Question

If the wavelengths of maximum intensity of radiation emitted by two black bodies A and B are \( 0.5 \, \mu\text{m} \) and \( 0.1 \, \text{mm} \) respectively, then ratio of the temperatures of the bodies A and B is

• A. 5
• B. 25
• C. 100
• D. 200

Hint:

Wien's Displacement Law: \( \lambda_m T = b \) (constant). This implies \( T_1 / T_2 = \lambda_{m2} / \lambda_{m1} \). Be careful with units of wavelength: \( 1 \, \mu\text{m} \) (micrometer) = \( 10^{-6} \, \text{m} \). \( 1 \, \text{mm} \) (millimeter) = \( 10^{-3} \, \text{m} \). Ensure wavelengths are in the same unit before taking the ratio.

Answer 1

The Correct Option is D

According to Wien's displacement law, the wavelength \( \lambda_m \) at which the intensity of radiation emitted by a black body is maximum is inversely proportional to its absolute temperature \(T\).
\( \lambda_m T = b \), where \(b\) is Wien's displacement constant (\( \approx 2.
898 \times 10^{-3} \, \text{m K} \)).
So, \( \lambda_m \propto \frac{1}{T} \), or \( T \propto \frac{1}{\lambda_m} \).
For two black bodies A and B: \( \lambda_{mA} T_A = b \) \( \lambda_{mB} T_B = b \) Therefore, \( \lambda_{mA} T_A = \lambda_{mB} T_B \).
The ratio of temperatures \( \frac{T_A}{T_B} = \frac{\lambda_{mB}}{\lambda_{mA}} \).
Given wavelengths: \( \lambda_{mA} = 0.
5 \, \mu\text{m} = 0.
5 \times 10^{-6} \, \text{m} \).
\( \lambda_{mB} = 0.
1 \, \text{mm} = 0.
1 \times 10^{-3} \, \text{m} = 1 \times 10^{-4} \, \text{m} \).
Now calculate the ratio: \[ \frac{T_A}{T_B} = \frac{1 \times 10^{-4} \, \text{m}}{0.
5 \times 10^{-6} \, \text{m}} \] \[ \frac{T_A}{T_B} = \frac{10^{-4}}{0.
5 \times 10^{-6}} = \frac{1}{0.
5} \times \frac{10^{-4}}{10^{-6}} = 2 \times 10^{-4 - (-6)} = 2 \times 10^{-4+6} = 2 \times 10^2 \] \[ \frac{T_A}{T_B} = 2 \times 100 = 200 \] The ratio of the temperatures \( T_A : T_B = 200:1 \).
So, the ratio is 200.
This matches option (4).