In an electron diffraction experiment, electrons are accelerated from rest through a potential difference VVV and are incident on a thin crystalline target.
Show that the de Broglie wavelength λ\lambdaλ of an electron accelerated from rest through a potential difference VVV is given by:
λ=h2meeV \lambda = \frac{h}{\sqrt{2m_e eV}} λ=2meeVhwhere hhh is Planck's constant, mem_eme is the electron mass, and eee is the elementary charge.
In a particular experiment, the crystalline target has an interatomic spacing of 0.12 nm0.12\text{ nm}0.12 nm. To produce a clear diffraction pattern, the de Broglie wavelength of the electrons must be equal to 40%40\%40% of this spacing.
Calculate the required potential difference VVV to 2 significant figures.
State and explain how the appearance of the diffraction pattern on the screen changes if the potential difference VVV is increased.
