,

⇒ \(p=p^{\prime} \text { or } \sqrt{2 \mathrm{Km}}=\sqrt{2 \mathrm{~K}^{\prime}(2 m)}\)

or K = 2K’ ……….1

From conservation of energy, K = K’ + E…………..2

Solving Eqs. (1) and (2), we get,\(\Delta \mathrm{E}=\frac{\mathrm{K}}{2}\)

Now the minimum value of E for a hydrogen atom is 10.2 eV. or E  10.2 eV’

⇒ \(\begin{aligned}
& \frac{K}{2} \geq 10.2 \\
& \mathrm{~K} \geq 20.4 \mathrm{eV}
\end{aligned}\)

Therefore, the minimum kinetic energy of moving hydrogen is 20.4 eV