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(D) According to Wien's displacement law, the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ of a blac

Vedclass · Accepted Answer

$4:3$

Vedclass · Answer

(D) According to Wien's displacement law, the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ of a black body is constant.$\lambda_m T = b$ (where $b$ is Wien's constant).Therefore, $T \propto \frac{1}{\lambda_m}$.Given:$\lambda_A = 3.9 	imes 10^{-7} 	ext{ m}$$\lambda_B = 5.2 	imes 10^{-7} 	ext{ m}$The ratio of temperatures is $\frac{T_A}{T_B} = \frac{\lambda_B}{\lambda_A}$.Substituting the values:$\frac{T_A}{T_B} = \frac{5.2 	imes 10^{-7}}{3.9 	imes 10^{-7}} = \frac{5.2}{3.9} = \frac{52}{39} = \frac{4}{3}$.Thus, the ratio is $4:3$.

Two stars $A$ and $B$ radiate maximum energy at wavelengths $3.9 \times 10^{-7} \text{ m}$ and $5.2 \times 10^{-7} \text{ m}$ respectively. The ratio of the temperature of star $A$ to that of star $B$ will be

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