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(D) According to Wien's Displacement Law,the product of the wavelength of maximum emitted energy $(\lambda_m)$ and the absolute temperature $(T)$ of a

Vedclass · Accepted Answer

$2.04$

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(D) According to Wien's Displacement Law,the product of the wavelength of maximum emitted energy $(\lambda_m)$ and the absolute temperature $(T)$ of a black body is constant.$\lambda_m T = b$ (constant)Given:$T_1 = 700 \ K$,$\lambda_{m1} = 4.08 \ \mu m$$T_2 = 1400 \ K$,$\lambda_{m2} = ?$Using the relation $\lambda_{m1} T_1 = \lambda_{m2} T_2$:$\lambda_{m2} = \frac{\lambda_{m1} T_1}{T_2}$$\lambda_{m2} = \frac{4.08 	imes 700}{1400}$$\lambda_{m2} = 4.08 	imes 0.5 = 2.04 \ \mu m$.

The wavelength of maximum emitted energy of a body at $700 \ K$ is $4.08 \ \mu m$. If the temperature of the body is raised to $1400 \ K$,the wavelength of maximum emitted energy will be ........ $\mu m$.

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