Question 763: Thermodynamics & KTG

Question

Ordinary bodies P and Q radiate maximum energy with wavelength difference $3 \mu \mathrm{~m}$. The absolute temperature of body P is four times that of Q. The wavelength at which body Q radiates maximum energy is

Accepted Answer

To solve the problem, we start by noting the given information:
The difference in the wavelengths at which bodies P and Q radiate maximum energy is $3 \mu 	ext{m}$. This can be expressed as:
$   \lambda_{\mathrm{Q}} - \lambda_{\mathrm{P}} = 3 \mu 	ext{m} \quad 	ext{.... (i)}   $
According to Wien's law, the relationship between the wavelength of maximum emission ($\lambda$) and temperature ($T$) for a black body is given by:
$   \lambda_P T_P = \lambda_Q T_Q   $
We are informed that the temperature $T_P$ of body P is four times that of body Q ($T_Q$), therefore:
$   \lambda_P \cdot 4T_Q = \lambda_Q \cdot T_Q   $
Simplifying the equation above gives us:
$   \lambda_Q = 4 \lambda_P \quad 	ext{.... (ii)}   $
Using equation (ii), we substitute back into equation (i):
Substitute $\lambda_Q = 4 \lambda_P$ into $\lambda_Q - \lambda_P = 3 \mu 	ext{m}$:
$   4\lambda_P - \lambda_P = 3 \mu 	ext{m}   $
Solving for $\lambda_P$:
$   3\lambda_P = 3 \mu 	ext{m} \implies \lambda_P = 1 \mu 	ext{m}   $
Finally, using $\lambda_P = 1 \mu 	ext{m}$ in the expression $\lambda_Q = 4 \lambda_P$:
Solve for $\lambda_Q$:
$   \lambda_Q = 4 	imes 1 \mu 	ext{m} = 4 \mu 	ext{m}   $
Thus, the wavelength at which body Q radiates maximum energy is $4 \mu 	ext{m}$.

Answer

2 $\mu$m

Answer

4 $\mu$m

Answer

6 $\mu$m

Answer

8 $\mu$m

Thermodynamics & KTG

Ordinary bodies P and Q radiate maximum energy with wavelength difference 3 \mu \mathrm{~m}. The absolute temperature of body P is four times that of Q. The wavelength at which body Q radiates maximum energy is

Choose the correct answer: