To determine if the voltmeter reading of 3.5 kV is consistent with the de Broglie wavelength of 0.02 nm, we first need to calculate the momentum of the electrons after being accelerated through the potential difference (V) using:

$KE = eV$

The kinetic energy (KE) gained by the electrons is equal to the electric charge ($e$) multiplied by the potential difference. Thus:

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

The momentum $p$ of the electrons can then be calculated using the relationship:

$p = \\sqrt{2m_e \\cdot KE}$

Substituting the mass of the electron ($m_e = 9.11 \times 10^{-31} ext{ kg}$):

$p = \\sqrt{2 \times (9.11 \times 10^{-31}) \times (5.6 \times 10^{-16})} = 1.10 \times 10^{-24} ext{ kg m/s}$

Now using the de Broglie equation:

\\lambda = \\rac{h}{p}

Substituting Planck's constant ($h = 6.63 \times 10^{-34} ext{ Js}$):

\\lambda = \\rac{6.63 \times 10^{-34}}{1.10 \times 10^{-24}} \approx 6.02 \times 10^{-10} ext{ m} = 0.0602 ext{ nm}

Since the calculated wavelength (0.0602 nm) does not match the given wavelength (0.02 nm), the voltmeter reading is not consistent.

Change 1: Increase the accelerating potential.

By increasing the potential difference across the electron beam, the energy and, therefore, the momentum of the electrons will increase. This results in a shorter de Broglie wavelength, which can lead to more distinct diffraction patterns as seen in the second experiment.

Change 2: Use a thinner metal foil.

Reducing the thickness of the metal foil will allow for greater interaction of the electrons with it, enhancing the diffraction effects and consequently improving the visibility of the fringes in the resultant diffraction pattern.