(B) According to Wien's displacement law,the product of the wavelength of maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ is a cons

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(B) According to Wien's displacement law,the product of the wavelength of maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ is a constant:$\lambda_m T = b$ (constant)Given:Initial temperature $T_1 = 6000\,K$Initial peak wavelength $\lambda_{m1} = 4800\,\mathring{A}$Final temperature $T_2 = 3000\,K$Using the relation $\lambda_{m1} T_1 = \lambda_{m2} T_2$:$4800 \times 6000 = \lambda_{m2} \times 3000$Solving for $\lambda_{m2}$:$\lambda_{m2} = \frac{4800 \times 6000}{3000}$$\lambda_{m2} = 4800 \times 2$$\lambda_{m2} = 9600\,\mathring{A}$

Class 11 Physics 10-2.Heat Transfer Spectral Emissive Power and Wien's Displacement Law

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Solar radiation emitted by the sun resembles that emitted by a black body at a temperature of $6000\,K$. Maximum intensity is emitted at a wavelength of about $4800\,\mathring{A}$. If the sun was cooled down from $6000\,K$ to $3000\,K$,then the peak intensity would occur at a wavelength of ........ $\mathring{A}$.

A
$4800$
B
$9600$
C
$2400$
D
$19200$

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10-2.Heat Transfer

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Spectral Emissive Power and Wien's Displacement Law

The wavelength of maximum emission of radiation from a body at $2000 \ K$ is $4 \ \mu m$. What is the wavelength of maximum emission of radiation from the body at $2400 \ K$ in $\mu m$?

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If wavelengths of maximum intensity of radiations emitted by the sun and the moon are $0.5 \times 10^{-6} \ m$ and $10^{-4} \ m$ respectively,the ratio of their temperatures is

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The graph of relative intensity versus wavelength for three perfectly black bodies at temperatures $T_1, T_2,$ and $T_3$ is shown below. The relationship between their temperatures is:

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$A$ black body at a temperature of $1640 \ K$ has the wavelength corresponding to maximum emission equal to $1.75 \ \mu m$. Assuming the moon to be a perfectly black body,the temperature of the moon,if the wavelength corresponding to maximum emission is $14.35 \ \mu m$,is ...... $K$.

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In a certain planetary system,it is observed that one of the celestial bodies having a surface temperature of $200 \; K$,emits radiation of maximum intensity near the wavelength $12 \; \mu m$. The surface temperature (in $K$) of a nearby star which emits light of maximum intensity at a wavelength $\lambda = 4800 \; \mathring A$ is

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The wavelength of maximum emission of radiation from a body at $2000 \ K$ is $4 \ \mu m$. What is the wavelength of maximum emission of radiation from the body at $2400 \ K$ in $\mu m$?

If wavelengths of maximum intensity of radiations emitted by the sun and the moon are $0.5 \times 10^{-6} \ m$ and $10^{-4} \ m$ respectively,the ratio of their temperatures is

The graph of relative intensity versus wavelength for three perfectly black bodies at temperatures $T_1, T_2,$ and $T_3$ is shown below. The relationship between their temperatures is:

$A$ black body at a temperature of $1640 \ K$ has the wavelength corresponding to maximum emission equal to $1.75 \ \mu m$. Assuming the moon to be a perfectly black body,the temperature of the moon,if the wavelength corresponding to maximum emission is $14.35 \ \mu m$,is ...... $K$.

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