Question

Two spherical bodies $A$ (radius $6$ cm) and $B$ (radius $18$ cm) are at temperature $T_1$ and $T_2$, respectively. The maximum intensity in the emission spectmm 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. 3
• C. 2
• D. 4

Answer 1

The correct option is A 9

${\lambda }_{m}T=$ constant

${\lambda }_A{T}_A={\lambda }_B{T}_B$

From Stephens law total energy radiated $E=\sigma A{T}^4$
or

$\frac{E_A}{E_B}={\left(\frac{R_A}{R_B}\right)}^2\times {\left(\frac{T_A}{T_B}\right)}^4$
or

$\frac{E_A}{E_B}={\left(\frac{R_A}{R_B}\right)}^2\times {\left(\frac{{\lambda }_A}{{\lambda }_B}\right)}^4$

$={\left(\frac{6}{18}\right)}^2\times {\left(\frac{1500}{50}\right)}^4$

$=9$