In a transmission electron microscope (TEM), electrons are emitted from a filament and accelerated from rest through an anode-cathode potential difference VVV before passing through a specimen.
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 TEM analysis of a novel two-dimensional molybdenum disulfide specimen, the atomic plane spacing is 0.16 nm0.16\text{ nm}0.16 nm. To achieve high-contrast diffraction rings, the de Broglie wavelength of the imaging electrons must be exactly 25%25\%25% of this atomic spacing.
Calculate the accelerating potential difference VVV required to produce these electrons. Give your answer to 2 significant figures.
State and explain how the diameter of the diffraction rings formed on the TEM's fluorescent screen would change if the accelerating potential difference VVV is increased.
26 exam-style questions on AQA A Level Physics Wave-particle duality (A-level only), covering Newton's corpuscular theory of light (A-level only), Significance of Young's double slits experiment (A-level only), Electromagnetic waves (A-level only), The discovery of photoelectricity (A-level only), Wave-particle duality (A-level only), and Electron microscopes (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.