What should be the order

Question:

What should be the order of arrangement of de-Broglie wavelength of electron $\left(\lambda_{e}\right)$, an $\alpha$-particle $\left(\lambda_{\alpha}\right)$ and proton $\left(\lambda_{p}\right)$ given that all have the same kinetic energy ?

  1. $\lambda_{\mathrm{e}}=\lambda_{\mathrm{p}}=\lambda_{\alpha}$

  2. $\lambda_{\mathrm{e}}<\lambda_{\mathrm{p}}<\lambda_{\alpha}$

  3. $\lambda_{\mathrm{e}}>\lambda_{\mathrm{p}}>\lambda_{\alpha}$

  4. $\lambda_{\mathrm{e}}=\lambda_{\mathrm{p}}>\lambda_{\alpha}$

Correct Option: , 3

Solution:

$\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \propto \frac{1}{\sqrt{\mathrm{m}}}$

$m_{\alpha}>m_{p}>m_{e}$

so $\lambda_{e}>\lambda_{p}>\lambda_{\alpha}$

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