JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2023
- Medium
- MCQ
The de Broglie wavelength of a molecule in a gas at room temperature \((300\,K )\) is \(\lambda_1\). If the temperature of the gas is increased to \(600\,K\), then the de Broglie wavelength of the same gas molecule becomes \(..............\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sqrt{2}} \lambda_1\)
Step-by-step Solution
Detailed explanation
From K.T.G.
\(v _{ RMS }=\sqrt{\frac{3 k _{ B } T }{ m }}\)
\(v _{ RMS } \propto \sqrt{ T }\)
\(\text { and } \frac{ h }{ m v_{ RMS }}=\lambda \text { i.e., } \lambda \propto \frac{1}{\sqrt{ T }}\)
\(\frac{\lambda_2}{\lambda_1}=\sqrt{\frac{ T _1}{ T _2}}=\sqrt{\frac{300}{600}}=\frac{1}{\sqrt{2}}\)
\(\lambda_2=\frac{\lambda_1}{\sqrt{2}}\)
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