Q139 / 150 · MHT CET 2020 (14 Oct Shift 1) · Physics

An electron and photon are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{\mathrm{p}}$ to $\lambda_{\mathrm{e}}$ is $\left[\mathrm{m}_{\mathrm{e}}=\right.$ mass of electron, $\mathrm{m}_{\mathrm{p}}=$ mass of proton $]$

  • A
    $\left(\frac{m_{p}}{m_{e}}\right)^{\frac{1}{2}}$
  • Correct answer
    B
    $\left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}$
  • C
    $\left(\frac{m_{e}}{m_{p}}\right)$
  • D
    $\left(\frac{m p}{m}\right)$

Solution

Both will have same kinetic energy $\mathrm{k}$. Momentum of electron $\quad P_{\mathrm{e}}=\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{k}}$ Momentum of proton $P_{p}=\sqrt{2 m_{p} k}$ $\begin{array}{l} \lambda_{p}=\frac{h}{p_{p}}, \lambda_{e}=\frac{h}{p_{e}} \\ \therefore \frac{\lambda_{p}}{\lambda_{e}}=\frac{P_{e}}{P_{p}}=\frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}=\left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}} \end{array}$

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

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