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:

The de-Broglie wavelength of a particle is given by the formula:

λ = h / p

where h is Planck's constant and p is the momentum of the particle. For a photon, the momentum is given by p = E / c, where E is the energy of the photon and c is the speed of light. Therefore, the wavelength of a photon is given by:

λ = h c / E

For an electron, the momentum is given by p = m v, where m is the mass of the electron and v is its velocity. Therefore, the wavelength of an electron is given by:

λ = h / (m v)

Given that the wavelengths of the electron and the photon are the same, we can equate the two expressions for λ to get:

h c / E = h / (m v)

Solving for E, we get:

E = h c m v / h = m v c

The kinetic energy (K.E.) of a particle is given by K.E. = 1/2 m v^2. Therefore, the K.E. of the electron is:

K.E.(electron) = 1/2 m (0.25c)^2 = 0.0625 m c^2

The K.E. of the photon is simply its energy E, which we found to be m v c = m (0.25c) c = 0.25 m c^2.

Therefore, the ratio of the K.E. of the electron to the K.E. of the photon is:

K.E.(electron) / K.E.(photon) = 0.0625 m c^2 / 0.25 m c^2 = 0.25

So, the ratio of the kinetic energy of the electron to the kinetic energy of the photon is 0.25.