Calculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Potential difference, V = 56 V

Planck’s constant, h = 6.6 x 10⁻³⁴ Js

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

Charge on an electron, e = 1.6 x 10⁻¹⁹ C

Ans (a).
At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as:

-1/2mv²=eV

v² =2eV/m

Therefore ,v = √ (2 x 1.6 x10⁻¹⁹ x 56)/ (9.1 x 10⁻³¹)
= √(19.69 x 10¹² )=4.44 x 10⁶ m/s

The momentum of each accelerated electron is given as:

p = mv = 9.1 x 10^-31 x 4.44 x 10⁶ = 4.04 x 10^-24 kg m s⁻¹ Therefore, the momentum of each electron is 4.04 x 10^-24 kg m s^-1.

Ans (b).

De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:

λ = 12.27 Aº/√V

= 12.27 x 10⁻¹⁰ /√56
= 0.1639 nm

Therefore, the de Broglie wavelength of each electron is 0.1639 nm.