An electron moving with speed $$v$$ and a photon moving with speed $$c$$, have the same D-Broglie wavelength. The ratio of the kinetic energy of the electron to that of a photon is:

We begin with the de-Broglie relation, which connects wavelength $$\lambda$$ with linear momentum $$p$$:

$$\lambda \;=\;\frac{h}{p}$$

Here $$h$$ is Planck’s constant. Because the electron and the photon are said to have the same de-Broglie wavelength, we can write

$$\lambda_{\text{electron}} \;=\;\lambda_{\text{photon}} \;=\;\lambda$$

Substituting the expression for each momentum in the de-Broglie formula, we obtain

For the electron (non-relativistic): $$p_{\text{e}} = m_{\text{e}}\,v$$, so $$\lambda = \dfrac{h}{m_{\text{e}}\,v}$$

For the photon: $$p_{\gamma} = \dfrac{h}{\lambda}$$ by direct inversion of the same formula.

Since the wavelengths are equal, we must have equal momenta:

$$p_{\text{e}} = p_{\gamma}$$

Therefore

$$m_{\text{e}}\,v = \frac{h}{\lambda}$$

but the right-hand side is simply $$p_{\gamma}$$, so we can keep the concise equality

$$p_{\gamma} = m_{\text{e}}\,v$$

Now we write the kinetic energy of each particle.

For the electron (classical, non-relativistic):

$$K_{\text{e}} \;=\;\frac{1}{2}\,m_{\text{e}}\,v^{2}$$

For the photon, the entire energy is kinetic and is given by the relativistic expression $$E = pc$$:

$$K_{\gamma} \;=\;p_{\gamma}\,c$$

We now form the required ratio:

$$\frac{K_{\text{e}}}{K_{\gamma}} \;=\;\frac{\dfrac{1}{2}\,m_{\text{e}}\,v^{2}}{p_{\gamma}\,c}$$

Substituting $$p_{\gamma} = m_{\text{e}}\,v$$ from the equality of momenta, we get

$$\frac{K_{\text{e}}}{K_{\gamma}} \;=\;\frac{\dfrac{1}{2}\,m_{\text{e}}\,v^{2}}{m_{\text{e}}\,v\,c}$$

Now the mass $$m_{\text{e}}$$ cancels out, leaving

$$\frac{K_{\text{e}}}{K_{\gamma}} \;=\;\frac{1}{2}\,\frac{v}{c}$$

Simplifying, we obtain

$$\frac{K_{\text{e}}}{K_{\gamma}} \;=\;\frac{v}{2c}$$

Hence, the correct answer is Option C.