(C) According to Wien's displacement law,$\lambda T = b$ (constant),so $T \propto \frac{1}{\lambda}$.Since the new wavelength is $\lambda' = \frac{\la

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(C) According to Wien's displacement law,$\lambda T = b$ (constant),so $T \propto \frac{1}{\lambda}$.Since the new wavelength is $\lambda' = \frac{\lambda}{3}$,the new temperature $T'$ becomes $T' = 3T$.The emissive power of a black body is given by Stefan-Boltzmann law: $E = \sigma T^4$.Therefore,the new emissive power $E'$ is:$E' = \sigma (T')^4 = \sigma (3T)^4 = 81 \sigma T^4$.Since $E = \sigma T^4$,we get $E' = 81 E$.

Class 11 Physics 10-2.Heat Transfer Radiation by Stefan's Boltzmann Law

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$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now,due to a change in temperature of that body,it radiates maximum energy at wavelength $\frac{\lambda}{3}$. At that new temperature,the emissive power is: (in $E$)

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A
$16$
B
$256$
C
$81$
D
$128$

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10-2.Heat Transfer

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Radiation by Stefan's Boltzmann Law

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$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now,due to a change in temperature of that body,it radiates maximum energy at wavelength $\frac{\lambda}{3}$. At that new temperature,the emissive power is: (in $E$)

Similar Questions

$A$ rectangular metal plate $8 \ cm \times 4 \ cm$ at $127^{\circ} C$ emits $E \ Js^{-1}$. If both length and breadth are halved and the temperature is raised to $327^{\circ} C$,the rate of emission is

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Two spheres of radii $4 \, m$ and $1 \, m$ have temperatures $2000 \, K$ and $4000 \, K$ respectively. Find the ratio of their emitted energy.

Two spheres of the same material have radii $1 \ m$ and $4 \ m$ and temperatures $4000 \ K$ and $2000 \ K$ respectively. The ratio of the energy radiated by them is .....

$A$ black body radiates energy at a rate of $E \, W/m^2$ at a temperature of $T \, K$. When the temperature is reduced to $T/2 \, K$,the radiated energy will be .....

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