Question

A black body has a temperature of 2900 K. The wavelength of peak emission ($λ_{max}$) of the radiation emitted from it is approximately equal to:

• A. 1.500 nm
• B. 1000 nm
• C. 3000 nm
• D. 6000 nm

Hint:

Wien’s Displacement Law is useful for determining the peak wavelength of radiation emitted by a black body at a given temperature.

Answer 1

The Correct Option is B

1. Use Wien's Displacement Law to calculate the wavelength of peak emission:

\( \lambda_{\text{max}} = \frac{b}{T} \)

Where:

• \( b = 2.898 \times 10^{-3} \, \text{m·K} \) (Wien's constant)
• \( T = 2900 \, \text{K} \) (Temperature of the black body)

2. Substituting the values:

\[ \lambda_{\text{max}} = \frac{2.898 \times 10^{-3}}{2900} \]

\( \lambda_{\text{max}} = 1.000 \times 10^{-6} \, \text{m} \)

3. Convert to nanometers:

\( \lambda_{\text{max}} = 1000 \, \text{nm} \)

The wavelength of peak emission for a black body at 2900 K is approximately 1000 nm, which corresponds to infrared radiation.