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Question

(A) According to Wien's displacement law, ${\lambda _m T = 	ext{constant}}$, so ${\lambda _1 T_1 = \lambda _2 T_2}$.Given ${\lambda _1 = \lambda _0}$

Vedclass · Accepted Answer

$256/81$

Vedclass · Answer

(A) According to Wien's displacement law, ${\lambda _m T = 	ext{constant}}$, so ${\lambda _1 T_1 = \lambda _2 T_2}$.Given ${\lambda _1 = \lambda _0}$ and ${\lambda _2 = \frac{3\lambda _0}{4}}$.Therefore, $\frac{T_2}{T_1} = \frac{\lambda _1}{\lambda _2} = \frac{\lambda _0}{3\lambda _0 / 4} = \frac{4}{3}$.According to the Stefan-Boltzmann law, the emissive power $E \propto T^4$.Thus, the ratio of emissive powers is $\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} ight)^4$.Substituting the values, $\frac{E_2}{E_1} = \left( \frac{4}{3} ight)^4 = \frac{256}{81}$.

The wavelength corresponding to the maximum energy emitted by a body at a certain temperature is ${\lambda _0}$. If the temperature of the body is increased such that the new maximum wavelength becomes $\frac{3{\lambda _0}}{4}$, then by what factor does the emissive power increase?

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