Question

A blackbody at \( 1227^\circ\text{C} \) emits radiation with maximum intensity at a wavelength of \( 5600\,\text{\AA} \). If the temperature of the body is increased by \( 1000^\circ\text{C} \), the maximum intensity will be at wavelength

• A. $3000\text{\AA}$
• B. $3360\text{\AA}$
• C. $5000\text{\AA}$
• D. $6600\text{\AA}$

Hint:

- $\lambda_{\max} \propto \frac{1}{T}$ - Higher temperature $\Rightarrow$ shorter wavelength

Answer 1

The Correct Option is B

Concept:
Wien’s displacement law: \[ \lambda_{\max} T = \text{constant} \Rightarrow \lambda_1 T_1 = \lambda_2 T_2 \]

Step 1: Convert temperatures to Kelvin.
\[ T_1 = 1227 + 273 = 1500\ \text{K} \] \[ T_2 = (1227 + 1000) + 273 = 2500\ \text{K} \]

Step 2: Apply Wien’s law.
\[ \lambda_1 T_1 = \lambda_2 T_2 \] \[ 5600 \times 1500 = \lambda_2 \times 2500 \]

Step 3: Solve for $\lambda_2$.
\[ \lambda_2 = \frac{5600 \times 1500}{2500} = 5600 \times \frac{3}{5} = 3360\text{\AA} \]

Step 4: Select correct option.
Closest correct value → \( 3360\text{\AA} \) \[ \Rightarrow \text{Answer: Option (B)} \]