1. Use equality of de-Broglie wavelengths

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

Given that the electron and photon have the same de-Broglie wavelength, $\lambda_e = \lambda_{ph}$ so, $\frac{h}{p_e} = \frac{h}{p_{ph}} \implies p_e = p_{ph}$

Hence, $\frac{p_e}{p_{ph}} = 1$

So options C and D are incorrect.

2. Kinetic energy of the electron

For an electron moving with speed $v$ (non-relativistic form intended here), $E_e = \frac{1}{2}mv^2$

Also its momentum is $p_e = mv$

So, $E_e = \frac{1}{2}p_ev$

3. Energy of the photon

For a photon, $E_{ph} = p_{ph}c$

4. Compare energies

Since $p_e = p_{ph}$, $\frac{E_e}{E_{ph}} = \frac{\frac{1}{2}p_ev}{p_{ph}c} = \frac{1}{2}\cdot \frac{v}{c}\cdot \frac{p_e}{p_{ph}}$

But $p_e = p_{ph}$, so $\frac{E_e}{E_{ph}} = \frac{v}{2c}$

Thus the correct option is: $\boxed{\frac{E_e}{E_{ph}} = \frac{v}{2c}}$

So option B is correct.

5. Check all options

• A: $\dfrac{E_e}{E_{ph}} = \dfrac{2c}{v}$ ❌
• B: $\dfrac{E_e}{E_{ph}} = \dfrac{v}{2c}$ ✅
• C: $\dfrac{p_e}{p_{ph}} = \dfrac{2c}{v}$ ❌ since $\dfrac{p_e}{p_{ph}}=1$
• D: $\dfrac{p_e}{p_{ph}} = \dfrac{v}{2c}$ ❌ since $\dfrac{p_e}{p_{ph}}=1$

Therefore, the correct answer is B.