Ordinary bodies $P$ and $Q$ radiate maximum energy with wavelength difference $3 \mu \mathrm{~m}$. The…

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
  1. $2 \mu \mathrm{~m}$
  2. $4 \mu \mathrm{~m}$
  3. $6 \mu \mathrm{~m}$
  4. $8 \mu \mathrm{~m}$

Solution

Given $\lambda_Q-\lambda_P=3 \mu \mathrm{~m}$...(i)
From Wien's law, $\begin{array}{ll} & \lambda_P T_P=\lambda_Q T_Q \\ \therefore \quad & \lambda_P 4 T_Q=\lambda_Q T_Q \\ & \Rightarrow \lambda_Q=4 \lambda_P ...(ii)\\ & 4 \lambda_P-\lambda_P=3 \mu \mathrm{~m} \\ & 3 \lambda_P=3 \mu \mathrm{~m} \\ & \lambda_P=1 \mu \mathrm{~m} \\ \therefore \quad & \lambda_Q=4 \mu \mathrm{~m} \end{array}$ ...(given, $\mathrm{T}_{\mathrm{P}}=4 \mathrm{~T}_{\mathrm{Q}}$ ) ...[From(i)] ...[From(ii)] ^

Asked in: MHT CET 2024 (10 May Shift 1)

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