8.2 Wave-particle duality — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Keep the mass in kilograms, the speed in metres per second and the wavelength in metres.

Warm-up: remember

A ball of mass 0.20 kg moves at $3.0\,\text{m}\,\text{s}^{-1}$ . Find its momentum.
Find the energy of a photon of wavelength 500 nm.
An electron is accelerated through $200\,\text{V}$ . Find its kinetic energy in joules.

Answers

$p = mv = 0.20 \times 3.0 = 0.60\,\text{kg}\,\text{m}\,\text{s}^{-1}$ .
$E = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{5.00 \times 10^{-7}} = 3.98 \times 10^{-19}\,\text{J}$ .
$E_k = eV = 1.60 \times 10^{-19} \times 200 = 3.2 \times 10^{-17}\,\text{J}$ .

If question 1 goes wrong, revisit the definition of momentum in Section 3.4. If question 2 goes wrong, use $E = \dfrac{hc}{\lambda}$ from Section 8.1. If question 3 goes wrong, the kinetic energy is the charge multiplied by the potential difference.

Practise with help

1. An electron moves at $2.0 \times 10^{6}\,\text{m}\,\text{s}^{-1}$ . Find its de Broglie wavelength. The mass of an electron is $9.11 \times 10^{-31}\,\text{kg}$ .

Hint

Find the momentum first, then divide the Planck constant by it.

Next step

Write $p = mv = 9.11 \times 10^{-31} \times 2.0 \times 10^{6}$ , then use $\lambda = \dfrac{h}{p}$ .

Full working

\begin{aligned} p &= mv = 9.11 \times 10^{-31} \times 2.0 \times 10^{6} = 1.82 \times 10^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1} \\ \lambda &= \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.82 \times 10^{-24}} = 3.6 \times 10^{-10}\,\text{m} \end{aligned}

The de Broglie wavelength is $3.6 \times 10^{-10}\,\text{m}$ .

2. Find the de Broglie wavelength of a proton moving at $1.0 \times 10^{6}\,\text{m}\,\text{s}^{-1}$ . The mass of a proton is $1.67 \times 10^{-27}\,\text{kg}$ .

Hint

The proton is much heavier than the electron, so expect a much shorter wavelength.

Next step

Use the same two steps as question 1, but with the proton mass.

Full working

\begin{aligned} p &= 1.67 \times 10^{-27} \times 1.0 \times 10^{6} = 1.67 \times 10^{-21}\,\text{kg}\,\text{m}\,\text{s}^{-1} \\ \lambda &= \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-21}} = 4.0 \times 10^{-13}\,\text{m} \end{aligned}

The proton's wavelength is $4.0 \times 10^{-13}\,\text{m}$ .

3. An electron has a de Broglie wavelength of $1.0 \times 10^{-10}\,\text{m}$ . Find its momentum.

Hint

Rearrange the de Broglie equation to make the momentum the subject.

Next step

Write $p = \dfrac{h}{\lambda}$ and substitute the two values.

Full working

p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{1.0 \times 10^{-10}} = 6.6 \times 10^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1}

The momentum is $6.6 \times 10^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1}$ .

4. An electron is accelerated through $500\,\text{V}$ . Find its de Broglie wavelength.

Hint

Find the kinetic energy in joules, then the momentum, then the wavelength.

Next step

Write $E_k = 1.60 \times 10^{-19} \times 500$ , then $p = \sqrt{2mE_k}$ .

Full working

\begin{aligned} E_k &= 1.60 \times 10^{-19} \times 500 = 8.0 \times 10^{-17}\,\text{J} \\ p &= \sqrt{2 \times 9.11 \times 10^{-31} \times 8.0 \times 10^{-17}} = 1.21 \times 10^{-23}\,\text{kg}\,\text{m}\,\text{s}^{-1} \\ \lambda &= \frac{6.63 \times 10^{-34}}{1.21 \times 10^{-23}} = 5.5 \times 10^{-11}\,\text{m} \end{aligned}

The wavelength is $5.5 \times 10^{-11}\,\text{m}$ .

5. Find the de Broglie wavelength of a 0.050 kg ball moving at $20\,\text{m}\,\text{s}^{-1}$ , and explain why this wavelength is never observed.

Hint

Use the same equation, with the ball's mass and speed.

Next step

Find the momentum, then divide the Planck constant by it. Compare the wavelength with the size of any gap you could make.

Full working

\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{0.050 \times 20} = 6.6 \times 10^{-34}\,\text{m}

The wavelength is about $10^{-34}\,\text{m}$ , far smaller than an atom. To show diffraction, a gap or a crystal must be about the same size as the wavelength. No such gap can be made, so the wave nature of the ball is never observed.

6. Describe the electron diffraction experiment, and explain how it shows that electrons have wave properties.

Hint

Think of the beam, the target and the pattern on the screen.

Next step

Name the beam, the graphite film or crystal, and the rings. Then link the ring spacing to a wavelength.

Full working

A beam of electrons is fired through a thin film of graphite, or at a crystal of nickel, and the electrons that pass through are recorded on a screen. The screen shows a pattern of concentric rings, with a dark centre. Only waves produce such a pattern when they pass through a regular array of gaps. The spacing of the rings matches the de Broglie wavelength, which is calculated from $\lambda = \dfrac{h}{p}$ . The experiment therefore shows that electrons behave as waves.

7. In an electron diffraction tube, the rings move closer together when the accelerating voltage is increased. Explain why.

Hint

Work through the chain: voltage, kinetic energy, momentum, wavelength, angle of diffraction.

Next step

Say what a larger voltage does to the kinetic energy and the momentum. Then say what a larger momentum does to the wavelength.

Full working

A larger accelerating voltage gives each electron more kinetic energy. More kinetic energy means more momentum, since $p = \sqrt{2mE_k}$ . The de Broglie wavelength is $\lambda = \dfrac{h}{p}$ , so a larger momentum gives a shorter wavelength. A shorter wavelength diffracts through a smaller angle, so the rings move closer together.