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 of the kinetic energy of electron in terms of mass $m$, speed of light $c$ and Planck’s constant.
A.$\dfrac{\lambda mc}{h}$
B.$\dfrac{hmc}{\lambda }$
C.$\dfrac{2hmc}{\lambda }$
D.$\dfrac{2\lambda mc}{h}$

Hint: The de-Broglie wavelength of the particle is given by the equation,
$\lambda =\dfrac{h}{mv}$
Which can be rearranged as,
$v=\dfrac{h}{m\lambda }$
And also energy of a photon is given by the equation,
${{E}_{P}}=\dfrac{hc}{\lambda }$

Complete step by step answer:
We know that electrons also have dual nature characteristics in which it can exhibit both particle nature as well as wave nature. The electron wave is having a wavelength $\lambda $ . This wavelength depends on how much the electron is carrying. This gives rise to the formula for the de-Broglie wavelength and it is generally called as de-Broglie relation and $\lambda $is known as the de-Broglie wavelength of an electron.
Here as we mentioned above
The de-Broglie wavelength of the particle is given by the equation,
$\lambda =\dfrac{h}{mv}$
Where $\lambda $is the de-Broglie wavelength, $m$is the mass of the electron and $c$ is the velocity of light. Which can be rearranged as,
$v=\dfrac{h}{m\lambda }$
Now energy of a photon is given by the equation,
${{E}_{P}}=\dfrac{hc}{\lambda }$
Now let us take the ratio of the energy of the electron to the kinetic energy of the electron.
Kinetic energy of electron with velocity u is
${{E}_{e}}=\dfrac{1}{2}m{{u}^{2}}$
Therefore the ratio will be,
$\dfrac{{{E}_{p}}}{{{E}_{e}}}=\dfrac{\dfrac{hc}{\lambda }}{\dfrac{1}{2}m{{u}^{2}}}=\dfrac{2hc}{\lambda m{{u}^{\begin{smallmatrix}
 2 \\

\end{smallmatrix}}}}$
Substituting the value v in this equation,
$v=u$
$\dfrac{{{E}_{p}}}{{{E}_{e}}}==\dfrac{2hc}{\lambda m{{\left( \dfrac{h}{m\lambda } \right)}^{2}}}=\dfrac{2\lambda mc}{h}$

Therefore the correct answer is option D.


Note:
An electron is having a wavelength in a standing wave inside an electron. Every electron is having a de-Broglie wavelength incorporated with it, which is related to its momentum. A resting electron in the reference frame has a momentum of zero and a frequency associated with its mass. So at rest electrons are actually a standing wave.