Q74 / 90 · JEE Main 2024 (08 Apr Shift 2) · Physics

A proton and an electron have the same de Broglie wavelength. If $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{e}}$ be the kinetic energies of proton and electron respectively, then choose the correct relation :

  • A
    $K_p>K_e$
  • Correct answer
    B
    $\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{e}}$
  • C
    $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}$
  • D
    $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}{ }^2$

Solution

De Broglie wavelength of proton \& electron $=\lambda$ $\begin{aligned} & \because \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \therefore \mathrm{p}_{\text {proton }}=\mathrm{p}_{\text {electron }} \\ & \because \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{~m}} \\ & \therefore \mathrm{KE}_{\text {proton }} < \mathrm{KE}_{\text {electron }} \\ & {\left[\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{c}}\right]}\end{aligned}$

Hence, the correct answer is option: 'B'.

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