Question

A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be observed at :

• A. 7500 Å
• B. 1500 Å
• C. 6000 Å
• D. 3000 Å

Answer 1

The Correct Option is D

To solve this question, we need to use Wien's Displacement Law, which states that the wavelength of maximum intensity (\( \lambda_{\text{max}} \)) is inversely proportional to the absolute temperature (T) of a black body. The law is given by the formula:

\lambda_{\text{max}} \cdot T = b

where \( b \) is Wien's displacement constant, approximately equal to 2898 µm·K.

1. First, convert the initial temperature from Celsius to Kelvin:
   • \( T_1 = 1227^\circ C + 273 = 1500 \, \text{K} \)
2. Given that the wavelength of maximum intensity at this temperature is 5000 Å, we can express Wien's Law for this case as:
   • \( \lambda_{\text{max1}} \cdot T_1 = b \)
   • \( 5000 \, \text{Å} \cdot 1500 \, \text{K} = b \)
3. Now, increase the temperature by 1000°C and convert it to Kelvin:
   • \( T_2 = (1227^\circ C + 1000^\circ C) + 273 = 2500 \, \text{K} \)
4. Apply Wien's Displacement Law for the new temperature:
   • \( \lambda_{\text{max2}} \cdot T_2 = b \)
   • Substitute for the constant \( b \) from the first condition:
   • \( \lambda_{\text{max2}} \cdot 2500 \, \text{K} = 5000 \, \text{Å} \cdot 1500 \, \text{K} \)
5. Solve for \( \lambda_{\text{max2}} \):
   • \( \lambda_{\text{max2}} = \frac{{5000 \, \text{Å} \cdot 1500}}{{2500}} \)
   • \( \lambda_{\text{max2}} = 3000 \, \text{Å} \)

Therefore, when the temperature of the black body is increased to 2500 K, the maximum intensity of radiation will be observed at a wavelength of 3000 Å. Thus, the correct answer is 3000 Å.

This solution is consistent with Wien's Displacement Law, which ensures that as the temperature of a body increases, the wavelength at which it emits most intensely shifts towards the shorter end of the spectrum (reflecting higher energy radiation).