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:

Dual Nature of Radiation PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q62 with Solution

View written solutionFree

Correct answer: C

For any particle,

\lambda = \frac{h}{p}

If the electron and photon have the same wavelength, then their momenta are equal:

p_e = p_\gamma

Given electron speed:

v = 25\%\, c = \frac{c}{4}

For this standard JEE-level question, we use non-relativistic momentum and kinetic energy for the electron:

p_e = mv

K_e = \frac{1}{2}mv^2

For a photon,

E_\gamma = p_\gamma c

Since a photon has no rest mass, its kinetic energy is its total energy:

K_\gamma = p_\gamma c

Let the common momentum be $p$ . Then for the electron,

p = mv Rightarrow m = \frac{p}{v}

So,

K_e = \frac{1}{2}mv^2 = \frac{1}{2}\left(\frac{p}{v}\right)v^2 = \frac{1}{2}pv

For the photon,

K_\gamma = pc

Therefore,

\frac{K_e}{K_\gamma} = \frac{\frac{1}{2}pv}{pc} = \frac{v}{2c}

Now substitute $v = \frac{c}{4}$ :

\frac{K_e}{K_\gamma} = \frac{1}{2}\cdot\frac{1}{4} = \frac{1}{8}

\frac{\text{K.E. of electron}}{\text{K.E. of photon}} = \frac{1}{8}

So the correct option is C.