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NEETPhysicsDual Nature of Matter

The de-Broglie wavelength of a thermal neutron at 27^ C is ₁ . At a temperature of 927^ C , the de-Broglie wavelength of a thermal neutron is ₂ . The ratio ₁ : ₂ is:

Options

  1. A1 : 2
  2. B4 : 1
  3. C2 : 1
  4. D1 : 4

Correct answer

C. 2 : 1

Step-by-step solution

The kinetic energy of a thermal neutron at absolute temperature T is given by K = 3 2 kT . The de-Broglie wavelength is = h 2mK = h 3mkT . This implies that 1 T . Converting the given temperatures to Kelvin: T₁ = 27^ C + 273 = 300 K T₂ = 927^ C + 273 = 1200 K Taking the ratio of the wavelengths: ₁ ₂ = T₂ T₁ = 1200 300 = 4 = 2 Thus, the ratio ₁ : ₂ is 2 : 1 . Answer: 2 : 1

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