 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2018/paper-2/section-a-ii/14/i/solution

For a nonrelativistic electron,

λ = \frac{h}{p} = \frac{h}{2 m _{e} E}, E = 20 (1.6 \times 1 0^{- 19}) = 3.2 \times 1 0^{- 18} J .

Using the supplied constants gives

λ ≃ 2.75 \times 1 0^{- 10} m = 0.275 nm .

This is comparable with atomic lattice spacings. A suitable crystalline surface or sufficiently thin crystal can therefore diffract the electron beam; the reciprocal-lattice geometry and scattering conditions determine the actual angles. A grating or slit can also diffract matter waves if its scale and coherence conditions are suitable. The kinetic energy is tiny compared with the electron rest energy, so relativistic corrections are negligible at the stated precision.

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