Calculations with the de Broglie equation

Overview

Recall questions applying λ = h/mv to an electron, a proton, a neutron, a cricket ball and a walking person, working back from a wavelength to a speed or a momentum, the derivation of λ = h/√(2meV) for an accelerated electron, and why about 100 V is a suitable accelerating voltage for a demonstration.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   Calculate the de Broglie wavelength of an electron travelling at 1.0 × 10⁷ m s⁻¹.

   Model answer

   λ = h/mv = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 1.0 × 10⁷) = 7.3 × 10⁻¹¹ m, and 6.63 × 10⁻³⁴ ÷ 7.28 × 10⁻¹¹ returns a momentum of 9.11 × 10⁻²⁴ kg m s⁻¹ as required.

2. 2

   An electron is accelerated from rest through a potential difference of 100 V. Calculate its de Broglie wavelength.

   Model answer

   Its kinetic energy is eV = 1.60 × 10⁻¹⁷ J, so v = √(2 × 1.60 × 10⁻¹⁷ / 9.11 × 10⁻³¹) = 5.93 × 10⁶ m s⁻¹ and λ = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 5.93 × 10⁶) = 1.2 × 10⁻¹⁰ m, and ½ × 9.11 × 10⁻³¹ × (5.93 × 10⁶)² returns 1.60 × 10⁻¹⁷ J.

3. 3

   An electron is accelerated from rest through 2500 V. Calculate its de Broglie wavelength.

   Model answer

   Its kinetic energy is 1.60 × 10⁻¹⁹ × 2500 = 4.0 × 10⁻¹⁶ J, so v = √(2 × 4.0 × 10⁻¹⁶ / 9.11 × 10⁻³¹) = 2.96 × 10⁷ m s⁻¹ and λ = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 2.96 × 10⁷) = 2.5 × 10⁻¹¹ m, and ½ × 9.11 × 10⁻³¹ × (2.96 × 10⁷)² returns 4.0 × 10⁻¹⁶ J.

4. 4

   Calculate the de Broglie wavelength of a proton travelling at 2.0 × 10⁶ m s⁻¹.

   Model answer

   λ = 6.63 × 10⁻³⁴ / (1.67 × 10⁻²⁷ × 2.0 × 10⁶) = 2.0 × 10⁻¹³ m, and 6.63 × 10⁻³⁴ ÷ 1.99 × 10⁻¹³ returns a momentum of 3.34 × 10⁻²¹ kg m s⁻¹.

5. 5

   A neutron of mass 1.67 × 10⁻²⁷ kg moves at 3.0 × 10³ m s⁻¹. Calculate its de Broglie wavelength and comment on whether it could be diffracted by a crystal.

   Model answer

   λ = 6.63 × 10⁻³⁴ / (1.67 × 10⁻²⁷ × 3.0 × 10³) = 1.3 × 10⁻¹⁰ m, which is comparable with the spacing of atoms in a crystal, so such neutrons are diffracted strongly, and 6.63 × 10⁻³⁴ ÷ 1.32 × 10⁻¹⁰ returns a momentum of 5.0 × 10⁻²⁴ kg m s⁻¹.

6. 6

   Calculate the de Broglie wavelength of a cricket ball of mass 0.16 kg moving at 30 m s⁻¹.

   Model answer

   λ = 6.63 × 10⁻³⁴ / (0.16 × 30) = 6.63 × 10⁻³⁴ / 4.8 = 1.4 × 10⁻³⁴ m, which is vastly smaller than an atomic nucleus, so no wave behaviour could ever be detected.

Other ways to work this

The same content, four other ways.

What is inside Levo

Six ways to work the same material.