Heat And Thermodynamics Ques 66
- Two spherical stars $A$ and $B$ emit black body radiation. The radius of $A$ is 400 times that of $B$ and $A$ emits $10^{4}$ times the power emitted from $B$. The ratio $\left(\frac{\lambda _A}{\lambda _B}\right)$ of their wavelengths $\lambda _A$ and $\lambda _B$ at which the peaks occur in their respective radiation curves is
(2015 Adv.)
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Solution:
Formula:
- Power, $P=\left(\sigma T^{4} A\right)=\sigma T^{4}\left(4 \pi R^{2}\right)$
or, $\quad P \propto T^{4} R^{2}$
According to Wien’s law,
$$ \lambda \propto \frac{1}{T} $$
( $\lambda$ is the wavelength at which peak occurs)
$\therefore$ Eq. (i) will become,
$$ \begin{gathered} P \propto \frac{R^{2}}{\lambda^{4}} \text { or } \lambda \propto\left[\frac{R^{2}}{P}\right]^{1 / 4} \\ \Rightarrow \quad \frac{\lambda _A}{\lambda _B}=\left[\frac{R _A}{R _B}\right]^{1 / 2}\left[\frac{P _B}{P _A}\right]^{1 / 4}=[400]^{1 / 2}\left[\frac{1}{10^{4}}\right]^{1 / 4}=2 \end{gathered} $$
BETA
