An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of electron

Wavelength of an electron (λ_e) and a photon (λ_p), λ_e = λ_p = 1 nm = 1 x 10⁻⁹ m

Planck’s constant, h = 6.63 x 10^-34 Js

Ans (a).

The momentum of an elementary particle is given by de Broglie relation:

λ = h/p   => p = h/λ

It is clear that momentum depends only on the wavelength of the particle. Since the wavelength of an electron and photon are equal, both have an equal momentum.

Therefore , p = 6.63 x 10⁻³⁴ / ( 1 x 10⁻⁹)

= 6.63 x 10⁻²⁵ kg ms⁻¹

Ans (b).

The energy of a photon is given by the relation:

E = hc/λ

Where, Speed of light, c = 3 x 10⁸ m/s ,

Therefore, E = (6.63 x 10⁻³⁴ ) x (3 x 10⁸ /(1 x 10⁻⁹) (1.6 x 10⁻¹⁹ )

=1243.1 eV = 1.243 kev

Therefore, the energy of the photon is 1.243 keV.

Ans (c).
The kinetic energy (K) of an electron having momentum p, is given by the relation:

K = 1/2 p²/m

Where,

m = Mass of the electron = 9.1 x 10^-31 kg

p = 6.63 x 10^-25 kg m s^-1

Therefore , K = 1/2  ( 6.63 x 10^-25 )² /( 9.1 x 10^-31)  = 2.415 x 10⁻¹⁹ J

= (2.415 x 10⁻¹⁹) /(1.6 x 10⁻¹⁹)  = 1.51 eV

Hence, the kinetic energy of the electron is 1.51 eV.