MHT CET · Physics · Thermal Properties of Matter
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
- A \(2 \mu \mathrm{~m}\)
- B \(4 \mu \mathrm{~m}\)
- C \(6 \mu \mathrm{~m}\)
- D \(8 \mu \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(4 \mu \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
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)]
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)]
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