The intensity of radiation emitted by the sun | vedclass

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(B) According to Wien's displacement law,the product of the wavelength corresponding to maximum intensity $(\lambda_{\max})$ and the absolute temperat

Vedclass · Accepted Answer

$0.69$

Vedclass · Answer

(B) According to Wien's displacement law,the product of the wavelength corresponding to maximum intensity $(\lambda_{\max})$ and the absolute temperature $(T)$ of a black body is constant.$\lambda_{\max} T = b$ (constant)Therefore,$T \propto \frac{1}{\lambda_{\max}}$.Let $T_S$ and $\lambda_S$ be the temperature and wavelength of the sun,and $T_N$ and $\lambda_N$ be the temperature and wavelength of the north star.$\frac{T_S}{T_N} = \frac{(\lambda_N)_{\max}}{(\lambda_S)_{\max}}$Given $(\lambda_S)_{\max} = 510 \ nm$ and $(\lambda_N)_{\max} = 350 \ nm$.$\frac{T_S}{T_N} = \frac{350}{510} = \frac{35}{51} \approx 0.686 \approx 0.69$.

The intensity of radiation emitted by the sun has its maximum value at a wavelength of $510 \ nm$ and that emitted by the north star has the maximum value at $350 \ nm$. If these stars behave like black bodies,then the ratio of the surface temperature of the sun and north star is

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