A black body has a maximum wavelength {lambda | vedclass

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(B) According to Wien's Displacement Law, the product of the wavelength corresponding to maximum spectral emissive power and the absolute temperature

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$\frac{2}{3}{\lambda _m}$

Vedclass · Answer

(B) According to Wien's Displacement Law, the product of the wavelength corresponding to maximum spectral emissive power and the absolute temperature is constant.${\lambda _m} T = 	ext{constant}$Therefore, ${\lambda _{m1}} T_1 = {\lambda _{m2}} T_2$Given:$T_1 = 2000 \; K$$T_2 = 3000 \; K$${\lambda _{m1}} = {\lambda _m}$Substituting the values:${\lambda _m} 	imes 2000 = {\lambda _{m2}} 	imes 3000$${\lambda _{m2}} = \frac{2000}{3000} 	imes {\lambda _m}$${\lambda _{m2}} = \frac{2}{3}{\lambda _m}$

$A$ black body has a maximum wavelength ${\lambda _m}$ at a temperature of $2000 \; K$. Its corresponding wavelength at a temperature of $3000 \; K$ will be:

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