Q133 / 160 · WBJEE 2012 · Physics

The de-Broglie wavelength of an electron moving with a velocity $c / 2$ ( $c=$ velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is

  • A
    $1: 4$
  • Correct answer
    B
    $1: 2$
  • C
    $1: 1$
  • D
    $2: 1$

Solution

de-Broglie wavelength $$ \lambda=\frac{h}{m v} $$ Here, $\lambda_{e}=\frac{h}{m_{e} \frac{c}{2}}$ and $\lambda_{p}=\frac{h}{m_{p} c}$ Given, $\quad \lambda_{e}=\lambda_{B}$ So, $\quad \frac{h}{m_{e} \frac{c}{2}}=\frac{h}{m_{p} c}$ $\frac{m_{e}}{m_{p}}=2$ Ratio of KE $$ \begin{array}{l} \frac{K_{e}}{K_{p}}=\frac{\frac{1}{2} m_{e} v_{e}^{2}}{\frac{1}{2} m_{p} v_{p}^{2}} \\ \frac{K_{e}}{K_{p}}=\frac{1}{2} \end{array} $$

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

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