Question

Two black spheres P & Q have radii in the ratio 4 : 3. The wavelength of maximum intensity of radiation are in the ratio 4 : 5 respectively. The ratio of radiated power by P to Q is ______.

• A. $\frac{625}{144}$
• B. $\frac{125}{81}$
• C. $\frac{25}{9}$
• D. $\frac{5}{3}$

Hint:

Combine Wien's Law and Stefan's Law into one powerful super-ratio: $\frac{E_1}{E_2} = \left( \frac{R_1}{R_2} \right)^2 \left( \frac{\lambda_2}{\lambda_1} \right)^4$. This skips the intermediate temperature step entirely!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to determine the ratio of the total radiated power (emissive power) of two black body spheres using their given radii and the peak emission wavelengths.

Step 2: Detailed Explanation:
We are given two ratios:
Radii ratio: $\frac{R_P}{R_Q} = \frac{4}{3}$
Wavelength of max intensity ratio: $\frac{\lambda_P}{\lambda_Q} = \frac{4}{5}$
1. Find the Temperature Ratio:
According to Wien's Displacement Law, the wavelength of maximum intensity is inversely proportional to the absolute temperature:
$\lambda_{\text{max}} T = b \implies T \propto \frac{1}{\lambda_{\text{max}}}$
Therefore, the ratio of their temperatures is the inverse of their wavelength ratio:
$\frac{T_P}{T_Q} = \frac{\lambda_Q}{\lambda_P} = \frac{5}{4}$
2. Find the Radiated Power Ratio:
According to the Stefan-Boltzmann Law, the total power ($E$) radiated by a black body sphere is proportional to its surface area ($A = 4\pi R^2$) and the fourth power of its absolute temperature ($T^4$):
$E = \sigma A T^4 \propto R^2 T^4$
Set up the ratio for the two spheres P and Q:
$\frac{E_P}{E_Q} = \left( \frac{R_P}{R_Q} \right)^2 \times \left( \frac{T_P}{T_Q} \right)^4$
Substitute the known ratios:
$\frac{E_P}{E_Q} = \left( \frac{4}{3} \right)^2 \times \left( \frac{5}{4} \right)^4$
Expand the powers:
$\frac{E_P}{E_Q} = \left( \frac{16}{9} \right) \times \left( \frac{625}{256} \right)$
Simplify the fraction by cross-canceling 16 and 256 (since $256 = 16 \times 16$):
$\frac{E_P}{E_Q} = \frac{1}{9} \times \frac{625}{16}$
$\frac{E_P}{E_Q} = \frac{625}{9 \times 16}$
$\frac{E_P}{E_Q} = \frac{625}{144}$

Step 3: Final Answer:
The ratio of radiated power is $\frac{625}{144}$, matching option (a).