(C) According to Wien's displacement law, $\lambda_{m} T = \text{constant}$, where $\lambda_{m}$ is the wavelength corresponding to maximum energy and

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(C) According to Wien's displacement law, $\lambda_{m} T = \text{constant}$, where $\lambda_{m}$ is the wavelength corresponding to maximum energy and $T$ is the absolute temperature.Therefore, $\lambda_{1} T_{1} = \lambda_{2} T_{2}$.Given $\lambda_{1} = 11 \times 10^{-5} \ cm$ and $\lambda_{2} = 5.5 \times 10^{-5} \ cm$.We are given $T_{1} = n T_{2}$.Substituting the values into the equation: $(11 \times 10^{-5}) \times (n T_{2}) = (5.5 \times 10^{-5}) \times T_{2}$.$n = \frac{5.5 \times 10^{-5}}{11 \times 10^{-5}} = \frac{5.5}{11} = 0.5$.

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

Medium

The maximum energy in the thermal radiation from a hot source occurs at a wavelength of $11 \times 10^{-5} \ cm$. According to Wien's law, the temperature of the source (on Kelvin scale) will be $n$ times the temperature of another source (on Kelvin scale) for which the wavelength at maximum energy is $5.5 \times 10^{-5} \ cm$. The value of $n$ is

A
$2$
B
$4$
C
$0.5$
D
$1$

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10-2.Heat Transfer

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Spectral Emissive Power and Wien's Displacement Law

Three black discs $x, y, z$ have radii $1 \ m, 2 \ m$ and $3 \ m$ respectively. The wavelengths corresponding to maximum intensity are $200 \ nm, 300 \ nm$ and $400 \ nm$ respectively. The relation between emissive power $E_x, E_y$ and $E_z$ is:

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MediumAP EAMCET 2020

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At a temperature of $2000 \; K$,the maximum energy of a black body is emitted at a wavelength of $14 \; \mu m$. When its temperature is reduced to $1000 \; K$,the wavelength corresponding to the maximum energy emitted is ....... $\mu m$.

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Black bodies $A$ and $B$ radiate maximum energy with wavelength difference $4 \mu m$. The absolute temperature of body $A$ is $3$ times that of $B$. The wavelength at which body $B$ radiates maximum energy is (in $\mu m$)

MediumMHT CET 2025

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The surface temperature of the sun which has maximum energy emission at $500 \,nm$ is $6000 \,K$. The temperature of a star which has maximum energy emission at $400 \,nm$ will be: (in $\,K$)

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Three black discs $x, y, z$ have radii $1 \ m, 2 \ m$ and $3 \ m$ respectively. The wavelengths corresponding to maximum intensity are $200 \ nm, 300 \ nm$ and $400 \ nm$ respectively. The relation between emissive power $E_x, E_y$ and $E_z$ is:

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