Figure 3 shows part of the apparatus used to investigate electron diffraction - AQA - A-Level Physics - Question 3 - 2019 - Paper 7

To determine the consistency of the voltmeter reading with the de Broglie wavelength, we first convert the potential difference into kinetic energy:

$$KE = eV = (1.6 \times 10^{-19} \, \text{C})(3.5 \times 10^{3} \, \text{V}) = 5.6 \times 10^{-16} \, \text{J}$$

The kinetic energy can be related to the momentum ($p$) of the electron:

$$KE = \frac{p^2}{2m} \Rightarrow p = \sqrt{2m \times KE}$$

Using the mass of an electron, $m = 9.11 \times 10^{-31} \, \text{kg}$:

$$p = \sqrt{2 \times (9.11 \times 10^{-31} \, \text{kg}) \times (5.6 \times 10^{-16} \, \text{J})} \approx 1.52 \times 10^{-24} \, \text{kg m/s}$$

Now, we can find the de Broglie wavelength:

$$\lambda = \frac{h}{p} = \frac{(6.63 \times 10^{-34} \, \text{Js})}{(1.52 \times 10^{-24} \, \text{kg m/s})} \approx 4.36 \times 10^{-10} \, \text{m} = 0.436 \, \text{nm}$$

Since 0.436 nm is significantly greater than 0.02 nm, the voltmeter reading is not consistent with a de Broglie wavelength of 0.02 nm.

1. Decrease the acceleration voltage: By lowering the voltage across the electron beam, the energy and hence the momentum of the electrons will decrease. This results in a longer de Broglie wavelength, which can lead to a wider diffraction pattern, as observed in the second experiment.

2. Increase the distance between the metal foil and the photographic film: Increasing this distance allows the diffraction pattern to spread out more, producing a result similar to that seen in the second experiment.

These changes could effectively alter the diffraction patterns displayed in Figure 4.