The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $$25\%$$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

We need to find the ratio of kinetic energies of an electron and a photon that have the same de Broglie wavelength, given that the electron's velocity is 25% of the speed of light.

For the electron (using classical kinetic energy since $$v = 0.25c$$):

$$KE_e = \frac{1}{2}m_e v_e^2$$

We can also write this as: $$KE_e = \frac{p_e v_e}{2}$$ (since $$p_e = m_e v_e$$).

For the photon:

$$KE_{ph} = E_{ph} = pc = \frac{h}{\lambda} \cdot c = \frac{hc}{\lambda}$$

Since both have the same de Broglie wavelength $$\lambda$$:

For the electron: $$\lambda = \frac{h}{p_e}$$, so $$p_e = \frac{h}{\lambda}$$

For the photon: $$\lambda = \frac{h}{p_{ph}}$$, so $$p_{ph} = \frac{h}{\lambda}$$

Therefore, $$p_e = p_{ph}$$ (equal momenta).

$$\frac{KE_e}{KE_{ph}} = \frac{p_e v_e / 2}{p_{ph} \cdot c}$$

Since $$p_e = p_{ph}$$, these cancel:

$$\frac{KE_e}{KE_{ph}} = \frac{v_e}{2c} = \frac{0.25c}{2c} = \frac{1}{8}$$

The correct answer is Option (2): $$\frac{1}{8}$$.