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JEE Main 2021 (24 Feb, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2021 (24 Feb, Shift 2)

Q97JEE Main 2021 · 24 FEB · SHIFT 2Easy

Matter Waves: de Broglie Wavelength

The de-Broglie wavelength of a proton and α\alpha-particle are equal. The ratio of their velocities is:

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Answer: (c)

  1. Equal λ\lambda means equal momentum: mpvp=mαvαm_p v_p = m_\alpha v_\alpha.
  2. vpvα=mαmp=4\dfrac{v_p}{v_\alpha} = \dfrac{m_\alpha}{m_p} = 4
  3. So the answer is (c) 4:14 : 1.
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