What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

Kinetic energy of the electron, Ek = 120 eV

Planck’s constant, h = 6.6 x 10⁻³⁴Js

Mass of an electron, m = 9.1 x 10⁻³¹ kg

Charge on an electron, e = 1.6 x 10⁻¹⁹ C
Ans (a).

For the electron, we can write the relation for kinetic energy as:

Ek = 1/2 mv²

Where, v = Speed of the electron

Therefore, v =√(2Ek/m) =√( 2 x 120 x 1.6 x 10 x 120 )/(9.1 x 10⁻³¹ )

=√ (42.198 x 10¹²)= 6.496 x 10⁶ m/s

Momentum of the electron, p = mv = 9.1 x 10⁻³¹ x 6.496 x 10⁶

=5.91  x 10²⁴ kg ms⁻¹

Therefore, the momentum of the electron is 5.91 x 10^-24 kg m s^-1.

Ans (b).

Speed of the electron, v = 6.496 x 10⁶ m/s

Ans (c).

De Broglie wavelength of an electron having a momentum p, is given as:

λ = h/p  = ( 6.6 x 10⁻³⁴) /( 5.91 x 10^-24) = 1.116 x 10⁻¹⁰m

= 0.112 nm

Therefore, the de Broglie wavelength of the electron is 0.112 nm.