Q103 / 150 · MHT CET 2026 (15 April Shift 2) · Physics
The energy spectrum of a black body exhibits a maximum around a wavelength $\lambda$. The temperature of a black body is now changed such that the energy is maximum at wavelength $2\lambda/3$. The power radiated by the black body will now increase by a factor of
- A$\dfrac{256}{81}$
- Correct answerB$\dfrac{81}{16}$
- C$\dfrac{162}{41}$
- D$\dfrac{16}{81}$
Solution
Using Wien's displacement law, $\lambda_{m} T = b$
$\lambda_{1} T_{1} = \lambda_{2} T_{2}$
$\lambda T_{1} = \left(\dfrac{2\lambda}{3}\right) T_{2}$
$T_{2} = \dfrac{3}{2} T_{1}$
According to Stefan-Boltzmann law, the power radiated by a black body is $P = \sigma A T^{4}$
$P \propto T^{4}$
$\dfrac{P_{2}}{P_{1}} = \left(\dfrac{T_{2}}{T_{1}}\right)^{4}$
$\dfrac{P_{2}}{P_{1}} = \left(\dfrac{3}{2}\right)^{4} = \dfrac{81}{16}$
Answer: $\dfrac{81}{16}$
$\lambda_{1} T_{1} = \lambda_{2} T_{2}$
$\lambda T_{1} = \left(\dfrac{2\lambda}{3}\right) T_{2}$
$T_{2} = \dfrac{3}{2} T_{1}$
According to Stefan-Boltzmann law, the power radiated by a black body is $P = \sigma A T^{4}$
$P \propto T^{4}$
$\dfrac{P_{2}}{P_{1}} = \left(\dfrac{T_{2}}{T_{1}}\right)^{4}$
$\dfrac{P_{2}}{P_{1}} = \left(\dfrac{3}{2}\right)^{4} = \dfrac{81}{16}$
Answer: $\dfrac{81}{16}$
Hence, the correct answer is option: 'B'.