JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Modern Physics

An electron of mass

m

and a photon have same energy

E

. The ratio of wavelength of electron to that of photon is: (c being the velocity of light)

(A) $\frac{1}{c} (\frac{E}{2 m})^{1/2}$ ✓ Correct
(B) $(\frac{E}{2 m})^{1/2}$
(C) $c (2 m E)^{1/2}$
(D) $\frac{1}{c} (\frac{2 m}{E})^{1/2}$

Answer: (A)

Solution

A photon of energy

E

has

λ_{p} = \frac{h c}{E}

.
The electron carries

E

as kinetic energy, so its momentum is

p = 2 m E

and its de Broglie wavelength is

λ_{e} = \frac{h}{2 m E}

.
Dividing,

\frac{λ _{e}}{λ _{p}} = \frac{h}{2 m E} \cdot \frac{E}{h c} = \frac{E}{c 2 m} = \frac{1}{c} \frac{E}{2 m}

The ratio of the electron wavelength to the photon wavelength is

\frac{1}{c} (\frac{E}{2 m})^{1/2}

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option

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