Two spherical bodies $A$ (radius $6 \mathrm{~cm}$ ) and $B$ (radius $18 \mathrm{~cm}$ ) are at temperatures $T_1$ and $T_2$, respectively. The maximum intensity in the emission spectrum of $A$ is at $500 \mathrm{~nm}$ and in that of $B$ is at $1500 \mathrm{~nm}$. Considering them to be black…

Question

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

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } \lambda_m \propto \frac{1}{T} \ & 	herefore \frac{\lambda_A}{\lambda_B}=\frac{T_B}{T_A}=\frac{500}{1500}=\frac{1}{3} \ & E \propto T^4 A 	ext { (where } A=	ext { surface area } \ & =4 \pi R^2 	ext { ) } \ & 	herefore E \propto T^4 R^2 \ & \frac{E_A}{E_B}=\left(\frac{T_A}{T_B}ight)^4\left(\frac{R_A}{R_B}ight)^2 \ & =(3)^4\left(\frac{6}{18}ight)^2=9 \end{aligned} $ $	herefore$ Answer is 9

Two spherical bodies $A$ (radius $6 \mathrm{~cm}$ ) and $B$ (radius $18 \mathrm{~cm}$ ) are at temperatures…

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