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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 \(..............\).

  1. A \(\frac{1}{\sqrt{2}} \lambda_1\)
  2. B \(2 \lambda_1\)
  3. C \(\frac{1}{2} \lambda_1\)
  4. D \(\sqrt{2} \lambda_1\)
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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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