Q100 / 120 · VITEEE 2011 · Physics

An electron accelerated under a potential difference $\mathrm{V}$ volt has a certain wavelength $\lambda$. Mass of proton is some 2000 times of the mass of the electron. If the proton has to have the same wavelength $\lambda$, then it will have to be accelerated under a potential difference of

  • A
    $\mathrm{V}$ volt
  • B
    $2000 \mathrm{~V}$ volt
  • Correct answer
    C
    $\frac{\mathrm{V}}{2000}$ volt
  • D
    $\sqrt{2000} \mathrm{~V}$ volt

Solution

According to question $\lambda_{e}=\lambda_{P}$ $\begin{aligned} & \text { or } \frac{h}{\sqrt{2 m_{e} Q_{e} V}}=\frac{h}{\sqrt{2 m_{p} Q_{p} V_{p}}} \\ \therefore \quad & m_{e} Q_{e} V=m_{p} Q_{p} V_{p} \\ \therefore \quad & V_{p}=\left(\frac{m_{e}}{m_{p}}\right)\left(\frac{Q_{e}}{Q_{p}}\right) V \\ &=\left(\frac{1}{2000}\right)(1) V \\ &=\frac{V}{2000} v o l t \end{aligned}$

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

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