Question

A black body has maximum wavelength \( \lambda_m \) at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be

• A. \( \frac{4\lambda_m}{9} \)
• B. \( \frac{2\lambda_m}{3} \)
• C. \( \frac{3\lambda_m}{2} \)
• D. \( \frac{9}{4}\lambda_m \)

Hint:

Wien’s law: \(\lambda_{\text{max}} \propto \frac{1}{T}\). Higher temperature shifts the peak to shorter wavelengths. Remember to use absolute temperature (Kelvin) only.

Answer 1

The Correct Option is B

Step 1: Understand the problem.
A black body has its peak (most intense) wavelength $\lambda_m$ at $2000$ K. We heat it to $3000$ K and ask for the new peak wavelength.

Step 2: Use Wien's law.
Wien's displacement law says the peak wavelength times the absolute temperature is a constant: $\lambda_{max} T = \text{constant}$.

Step 3: Write it for both temperatures.
So $\lambda_1 T_1 = \lambda_2 T_2$, giving $\lambda_2 = \lambda_1 \dfrac{T_1}{T_2}$.

Step 4: Insert the temperatures.
$\lambda_2 = \lambda_m \times \dfrac{2000}{3000}$.

Step 5: Simplify the fraction.
$\dfrac{2000}{3000} = \dfrac{2}{3}$, so $\lambda_2 = \dfrac{2\lambda_m}{3}$.

Step 6: State the answer.
Hotter bodies peak at shorter wavelengths, so the peak shrinks to $\frac{2}{3}\lambda_m$, which is option (B).
\[ \boxed{\lambda_2 = \frac{2\lambda_m}{3}} \]