Easy MCQ +4 / -1 PYQ · JEE Mains 2023

An $\alpha$-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:

  1. A ${\lambda _\alpha } > {\lambda _p} < {\lambda _e}$
  2. B ${\lambda _\alpha } > {\lambda _p} > {\lambda _e}$
  3. C ${\lambda _\alpha } = {\lambda _p} = {\lambda _e}$
  4. D ${\lambda _\alpha } < {\lambda _p} < {\lambda _e}$ Correct answer

Solution

$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m k}}$ <br/><br/> So, $\lambda \propto \frac{1}{\sqrt{m}}$ <br/><br/> So, $\lambda_{e}>\lambda_{p}>\lambda_{\alpha}$

About this question

Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis

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