JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Modern Physics

WhatsApp Email

The de - Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V . What should be the ratio of their wavelengths?

(m_{p} = 1.00727 u, m_{e} = 0.00055 u)

(A) $43 : 1$ ✓ Correct
(B) $41.1 : 1$
(C) $(1860)^{2} : 1$
(D) $1860 : 1$

Answer: (A)

Solution

A particle of charge

q

and mass

m

accelerated from rest through a potential difference

V

acquires momentum

p = 2 m q V

, so its de Broglie wavelength is

λ = \frac{h}{p} = \frac{h}{2 m q V}

The electron and the proton carry charges of equal magnitude and were accelerated through the same

100 V

, so only the masses distinguish them:

\frac{λ _{e}}{λ _{p}} = \frac{m _{p}}{m _{e}} = \frac{1.00727}{0.00055} = 1831 = 42.8

The wavelengths of the electron and the proton are in the ratio

43 : 1

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Modern Physics →

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Modern Physics

Solution

More JEE Main questions on Modern Physics