1. Use equality of de-Broglie wavelengths

For any particle/wave, $\lambda = \frac{h}{p}$

Given that the electron and the photon have the same wavelength, their momenta must be equal: $p_e = p_\gamma$

2. Momentum of the electron

For a non-relativistic electron moving with speed $v$, $p_e = mv$

So, $p_\gamma = mv$

3. Kinetic energy of the electron

$K_e = \frac{1}{2}mv^2$

Using $p_e = mv$, $K_e = \frac{1}{2}pv$

4. Kinetic energy of the photon

A photon has no rest mass, and its energy is $E_\gamma = p_\gamma c$

This is the energy associated with the photon. In such questions, this is taken as the photon's kinetic energy: $K_\gamma = pc$

5. Take the ratio

Since $p_e = p_\gamma = p$, $\frac{K_e}{K_\gamma} = \frac{\frac{1}{2}pv}{pc} = \frac{v}{2c}$

6. Match with options

$\frac{K_e}{K_\gamma} = \frac{v}{2c}$

So the correct option is: C: $\dfrac{v}{2c}$