Question:
Black bodies A and B radiate maximum energy with wavelength difference $4\mu \text{ m}$. The absolute temperature of body A is 3 times that of B. The wavelength at which body $B$ radiates maximum energy is
Black bodies A and B radiate maximum energy with wavelength difference $4\mu \text{ m}$. The absolute temperature of body A is 3 times that of B. The wavelength at which body $B$ radiates maximum energy is
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Higher temperature implies shorter wavelength of peak emission.
Updated On: Apr 26, 2026
- $4\mu \text{ m}$
- $6\mu \text{ m}$
- $2\mu \text{ m}$
- $8\mu \text{ m}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Wien's Displacement Law: $\lambda_{max} T = \text{constant}$. Thus, $\lambda_A T_A = \lambda_B T_B$.
Step 2: Analysis
Given $T_A = 3 T_B$. Therefore, $\lambda_A (3 T_B) = \lambda_B T_B \implies \lambda_A = \frac{\lambda_B}{3}$.
Also given $|\lambda_B - \lambda_A| = 4\mu\text{m}$. Since $T_A>T_B$, $\lambda_B>\lambda_A$.
Step 3: Calculation
$\lambda_B - \frac{\lambda_B}{3} = 4 \implies \frac{2\lambda_B}{3} = 4$
$\lambda_B = \frac{4 \times 3}{2} = 6\mu\text{m}$.
Step 4: Conclusion
Hence, the wavelength for body B is $6\mu\text{m}$.
Final Answer: (B)
Wien's Displacement Law: $\lambda_{max} T = \text{constant}$. Thus, $\lambda_A T_A = \lambda_B T_B$.
Step 2: Analysis
Given $T_A = 3 T_B$. Therefore, $\lambda_A (3 T_B) = \lambda_B T_B \implies \lambda_A = \frac{\lambda_B}{3}$.
Also given $|\lambda_B - \lambda_A| = 4\mu\text{m}$. Since $T_A>T_B$, $\lambda_B>\lambda_A$.
Step 3: Calculation
$\lambda_B - \frac{\lambda_B}{3} = 4 \implies \frac{2\lambda_B}{3} = 4$
$\lambda_B = \frac{4 \times 3}{2} = 6\mu\text{m}$.
Step 4: Conclusion
Hence, the wavelength for body B is $6\mu\text{m}$.
Final Answer: (B)
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