JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Atomic Structure

The wavelength of electrons accelerated from rest through a potential difference of 40 kV is

x \times 1 0^{- 12} m

. The value of

x

is ______. (Nearest Integer)
Given: Mass of electron

= 9.1 \times 1 0^{- 31} kg

Charge on an electron

= 1.6 \times 1 0^{- 19} C

Planck's constant

= 6.63 \times 1 0^{- 34} Js

Answer: 6

Solution

An electron accelerated from rest through a potential difference

V

gains kinetic energy

e V

, so its momentum is

p = 2 m e V

and the de Broglie wavelength is

λ = \frac{h}{2 m e V}

With

V = 4.0 \times 1 0^{4}

2 m e V = 2 (9.1 \times 1 0^{- 31}) (1.6 \times 1 0^{- 19}) (4.0 \times 1 0^{4}) = 1.165 \times 1 0^{- 44}

2 m e V = 1.079 \times 1 0^{- 22}

kg m s

^{- 1}

and

λ = \frac{6.63 \times 1 0 ^{- 34}}{1.079 \times 1 0 ^{- 22}} = 6.14 \times 1 0^{- 12} m

To the nearest integer,

x = 6

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer