Q4 / 155 · WBJEE 2022 · Chemistry

The de-Broglie wavelength $(\lambda)$ for electron (e), proton $(p)$ and $\mathrm{He}^{2+}$ ion $(\alpha)$ are in the following order. (Speed of e, p and $\alpha$ are the same)

  • A
    $\alpha>p>e$
  • Correct answer
    B
    $e>p>\alpha$
  • C
    $e>\alpha>p$
  • D
    $\alpha < p>e$

Solution

$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\because$ speed are same $\begin{aligned} &\therefore \lambda \propto \frac{1}{\mathrm{~m}} \\ &\therefore \lambda_e>\lambda_p>\lambda_\alpha \text { as } \quad \mathrm{m}_\alpha>\mathrm{m}_{\mathrm{p}}>\mathrm{m}_e \end{aligned}$

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

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