Question

A black body radiates maximum energy at wavelength '\(λ\)' at temperature \(T_1\) and its emissive power is E. When the temperature of the body is changed to \(T_2\), it radiates maximum energy at wavelength \(\frac{2λ}{3}\), then the emissive power will become

• A. \(\frac{99}{16}\,E\)
• B. \(\frac{81}{16}\,E\)
• C. \(\frac{63}{16}\,E\)
• D. \(\frac{45}{16}\,E\)

Hint:

Wien's law: lambda_m times T is constant. Stefan's law: E is proportional to T^4.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The wavelength of maximum emission depends inversely on temperature (Wien's displacement law), and the emissive power of a black body goes as the fourth power of temperature (Stefan-Boltzmann law).

Step 2: Key Formula or Approach:
1. \(\lambda_mT = \text{constant}\).
2. \(E \propto T^4\).

Step 3: Detailed Explanation:
The wavelength changes from \(\lambda\) to \(\dfrac{2\lambda}{3}\):
\[ \lambda T_1 = \frac{2\lambda}{3}T_2 \Rightarrow \frac{T_2}{T_1} = \frac32 \]
Then the emissive power ratio is
\[ \frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac32\right)^4 = \frac{81}{16} \]
So \(E_2 = \dfrac{81}{16}E\). The other options have numerators 99, 63 and 45, which are not fourth powers of simple ratios.

Final Answer:
The new emissive power is \(\dfrac{81}{16}E\), option (B). \[ \boxed{\frac{81}{16}E \text{ (B)}} \]