A black body is at a temperature of 5780 text | vedclass

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(A) According to Wien's displacement law, the wavelength $\lambda_m$ at which the spectral emissive power of a black body is maximum is given by $\lam

Vedclass · Accepted Answer

$U_1  U_3$

Vedclass · Answer

(A) According to Wien's displacement law, the wavelength $\lambda_m$ at which the spectral emissive power of a black body is maximum is given by $\lambda_m = \frac{b}{T}$.Given $b = 2.89 	imes 10^6 	ext{ nm K}$ and $T = 5780 	ext{ K}$.$\lambda_m = \frac{2.89 	imes 10^6}{5780} \approx 500 	ext{ nm}$.Since the peak emission occurs at $\lambda_m = 500 	ext{ nm}$, the energy density $U_2$ at $500 	ext{ nm}$ is the maximum.For wavelengths smaller than $\lambda_m$ $(300 	ext{ nm})$ and larger than $\lambda_m$ $(900 	ext{ nm})$, the energy density is lower than the peak value.Thus, $U_2$ is the maximum, and comparing $U_1$ and $U_3$, the curve for black body radiation is steeper on the shorter wavelength side.Therefore, $U_1  U_3$.

$A$ black body is at a temperature of $5780 \text{ K}$. The energy of radiation emitted by the body at wavelength $300 \text{ nm}$ is $U_1$, at wavelength $500 \text{ nm}$ is $U_2$, and at wavelength $900 \text{ nm}$ is $U_3$. Wien's constant $b = 2.89 \times 10^6 \text{ nm K}$. This shows that:

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