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The power radiated by a black body is PP and it radiates maximum energy at wavelength λ0\lambda_0. Temperature of the black body is now changed so that it radiates maximum energy at the wavelength 34λ0\frac{3}{4}\lambda_0. The power radiated by it now becomes nPnP. The value of nn is:

A

34\frac{3}{4}

B

43\frac{4}{3}

C

25681\frac{256}{81}

D

81256\frac{81}{256}

Step-by-Step Solution

According to Wien's displacement law, the wavelength corresponding to maximum spectral emissive power is inversely proportional to the absolute temperature of the black body, i.e., λmT=constant\lambda_m T = \text{constant}. Given initial wavelength λ1=λ0\lambda_1 = \lambda_0 at temperature T1T_1, and final wavelength λ2=34λ0\lambda_2 = \frac{3}{4}\lambda_0 at temperature T2T_2. Therefore, λ0T1=34λ0T2    T2=43T1\lambda_0 T_1 = \frac{3}{4}\lambda_0 T_2 \implies T_2 = \frac{4}{3}T_1. According to the Stefan-Boltzmann law, the total power radiated by a black body is directly proportional to the fourth power of its absolute temperature (PT4P \propto T^4). So, the ratio of final power to initial power is P2P1=(T2T1)4=(43)4=25681\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{4}{3}\right)^4 = \frac{256}{81}. Given P1=PP_1 = P and P2=nPP_2 = nP, we have nPP=25681    n=25681\frac{nP}{P} = \frac{256}{81} \implies n = \frac{256}{81}.

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