Question 676: Thermodynamics & KTG

Question

Black bodies A and B radiate maximum energy with wavelength difference $4 \mu \mathrm{~m}$. The absolute temperature of body A is 3 times that of B. The wavelength at which body $B$ radiates maximum energy is

Accepted Answer

Black bodies A and B radiate maximum energy with wavelength difference $4 \mu m$.
Temperatures are related:
$ T_A = 3 T_B $
We need the wavelength at which body B radiates maximum energy.
Step 1: Use WienÃ¢ÂÂs displacement law
$ \lambda_{	ext{max}} T = b \quad (	ext{WienÃ¢ÂÂs constant}) $
So:
$ \lambda_A T_A = \lambda_B T_B $
$ \lambda_A (3 T_B) = \lambda_B T_B $
$ \lambda_A = \frac{\lambda_B}{3} $
Step 2: Use the wavelength difference condition
They differ by $4 \mu m$:
$ \lambda_B - \lambda_A = 4 $
Substitute $\lambda_A = \frac{\lambda_B}{3}$:
$ \lambda_B - \frac{\lambda_B}{3} = 4 $
$ \frac{2}{3}\lambda_B = 4 $
$ \lambda_B = 6 \, \mu m $
Ã¢ÂÂ Final Answer
The wavelength at which body B radiates maximum energy is
$ \boxed{6 \, \mu m} $
Correct Option: B

Answer

$4 \mu \mathrm{~m}$

Answer

$6 \mu \mathrm{~m}$

Answer

$2 \mu \mathrm{~m}$

Answer

$8 \mu \mathrm{~m}$

Thermodynamics & KTG

Black bodies A and B radiate maximum energy with wavelength difference 4 \mu \mathrm{~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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