Question

Two spherical bodies $A$ (radius $6 \,cm$ ) and $B$ (radius $18\, cm$ ) are at temperatures $T_{1}$ and $T_{2}$, respectively. The maximum intensity in the emission spectrum of $A$ is at $500\, nm$ and in that of $B$ is at $1500\, nm$. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by $A$ to that of $B ?$

• 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 $\frac{T_{A}}{T_{B}}=\frac{\left(\lambda_{m}\right)_{B}}{\left(\lambda_{m}\right)_{A}}=\frac{1500 \,nm }{500 \,nm }$ 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} $
$\left[\right.$ Here, $\left.A=4 \pi R^{2}\right]$
$\therefore \frac{E_{A}}{E_{B}}=\left(\frac{R_{A}}{R_{B}}\right)^{2}\left(\frac{T_{A}}{T_{B}}\right)^{4}$
$=\left(\frac{6 \,cm }{18 \,cm }\right)^{2}(3)^{4}=9 $ ( Using (i))