A black body at 1227^{circ}C emits radiations | vedclass

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(B) According to $Wien's$ displacement law,the product of the wavelength of maximum intensity $(\lambda_{\max})$ and the absolute temperature $(T)$ is

Vedclass · Accepted Answer

$3000$

Vedclass · Answer

(B) According to $Wien's$ displacement law,the product of the wavelength of maximum intensity $(\lambda_{\max})$ and the absolute temperature $(T)$ is constant:$\lambda_{\max} T = b$ (constant)Given:Initial temperature $T_1 = 1227^{\circ}C = 1227 + 273 = 1500\;K$Initial wavelength $\lambda_{\max 1} = 5000\;\mathring{A}$Final temperature $T_2 = 1227^{\circ}C + 1000^{\circ}C = 2227^{\circ}C = 2227 + 273 = 2500\;K$Using the relation $\lambda_{\max 1} T_1 = \lambda_{\max 2} T_2$:$\lambda_{\max 2} = \frac{\lambda_{\max 1} T_1}{T_2}$$\lambda_{\max 2} = \frac{5000 	imes 1500}{2500}$$\lambda_{\max 2} = 5000 	imes 0.6 = 3000\;\mathring{A}$Therefore,the maximum intensity will be observed at $3000\;\mathring{A}$.

$A$ black body at $1227^{\circ}C$ emits radiations with maximum intensity at a wavelength of $5000\;\mathring{A}$. If the temperature of the body is increased by $1000^{\circ}C$,the maximum intensity will be observed at ...... $\mathring{A}$.

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