Question

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

• A. 9
• B. 12
• C. 3
• D. 6

Answer 1

Answer: (A)

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