Question

A light source having its intensity peak at the wavelength 289.8 nm is calibrated as 10,000 K which is the temperature of an equivalent black body radiation. Considering the same calibration, the temperature of light source (in K) having its intensity peak at the wavelength 579.6 nm (rounded off to the nearest integer) is \(\underline{\hspace{2cm}}\).

Hint:

Use Wien's displacement law \( \lambda_{\text{peak}} T = b \) to find the temperature corresponding to the peak wavelength.

Answer 1

Correct Answer: 5000

The temperature of a black body radiation source is related to the wavelength of peak intensity by Wien's displacement law: \[ \lambda_{\text{peak}} T = b, \] where \( b = 2.898 \times 10^6 \, \text{nm} \cdot \text{K} \), \( \lambda_{\text{peak}} \) is the wavelength at peak intensity, and \( T \) is the temperature. For the first light source: \[ T_1 = \frac{b}{\lambda_1} = \frac{2.898 \times 10^6}{289.8} = 10,000 \, \text{K}. \] For the second light source with \( \lambda_2 = 579.6 \, \text{nm} \): \[ T_2 = \frac{b}{\lambda_2} = \frac{2.898 \times 10^6}{579.6} = 5000 \, \text{K}. \] Thus, the temperature of the second light source is 5000 K.