NEET Physics Class 11 Chapter 8 Stefan-Boltzmann’S Law

NEET Physics Class 11 Chapter 8 Stefan-Boltzmann’S Law

According to this law, the amount of radiation emitted per unit of time from area A of a black body at absolute temperature T is directly proportional to the fourth power of the temperature.

u = σ A T4

where σ is Stefan’s constant = 5.67 x 10-8 W/m2 k4

A body that is not black absorbs and hence emits less radiation than that given by the equation.

For such a body, u e A T4

where e = emissivity (which is equal to absorptive power) which lies between 0 to 1

With the surroundings of temperature T0, net energy radiated by an area A per unit of time.

⇒ \(\Delta u=u-u_0=e \quad \sigma A\left(T^4-T_0^4\right)\)

Question 2. A black body at 2000K emits radiation with λm= 1250 nm. If the radiation coming from the star SIRUS λmis 71 nm, then the temperature of this star is …….
Answer:

Using Wien’s displacement law

⇒ \(\frac{T_2}{T_1}=\frac{\left(\lambda_m\right)_1}{\left(\lambda_m\right)_2}\)

∴ \(\mathrm{T}_2=\frac{2000 \times 1250 \times 10^{-3}}{71 \times 10^{-9}}\)

T2 = 35.211 K

Question 3. At 1600 K maximum radiation is emitted at a wavelength of 2µm. Then the corresponding wavelength at 2000 K will be –
Answer:

Using T1= T2

∴ \(\frac{\lambda_{m_2} T_1}{T_2}\)

∴ \(\frac{2 \times 10^{-6} \times 1600}{2000}\)

= 1.6 µm

Question 4. If the temperature of a body is increased by 50%, then the increase in the amount of radiation emitted by it will be
Answer: Percentage increase in the amount of radiation emitted

∴ \(\frac{E_2-E_1}{E_1} \times 100\)

⇒ \(\frac{\left(1.5 T_1\right)^4-T_1^4}{T_1^4} \times 100\)

⇒ \(\frac{E_2-E_1}{E_1} \times 100\)

⇒ \(\left[(1.5)^4-1\right] \times 10\)

⇒ \(\frac{E_2-E_1}{E_1} \times 100\)

= 400%

Question 5. A blackened platinum wire of length 5cm and perimeter 0.02 cm is maintained at a temperature of 300K. Then at what rate the wire is losing its energy? (Take σ = 57 × 10-8units)
Solution :

The rate of radiation heat loss is

⇒ \(\frac{d Q}{d t}=\mathrm{eA} \sigma \mathrm{T}^4 \text { (watts) }\)

for blackened surface e = 1

and A = (2πr)l = Perimeter × length

∴ A = 0.02 × 5 × 10-4

Thus

⇒ \(\frac{d Q}{d t}\)= 0.02 × 5 × 10-4× 5.7 × 10-8× (300)4

⇒ \(\frac{d Q}{d t}\)= 46.2W

Question 6. A hot black body emits energy at the rate of 16 J m-2 s-1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then find the value of energy radiated in Jm-2 s-1.
Answer:

Wein’s displacement law is :

λm.T = b i.e. \(\mathrm{T} \propto \frac{1}{\lambda_m}\)

Here, λm becomes half.

∴ Temperature doubles.

Also E = σT4

⇒ \(\frac{e_1}{e_2}=\left(\frac{T_1}{T_2}\right)^4\)

⇒ \(\mathrm{E}_2=. \mathrm{E}_1 \frac{e_1}{e_2}=\left(\frac{T_1}{T_2}\right)^4 \mathrm{e}_1=(2)^4 .16\)

= 16.16 = 256 J m-2 s-1

 

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