The de-Broglie wavelength of an electron moving with a velocity c3(c=3×108 m/s) is equal to the wavelength of photon. The ratio of the kinetic energies of electron and photon is

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 an electron moving with velocity v, its momentum p is given by:

p = m * v

where m is the mass of the electron. Substituting this into the de-Broglie wavelength formula gives:

λ = h / (m * v)

For a photon, its wavelength is related to its energy E by the formula:

E = h * c / λ

where c is the speed of light.

Given that the de-Broglie wavelength of the electron is equal to the wavelength of the photon, we can equate the two expressions for λ to get:

h / (m * v) = h * c / E

Solving for E gives:

E = m * v * c

The kinetic energy of the electron is given by:

K = 1/2 * m * v^2

The ratio of the kinetic energy of the electron to the energy of the photon is therefore:

K / E = (1/2 * m * v^2) / (m * v * c) = v / (2c)

Given that v = c/3, the ratio K / E is therefore:

K / E = (c/3) / (2c) = 1/6

So the ratio of the kinetic energies of the electron and the photon is 1:6.