A black body radiates maximum energy at wavelength $^{\prime} \lambda^{\prime}$ and its emissive power is ${…

A black body radiates maximum energy at wavelength $^{\prime} \lambda^{\prime}$ and its emissive power is ${ }^{\prime} \mathrm{E}^{\prime}$. Now, due to change in temperature of that body, it radiates maximum energy at wavelength $\frac{2 \lambda}{3}$. At that temperature, emissive power is
  1. $\frac{27 \mathrm{E}}{16}$
  2. $\frac{81 \mathrm{E}}{16}$
  3. $\frac{91 \mathrm{E}}{16}$
  4. $\frac{54 \mathrm{E}}{16}$

Solution

$\lambda_{\max } \mathrm{T}_{1}=\lambda_{2 \max } \mathrm{T}_{2} \quad \quad \mathrm{E}=\sigma \mathrm{A} \mathrm{T}_{1}^{4}$ $\mathrm{T}_{2}=\frac{\lambda_{1} \mathrm{~T}_{1}}{\lambda_{2}}$ $\frac{E_{2}}{E_{1}}=\frac{T_{2}^{4}}{T_{1}^{4}}$ $\therefore \mathrm{E}_{2}=\mathrm{E}_{1} \times \frac{\lambda_{1}^{4} \mathrm{~T}_{1}^{4}}{\lambda_{2}^{4}} \times \frac{1}{\mathrm{~T}_{1}^{4}}=\mathrm{E} \times \frac{\lambda_{1}^{4}}{\frac{16}{81} \lambda_{1}^{4}}=\frac{81}{16} \mathrm{E}$ .

Asked in: MHT CET 2020 (20 Oct Shift 2)

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