JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Modern Physics

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:

(A) $\frac{1}{1}$
(B) $\frac{1}{8}$ ✓ Correct
(C) $\frac{8}{1}$
(D) $\frac{1}{4}$

Answer: (B)

Solution

For the photon,

λ_{p h} = \frac{h c}{E _{p h}}

. For the electron,

λ_{e} = \frac{h}{m _{e} v _{e}} = \frac{h v _{e}}{2 K _{e}}

(using

K_{e} = \frac{1}{2} m_{e} v_{e}^{2}

).

Equal wavelengths (

λ_{e} = λ_{p h}

) with

v_{e} = 0.25 c

\frac{h v _{e}}{2 K _{e}} = \frac{h c}{E _{p h}} ⟹ \frac{K _{e}}{E _{p h}} = \frac{v _{e}}{2 c} = \frac{0.25 c}{2 c} = \frac{1}{8}

Key confirmed: B.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option

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