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Derive an Expression for de-Broglie wavelength in terms of kinetic energy and acceleration potential.

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Consider an electron (mass m, charge e) accelerated from rest through a potential V. The kinetic energy K of the electron equals the work done (eV ) on it by the electric field: 

K = e V ------- (1)

The de Broglie wavelength λ of the electron is then 

Substituting the numerical values of h, m, e, we get,

where V is the magnitude of accelerating potential in volts. For a 120 V accelerating potential, Eq. (1) gives l = 0.112 nm. This wavelength is of the same order as the spacing between the atomic planes in crystals. This suggests that matter waves associated with an electron could be verified by crystal diffraction experiments analogous to X-ray diffraction.

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