8.2 Wave-particle duality — Practise

1. The de Broglie wavelength

Answer: $p = mv = 1.82 \times 10^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1}$, so $\lambda = \dfrac{h}{p} = 3.6 \times 10^{-10}\,\text{m}$.

2. The momentum

Answer: $p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}}{1.0 \times 10^{-10}} = 6.6 \times 10^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1}$.

3. The de Broglie wavelength

Answer: $E_k = 8.0 \times 10^{-17}\,\text{J}$, then $p = \sqrt{2mE_k} = 1.21 \times 10^{-23}\,\text{kg}\,\text{m}\,\text{s}^{-1}$, so $\lambda = \dfrac{h}{p} = 5.5 \times 10^{-11}\,\text{m}$.

4. The speed

Answer: From $\lambda = \dfrac{h}{mv}$, $v = \dfrac{h}{m\lambda} = \dfrac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.0 \times 10^{-10}} = 3.6 \times 10^{6}\,\text{m}\,\text{s}^{-1}$.

5. The de Broglie wavelength

Answer: $\lambda = \dfrac{h}{p} = \dfrac{6.63 \times 10^{-34}}{0.050 \times 20} = 6.6 \times 10^{-34}\,\text{m}$. This is far smaller than an atom, so it is never observed.

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

• A beam of electrons is fired through a thin film of graphite, or at a crystal of nickel.
• The regular spacing of the atoms in the target acts like a diffraction grating.
• A pattern of concentric rings is recorded on a screen, which only waves can produce.
• The ring spacing matches the de Broglie wavelength, $\lambda = h/p$, so electrons behave as waves.

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

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

8. The de Broglie wavelength

Answer: $\lambda = \dfrac{h}{mv} = \dfrac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.0 \times 10^{6}} = 4.0 \times 10^{-13}\,\text{m}$. It is much shorter than an electron wavelength at the same speed, because the proton is heavier.

9. The momentum

Answer: $p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}}{5.00 \times 10^{-7}} = 1.3 \times 10^{-27}\,\text{kg}\,\text{m}\,\text{s}^{-1}$.

10. The potential difference

Answer: $p = \dfrac{h}{\lambda} = 1.33 \times 10^{-23}\,\text{kg}\,\text{m}\,\text{s}^{-1}$, then $E_k = \dfrac{p^2}{2m} = 9.65 \times 10^{-17}\,\text{J}$, so $V = \dfrac{E_k}{e} = 6.0 \times 10^{2}\,\text{V}$.

11. The wavelength

Answer: To three significant figures the wavelength is $5.49 \times 10^{-11}\,\text{m}$.

12. The percentage uncertainty

Answer: Momentum is $p = mv$, a product. The percentage uncertainties add: $0.10\% + 3.0\% = 3.1\%$. With a momentum of $1.822 \times 10^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1}$, the absolute uncertainty is $5.6 \times 10^{-26}\,\text{kg}\,\text{m}\,\text{s}^{-1}$.

13. A student must find the de Broglie wavelength of an electron. State which two quantities you would find first, and give the equations you would use.

• Find the kinetic energy first, from the accelerating voltage with $E_k = eV$.
• Then find the momentum from $p = \sqrt{2mE_k}$.
• Then use the de Broglie equation, $\lambda = h/p$, to find the wavelength.

Answer: Find the kinetic energy with $E_k = eV$, then the momentum with $p = \sqrt{2mE_k}$, then the wavelength with $\lambda = \dfrac{h}{p}$.