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Three black discs $x, y, z$ have radii $1 \mathrm{~m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ respectively. The wavelength corresponding to maximum intensity are $200 \mathrm{~nm}$, $300 \mathrm{~nm}$ and $400 \mathrm{~nm}$ respectively. The relation between emissive power $E_x, E_y$ and $E_z$ is
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Verified Answer
The correct answer is:
$E_x$ is maximum
According to the Stefan's Law:
$\begin{aligned}
& E=\frac{\sigma A T^4}{A} \\
& \Rightarrow E \propto T^4
\end{aligned}$
According to the Wein's Law:
$\begin{aligned}
& T=\frac{k}{\lambda_m} \\
& \therefore E \propto \frac{A}{\lambda^4} \\
& E_x: E_y: E_z=\frac{1}{\lambda_x^4}: \frac{1}{\lambda^4}: \frac{1}{\lambda^4}=\frac{1}{200^4}: \frac{1}{300^4}: \frac{1}{400^4}
\end{aligned}$
The relation between emissive power $E_x, E_y$ and $E_z$ is as follows:
$E x>E y>E z$
$\begin{aligned}
& E=\frac{\sigma A T^4}{A} \\
& \Rightarrow E \propto T^4
\end{aligned}$
According to the Wein's Law:
$\begin{aligned}
& T=\frac{k}{\lambda_m} \\
& \therefore E \propto \frac{A}{\lambda^4} \\
& E_x: E_y: E_z=\frac{1}{\lambda_x^4}: \frac{1}{\lambda^4}: \frac{1}{\lambda^4}=\frac{1}{200^4}: \frac{1}{300^4}: \frac{1}{400^4}
\end{aligned}$
The relation between emissive power $E_x, E_y$ and $E_z$ is as follows:
$E x>E y>E z$
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