We know that the de-Broglie wavelength $$\lambda$$ of a particle is given by:

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

where:

• $$h$$ is Planck's constant,
• $$p$$ is the momentum of the particle.

For a photon (which has zero rest mass), its energy $$E$$ and momentum $$p$$ are related by the equation:

$$E = cp$$

and its de-Broglie wavelength $$\lambda$$ is given by:

$$\lambda = \frac{h}{E} \times \frac{1}{c}$$

Now, since the photon and electron are said to have the same de-Broglie wavelength:

$$\lambda_{electron} = \lambda_{photon}$$

$$\frac{h}{p_{electron}} = \frac{h}{E_{photon}} \times \frac{1}{c}$$

For the electron, its momentum $$p_{electron}$$ is given by:

$$p_{electron} = m_{e}v_{electron}$$

where:

• $$m_{e}$$ is the electron's rest mass,
• $$v_{electron}$$ is the electron's velocity.

The kinetic energy $$K.E.$$ of the electron is:

$$K.E._{electron} = \frac{1}{2}m_{e}v_{electron}^2$$

The question states that $$v_{electron}$$ is $$25\%$$ ($$0.25c$$) of the speed of light $$c$$. So we write:

$$v_{electron} = 0.25c$$

Plugging this into the kinetic energy formula, we get:

$$K.E._{electron} = \frac{1}{2}m_{e}(0.25c)^2 = \frac{1}{2}m_{e} \times \frac{1}{16}c^2$$

$$K.E._{electron} = \frac{1}{32}m_{e}c^2$$

For a photon, $$p_{photon} = \frac{E_{photon}}{c}$$ and hence its kinetic energy (which, for a photon, is simply its energy) is:

$$K.E._{photon} = E_{photon} = cp_{photon}$$

Now, comparing the kinetic energies:

$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{cp_{photon}}$$

Since $$p_{electron} = p_{photon}$$ (from the de-Broglie relation), we can replace $$p_{photon}$$ with $$p_{electron}$$ which is $$m_{e}v_{electron}$$:

$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}m_{e}c^2}{m_{e}v_{electron}c}$$

$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{\frac{1}{32}c}{0.25c}$$

$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times \frac{1}{0.25}$$

$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{32} \times 4$$

$$\frac{K.E._{electron}}{K.E._{photon}} = \frac{1}{8}$$

So the correct answer is Option C $$\frac{1}{8}$$.