Question

Two metal spheres of the same material have radii $r$ and $2r$ and they emit thermal radiation with maximum intensities at wavelengths $\lambda$ and $2\lambda$ respectively. The respective ratio of the radiant energy emitted by them per second will be

• A. $4:1$
• B. $1:4$
• C. $8:1$
• D. $1:8$

Answer 1

The correct option is A $4:1$

From Wien's Displacement Law, for maximum intensity,
${\lambda }_{m}T=$ constant
So, $\frac{T_1}{T_2}=\frac{{\lambda }_{m_2}}{{\lambda }_{m_1}}$
$\Rightarrow \frac{T_1}{T_2}=\frac{2\lambda }{\lambda }=2$
$\Rightarrow T_1=2T_2$
Also, power of energy radiated,
$P=\sigma A{e}_r{T}^4$
For the sphere of the same material,
$P\propto A{T}^4$
So, $\frac{P_1}{P_2}=\frac{A_1{T_1}^4}{A_2{T_2}^4}$
$\Rightarrow \frac{P_1}{P_2}=\frac{{r_1}^2{T_1}^4}{{r_2}^2{T_2}^4}$ $[A=4\pi r^2]$
$\Rightarrow \frac{P_1}{P_2}=\frac{r^2{\left(2T_2\right)}^4}{{\left(2r\right)}^2{T_2}^4}=4:1$

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