Q119 / 150 · MHT CET 2026 (11 April Shift 1) · Physics
The energy spectrum of a black body exhibits a maximum of wavelength $\lambda_0$. The temperature of the black body is now changed such that the energy is maximum at wavelength $\dfrac{3\lambda_0}{2}$. The ratio of power radiated by the black body will be
- A$\dfrac{16}{9}$
- Correct answerB$\dfrac{16}{81}$
- C$\dfrac{64}{27}$
- D$\dfrac{4}{3}$
Solution
Using Wien's displacement law, $\lambda_{m} T = \text{constant}$
$\lambda_1 T_1 = \lambda_2 T_2$
$\lambda_0 T_1 = \dfrac{3\lambda_0}{2} T_2$
$\dfrac{T_2}{T_1} = \dfrac{2}{3}$
According to Stefan-Boltzmann law, the power radiated by a black body is $P \propto T^4$
$\dfrac{P_2}{P_1} = \left(\dfrac{T_2}{T_1}\right)^4$
$\dfrac{P_2}{P_1} = \left(\dfrac{2}{3}\right)^4 = \dfrac{16}{81}$
Answer: $\dfrac{16}{81}$
$\lambda_1 T_1 = \lambda_2 T_2$
$\lambda_0 T_1 = \dfrac{3\lambda_0}{2} T_2$
$\dfrac{T_2}{T_1} = \dfrac{2}{3}$
According to Stefan-Boltzmann law, the power radiated by a black body is $P \propto T^4$
$\dfrac{P_2}{P_1} = \left(\dfrac{T_2}{T_1}\right)^4$
$\dfrac{P_2}{P_1} = \left(\dfrac{2}{3}\right)^4 = \dfrac{16}{81}$
Answer: $\dfrac{16}{81}$
Hence, the correct answer is option: 'B'.