Question

Two spherical stars, A and B emit black body radiation. The radius of star A is \(400\) times that of star B and star A emits power \(10^{4}\) times emitted from star B. \(\lambda _{A}\) and \(\lambda _{B}\) are the wavelengths at which the peak occurs in their respective radiation curves such that \(\lambda _{A}=x\lambda _{B}\). The value of \(x\) is

• A. \(\frac{1}{4}\)
• B. \(4\)
• C. \(2\)
• D. \(\frac{1}{2}\)

Hint:

Use Stefan's law to compare temperatures from power and radius, then use Wien's displacement law.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Stefan's law: power \(P=\sigma\,4\pi R^{2}T^{4}\). Wien's law: \(\lambda_{m}T=\text{constant}\).

Step 2: Find the temperature ratio
\(\dfrac{P_{A}}{P_{B}}=\left(\dfrac{R_{A}}{R_{B}}\right)^{2}\left(\dfrac{T_{A}}{T_{B}}\right)^{4}\)
\(10^{4}=400^{2}\left(\dfrac{T_{A}}{T_{B}}\right)^{4}\), so \(\left(\dfrac{T_{A}}{T_{B}}\right)^{4}=\dfrac{10^{4}}{16\times10^{4}}=\dfrac{1}{16}\), giving \(\dfrac{T_{A}}{T_{B}}=\dfrac{1}{2}\).

Step 3: Use Wien's law
\(\dfrac{\lambda_{A}}{\lambda_{B}}=\dfrac{T_{B}}{T_{A}}=2\), so \(x=2\).

Final Answer:
The value of x is 2. \[ \boxed{\text{(C) }2} \]