Q107 / 180 · JIPMER 2017 · Physics

The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. Find the ratio of energy of photon to the kinetic energy of electron in terms of mass $m$, speed of light $c$ and planck constant.

  • A
    $\frac{\lambda m c}{h}$
  • B
    $\frac{h m c}{\lambda}$
  • C
    $\frac{2 h m c}{\lambda}$
  • Correct answer
    D
    $\frac{2 \lambda m c}{h}$

Solution

The de-Broglie wavelength $h=\frac{h}{m v} \Rightarrow v=\frac{h}{m \lambda}...(i)$ $E_p=\frac{h c}{\lambda}$ (since $\lambda$ is same $)...(ii)$ Kinetic energy of photon $=\frac{E_p}{E_e}=\frac{h c / \lambda}{\frac{1}{2} 1 m u^2}=\frac{2 h c}{\lambda m v^2}$ Substituting value of $v$ from Eq. (i), we get $\frac{E_p}{E_e}=\frac{2 h c}{\lambda_m\left(\frac{h}{m \lambda}\right)^2}=\frac{2 \lambda m c}{h}$

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

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