JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
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 \(..............\).
- A \(\frac{1}{\sqrt{2}} \lambda_1\)
- B \(2 \lambda_1\)
- C \(\frac{1}{2} \lambda_1\)
- D \(\sqrt{2} \lambda_1\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sqrt{2}} \lambda_1\)
Step-by-step Solution
Detailed explanation
From K.T.G. \(v _{ RMS }=\sqrt{\frac{3 k _{ B } T }{ m }}\) \(v _{ RMS } \propto \sqrt{ T }\) \(\text { and } \frac{ h }{ m v_{ RMS }}=\lambda \text { i.e., } \lambda \propto \frac{1}{\sqrt{ T }}\)…
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