JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Modern Physics

An electron of mass

m_{e}

and a proton of mass

m_{p}

are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is :

(A) $\frac{m _{p}}{m _{e}}$
(B)1
(C) $\frac{m _{e}}{m _{p}}$
(D) $\frac{m _{p}}{m _{e}}$ ✓ Correct

Answer: (D)

Solution

A particle of charge

q

accelerated through a potential difference

V

acquires kinetic energy

q V

, so its momentum is

p = 2 m q V

and

λ = \frac{h}{p} = \frac{h}{2 m q V}

The electron and the proton carry charges of equal magnitude and cross the same potential difference, so

λ \propto \frac{1}{m}

and

\frac{λ _{e}}{λ _{p}} = \frac{m _{p}}{m _{e}}

The electron wavelength is

\frac{m _{p}}{m _{e}}

times the proton wavelength.

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option

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