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

To determine this, we first need to calculate the de Broglie wavelength using the provided voltage.

The kinetic energy ( ( KE )) of the electrons is given by: $KE = eV$ Substituting in the given values, where ( e ) is the charge of an electron (approximately ( 1.6 \times 10^{-19} ) C) and ( V = 3.5 \times 10^3 ) V: $KE = 1.6 \times 10^{-19} \times 3.5 \times 10^3 = 5.6 \times 10^{-16} \text{ J}$

The relationship between kinetic energy and momentum ( ( p )) is: $KE = \frac{p^2}{2m}$ Rearranging gives: $p = \sqrt{2m \times KE}$ Substituting the mass of an electron (approximately ( 9.11 \times 10^{-31} \text{ kg} )) yields: $p = \sqrt{2 \times 9.11 \times 10^{-31} \times 5.6 \times 10^{-16}} = 1.14 \times 10^{-24} \text{ kg m/s}$

Now applying de Broglie's equation: $\lambda = \frac{h}{p}$ where ( h ) is Planck's constant ( ( 6.63 \times 10^{-34} \text{ J s} )). Thus, $\lambda = \frac{6.63 \times 10^{-34}}{1.14 \times 10^{-24}} \approx 5.8 \times 10^{-10} ext{ m} = 0.058 ext{ nm}$

Since the calculated wavelength (0.058 nm) does not match with the given value of approximately 0.02 nm, the voltmeter reading is not consistent with a de Broglie wavelength of 0.02 nm.

First Change

Statement 1: Increase the accelerating voltage. Relevant Change: A higher accelerating voltage increases the energy of the electrons, which decreases the de Broglie wavelength, allowing for better resolution in diffraction patterns.

Second Change

Statement 2: Use a target with greater spacing between atoms. Relevant Change: Having a target with larger atomic spacing enables clearer diffraction patterns due to increased interference effects, which can be observed in the more pronounced rings in the second experiment.