Question

Two spherical bodies A (radius $6\text{cm})$ and B (radius $18\text{cm})$ are at temperatures $T_1$ and $T_2$ , respectively. The maximum intensity in the emission spectrum of A is at $500\text{nm}$ and in that of B is at $1500\text{nm}$. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B ?

Answer 1

According to Wein's displacement law
${\lambda }_{m}T=\text{constant}$
$\therefore ({\lambda }_m{)}_A{T}_A=({\lambda }_m{)}_B\left(T_B\right)$
$\frac{T_A}{T_B}=\frac{({\lambda }_m{)}_B}{({\lambda }_m{)}_A}=\frac{1500\text{nm}}{500\text{nm}}$
$\frac{T_A}{T_B}=3$
According to Stefan Boltzmann law, rate of energy radiated by a black body.
$E=\sigma A{T}^4=\sigma 4\pi R^2{T}^4[\text{Here},A=4\pi R^2]$
$\because \frac{E_A}{E_B}={\left(\frac{R_A}{R_B}\right)}^2{\left(\frac{T_A}{T_B}\right)}^4={\left(\frac{6\text{cm}}{18\text{cm}}\right)}^2(3{)}^4=9$