Q156 / 180 · COMEDK 2018 · Physics

If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is

  • A
    Zero
  • B
    less than that of a proton
  • Correct answer
    C
    more than that of a proton
  • D
    equal to that of a proton

Solution

Given, $\quad \lambda_{e}=\lambda_{p}$ Since, de-Broglie wavelength, $$ \begin{aligned} \lambda &=\frac{h}{m v}=\frac{h}{p}=\frac{h}{\sqrt{2 m K}} \\ \Rightarrow \quad \lambda &=\frac{h}{\sqrt{2 m K}} \end{aligned} $$ where, $K$ is kinetic energy. $\begin{array}{ll}\therefore \quad \frac{\lambda_{e}}{\lambda_{p}} & =\sqrt{\frac{m_{p}}{m_{e}}} \cdot \sqrt{\frac{K_{p}}{K_{e}}} \\ \Rightarrow \quad \quad \frac{\lambda_{e}}{\lambda_{e}} & =\sqrt{\frac{m_{p}}{m_{e}}} \cdot \sqrt{\frac{K_{p}}{K_{e}}} \\ \Rightarrow \quad \text { [From Eq. (i)] }\end{array}$ $\Rightarrow \quad \sqrt{\frac{K_{e}}{K_{p}}}=\sqrt{\frac{m_{p}}{m_{e}}}$ Squaring both sides, we get $$ \begin{array}{ll} & \frac{K_{e}}{K_{p}}=\frac{m_{p}}{m_{e}} \\ \text { since } & m_{p}>m_{e} \\ \therefore & K_{e}>K_{p} \end{array} $$

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

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