Let E_e and E_p represent kinetic energy of electron and photon...

The kinetic energy of a photon is given by:

$E_p = \dfrac{hc}{\lambda_p}$

The kinetic energy of an electron can be written in terms of its de-Broglie wavelength $\lambda_e$ and velocity $v$:

$E_e = \dfrac{1}{2} m v^2 = \dfrac{1}{2} (mv) v = \dfrac{1}{2} \left(\dfrac{h}{\lambda_e}\right) v$

Taking the ratio of the kinetic energies:

$\dfrac{E_p}{E_e} = \dfrac{\dfrac{hc}{\lambda_p}}{\dfrac{hv}{2\lambda_e}} = \dfrac{2c}{v} \times \dfrac{\lambda_e}{\lambda_p}$

Given that $v = \dfrac{c}{100}$ and $\lambda_p = 4\lambda_e$, we substitute these values into the ratio:

$\dfrac{E_p}{E_e} = \dfrac{2c}{\left(\dfrac{c}{100}\right)} \times \dfrac{\lambda_e}{4\lambda_e}$

$\dfrac{E_p}{E_e} = 200 \times \dfrac{1}{4} = 50$

Answer: $50$