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JEE Main 2024 (8 Apr, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2024 (8 Apr, Shift 1)

Q57JEE Main 2024 · 8 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:

(Assume h = 6.63 ×10−34 J s, me=9.0×10−31 kg\times 10^{-34}\text{ J s},\ \mathrm{m_e} = 9.0 \times 10^{-31}\text{ kg} and mp=1836\mathrm{m_p} = 1836 times me\mathrm{m_e})

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Answer: (c)

  1. Same λ=h/p\lambda = h/p means same momentum pp.
  2. K=p22m⇒K∝1mK = \dfrac{p^2}{2m} \Rightarrow K \propto \dfrac{1}{m} for the same pp.
  3. KpKe=memp=11836\dfrac{K_p}{K_e} = \dfrac{m_e}{m_p} = \dfrac{1}{1836}
  4. So the answer is (c) 1:18361 : 1836.
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