Question

The energy spectrum of a black body exhibits a maximum around a wavelength '\(λ\)'. The temperature of the black body is now changed such that the energy is maximum around a wavelength \(2λ/3\). The power radiated by the black body will now increase by a factor

• A. \(\frac{9}{4}\)
• B. \(\frac{81}{16}\)
• C. \(\frac{3}{2}\)
• D. \(\frac{27}{8}\)

Hint:

Wien's law gives the new temperature; power goes as T^4.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Wien's displacement law: \(\lambda_mT = \text{constant}\). Stefan-Boltzmann law: power radiated \(P \propto T^4\).

Step 2: New temperature:
\(\lambda_1T_1 = \lambda_2T_2\) with \(\lambda_2 = \dfrac{2\lambda}{3}\):
\[ T_2 = T_1\cdot\frac{\lambda}{2\lambda/3} = \frac32T_1 \]

Step 3: Power ratio:
\[ \frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac32\right)^4 = \frac{81}{16} \]

Step 4: Check:
Option (B). Option (A) \(\tfrac94\) is only \((T_2/T_1)^2\), and (C) is \(T_2/T_1\) itself. Option (D) \(\tfrac{27}{8}\) is \((T_2/T_1)^3\).

Final Answer:
T becomes 3/2 times, so power rises by (3/2)^4. \[ \boxed{\text{(B) }\dfrac{81}{16}} \]