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JEE Main202310 Apr 2023Morning ShiftPhysicsDual Nature of MatterActual

The de Broglie wavelength of a molecule in a gas at room temperature 300 K is λ 1 . If the temperature of the gas is increased to 600 K , then the de Broglie wavelength of the same gas molecule becomes

Options

  1. A1 2 λ 1
  2. B2 λ 1
  3. C1 2 λ 1
  4. D2 λ 1

Correct answer

C. 1 2 λ 1

Step-by-step solution

The root mean squared velocity of a gas is given by v = 3 R T M ⇒ v ∝ T Let T 1 = 300   K T 2 = 600   K Taking velocity ratios at the given temperatures, v 1 v 2 = T 1 T 2 = 300 600 = 1 2 The de Broglie wavelength is given by λ = h m v . So, λ ∝ 1 v The ratio of the wavelengths is λ 1 λ 2 = v 2 v 1 = 2 1 ⇒ λ 2 = 1 2 λ 1

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