Ratio of Total Energy Radiated by Two Black Bodies

To find the ratio of the rate of total energy radiated by spherical body $A$ to that of spherical body $B$, we use Wien's Displacement Law and Stefan-Boltzmann Law.

Step 1: Determine the ratio of temperatures using Wien's Displacement Law
Wien's Displacement Law states that the wavelength corresponding to the maximum intensity ($\lambda_m$) is inversely proportional to the absolute temperature ($T$) of a black body:
$\lambda_m T = b \quad (\text{a constant})$

For bodies $A$ and $B$:
$\lambda_A T_1 = \lambda_B T_2$

Given:

• Peak wavelength for $A$, $\lambda_A = 500\text{ nm}$
• Peak wavelength for $B$, $\lambda_B = 1500\text{ nm}$

Thus, the temperature ratio is:
$\frac{T_1}{T_2} = \frac{\lambda_B}{\lambda_A} = \frac{1500\text{ nm}}{500\text{ nm}} = 3$

Step 2: Determine the ratio of total energy radiated using Stefan-Boltzmann Law
The rate of total energy radiated per unit time by a black body of surface area $A$ and temperature $T$ is given by Stefan-Boltzmann Law:
$P = \sigma A T^4$

Since the bodies are spherical, the surface area is $A = 4\pi R^2$, where $R$ is the radius of the body. Thus:
$P = 4\pi \sigma R^2 T^4 \implies P \propto R^2 T^4$

The ratio of the rate of total energy radiated by $A$ to that of $B$ is:
$\frac{P_A}{P_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_1}{T_2}\right)^4$

Given:

• Radius of $A$, $R_A = 6\text{ cm}$
• Radius of $B$, $R_B = 18\text{ cm}$

Substitute the known values into the ratio:
$\frac{R_A}{R_B} = \frac{6\text{ cm}}{18\text{ cm}} = \frac{1}{3}$

Now, calculating $\frac{P_A}{P_B}$:
$\frac{P_A}{P_B} = \left(\frac{1}{3}\right)^2 \times (3)^4 = \frac{1}{9} \times 81 = 9$

The ratio of the rate of total energy radiated by A to that of B is 9.