"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:"

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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 $\frac{1}{8}$.

"

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:

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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:

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