Question

A black body radiates heat at temperatures ${T_1}$ and ${T_2}$ $\left( {{T_2} > {T_1}} \right)$. Find the frequency corresponding to the maximum energy.
A) More at ${T_1}$
B) More at ${T_2}$
C) Equal for ${T_1}$ and ${T_2}$
D) Independent of ${T_1}$ and ${T_2}$

Answer 1

Hint:Stefan-Boltzmann law gives the rate at which energy is radiated by a black body to be directly proportional to the fourth power of the temperature of the black body. The energy radiated can also be expressed as the product of Planck's constant and the frequency of the radiation. Thus we can determine the maximum frequency for the two temperatures.

Formulas used:
-The energy radiated per second by a black body is given by, $E = \sigma A{T^4}$ where $\sigma $ is Stefan’s constant, $A$ is the area of the black body and $T$ is the temperature of the black body.
-The energy of radiation is given by, $E = h\nu $ where $h$ is Planck's constant and $\nu $ is the frequency of the radiated energy.

Complete step by step answer.
Step 1: List the information given and express the energy radiated for the two temperatures based on Stefan-Boltzmann law.
It is given that the same black body is heated at two temperatures ${T_1}$ and ${T_2}$ .
It is also mentioned that ${T_2} > {T_1}$ .
Then according to Stefan-Boltzmann law, the energy radiated when the black body is heated at temperature ${T_1}$ can be expressed as ${E_1} = \sigma A{T_1}^4$ --------- (1)
where $\sigma $ is Stefan’s constant and $A$ is the area of the black body.
Similarly, the energy radiated when the black body is heated at temperature ${T_2}$ can be expressed as ${E_2} = \sigma A{T_2}^4$ --------- (2)
Then dividing equation (2) by (1) we get, $\dfrac{{{E_2}}}{{{E_1}}} = \dfrac{{\sigma A{T_2}^4}}{{\sigma A{T_1}^4}}$
$ \Rightarrow \dfrac{{{E_2}}}{{{E_1}}} = \dfrac{{{T_2}^4}}{{{T_1}^4}}$ -------- (3)
Step 2: Express the radiated energies at the two temperatures in terms of their frequencies.
The radiated energy ${T_1}$ can be expressed as ${E_1} = h{\nu _1}$ ---------- (4) where $h$ is Planck's constant and ${\nu _1}$ is the maximum frequency of the radiated energy ${E_1}$.
The radiated energy ${T_2}$ can be expressed as ${E_2} = h{\nu _2}$ ---------- (5) where $h$ is Planck's constant and ${\nu _2}$ is the maximum frequency of the radiated energy ${E_2}$.
Substituting equations (4) and (5) in equation (3) we get, $\dfrac{{h{\nu _2}}}{{h{\nu _1}}} = \dfrac{{{T_2}^4}}{{{T_1}^4}}$
$ \Rightarrow \dfrac{{{\nu _2}}}{{{\nu _1}}} = \dfrac{{{T_2}^4}}{{{T_1}^4}}$
As it is already mentioned that ${T_2} > {T_1}$ , we can conclude that ${\nu _2} > {\nu _1}$ .
So the maximum frequency is more at ${T_2}$ .

So the correct option is B.

Note:A perfectly black body will absorb all radiations that fall on it irrespective of their frequencies and emit the same radiations. The energy emitted by the black body depends on the nature of its surface, its surface area and its temperature. Here the same black body is heated at different temperatures so the surface area and its nature remain the same.