Question

The power radiated by a black body is $P$ and it radiates maximum energy at wavelength, $\lambda_0$. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $\frac{3}{4} \lambda_0$ , the power radiated by it becomes $nP$. The value of $n$ is

• A. $\frac{81}{256}$
• B. $\frac{3}{4}$
• C. $\frac{256}{81}$
• D. $\frac{4}{3}$

Answer 1

The Correct Option is C

1. The problem involves a black body, which is a perfect emitter and absorber of radiation, following the laws of black body radiation.
2. The maximum wavelength of radiation emitted by a black body at temperature $T$ obeys Wien's Displacement Law:
   \(\lambda_{\text{max}} \cdot T = b \)
   where $b$ is a constant.
3. According to Wien's Law:
   • Initially, $\lambda_0 T_1 = b$.
   • After changing the temperature, it becomes $\frac{3}{4} \lambda_0 T_2 = b$.
4. From these equations, we can equate them:
   \[ \lambda_0 T_1 = \frac{3}{4} \lambda_0 T_2 \] \] \cancel{\lambda_0} \Rightarrow T_1 = \frac{3}{4} T_2 \]
5. The power radiated, $P$, by a black body is given by Stefan-Boltzmann Law:
   \[ P \propto T^4 \]
   Therefore, we have:
   • Initial power: $P \propto T_1^4$
   • New power: $nP \propto T_2^4$
6. Substitute the relation $T_1 = \frac{3}{4} T_2$ into the power equation:
   \[ \frac{nP}{P} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{T_2}{\frac{3}{4}T_2}\right)^4 = \left(\frac{4}{3}\right)^4 \] \] \Rightarrow n = \left(\frac{4}{3}\right)^4 = \frac{256}{81} \]
7. Therefore, the value of $n$, or the factor by which the power has increased, is $\frac{256}{81}$.

Hence, the correct answer is $\frac{256}{81}$.