Question

A black body has maximum wavelength \( \lambda_m \) at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be

• A. \( \frac{4\lambda_m}{9} \)
• B. \( \frac{2\lambda_m}{3} \)
• C. \( \frac{3\lambda_m}{2} \)
• D. \( \frac{9}{4}\lambda_m \)

Hint:

Wien’s law: \(\lambda_{\text{max}} \propto \frac{1}{T}\). Higher temperature shifts the peak to shorter wavelengths. Remember to use absolute temperature (Kelvin) only.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the peak wavelength \(\lambda_m\) of a black body at 2000 K and asked to find the new peak wavelength when the temperature is raised to 3000 K. This is a direct application of Wien’s displacement law.

Step 2: Key Formula or Approach:
Wien’s displacement law states: \(\lambda_{\text{max}} T = \text{constant}\).

Step 3: Detailed Explanation:
Let \(\lambda_1 = \lambda_m\) at \(T_1 = 2000\ \text{K}\) and \(\lambda_2\) at \(T_2 = 3000\ \text{K}\). Then: \[ \lambda_1 T_1 = \lambda_2 T_2 \quad \Rightarrow \quad \lambda_2 = \lambda_1 \frac{T_1}{T_2}. \] Substitute the values: \[ \lambda_2 = \lambda_m \times \frac{2000}{3000} = \lambda_m \times \frac{2}{3} = \frac{2\lambda_m}{3}. \]

Step 4: Final Answer:
The wavelength at 3000 K is \(\frac{2\lambda_m}{3}\), which corresponds to option (B).