Question

Calculate:

1. the momentum and
2. de Broglie wavelength,

of an electron with kinetic energy of 80 eV.

Answer 1

Given: K = 80 eV = 128 × 10⁻¹⁹ J

i. Using relation:

P = `sqrt(2 Km)`

Where m = mass of electron = 9.1 × 10³¹ kg

Substituting the values:

p = `sqrt(2 xx (9 xx 10^-31 kg) xx (1.28 xx 10^-17 J)`

Calculating inside the square root:

p = `sqrt(2 xx 9 xx 1.28 xx 10^-48)`

= `sqrt(23.04 xx 10^-48)`

= 4.8 × 10⁻¹⁹ kg m/s

The de Broglie wavelength λ is given by the formula:

λ = `h/p`

Where is Planck’s constant = 6.6 × 10⁻³⁴ J s

Substituting the values:

λ = `(6.6 xx 10^-34 J s)/(4.8 xx 10^-24 kg m//s)`

Calculating the wavelength:

λ = 1.375 × 10⁻¹⁰ m

= 13 Å