An electron with speed $$v$$ and a photon with speed $$c$$ have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are $$E_e$$ and $$P_e$$ and that of photon are $$E_{ph}$$ and $$P_{ph}$$ respectively. Which of the following is correct?

An electron with speed $$v$$ and a photon with speed $$c$$ have the same de-Broglie wavelength. The de-Broglie wavelength is given by $$\lambda = \frac{h}{p}$$, so equating $$\lambda_e$$ and $$\lambda_{ph}$$ leads to

$$\frac{h}{P_e} = \frac{h}{P_{ph}} \quad\Longrightarrow\quad P_e = P_{ph}\,.$$

The kinetic energy of the electron can be written as $$E_e = \frac12 mv^2 = \frac{P_e v}{2}$$ (using $$P_e = mv$$), and the energy of the photon is $$E_{ph} = P_{ph}\cdot c\,. $$

Hence, the ratio of energies becomes

$$\frac{E_e}{E_{ph}} = \frac{\frac{P_e v}{2}}{P_{ph}\cdot c} = \frac{v}{2c}\,, $$

where we have used $$P_e = P_{ph}$$. Therefore, the correct answer is Option B.