JEE MainPhysicsToughHeat

Power radiated by a black body is P_0 and the wavelength
corresponding to the maximum energyis around $\lambda_{0}$ . On changing
the temperature of the black body, it is observed that the power
radiatedis increased to $\frac{256}{81}P_{0}$ . The shift in the
wavelength
corresponding to the maximum energy will be

Show answer and solution

P = \sigma{AT}^{4}

\lambda_{m}T = b

T = \frac{b}{\lambda_{m}} \Rightarrow P = \frac{\sigma{Ab}^{4}}{\lambda_{m}^{4}}

\Rightarrow P \propto \frac{1}{\lambda_{m}^{4}} \Rightarrow \frac{P_{i}}{P_{2}} = {(\frac{\lambda_{m_{2}}}{\lambda_{m_{4}}})}^{4}

\frac{P_{0}}{\frac{256}{81}P_{0}} = {(\frac{\lambda_{m_{2}}}{\lambda_{0}})}^{4} \Rightarrow \lambda_{m_{2}} = \frac{3}{4}\lambda_{0}

\Delta\lambda = \frac{3}{4}\lambda_{0} - \lambda_{0} = - \frac{\lambda_{0}}{4}

Source: Grass-G / P-08 Physics paper 23 October.docx

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