Question

Calculate the de Broglie wavelength of (a) a 0.613 keV electron (mass = 9.109 × 10^-31 kg), (b) a 0.613 keV photon, and (c) a 0.613 keV neutron (mass = 1.675 × 10^-27 kg).

Answer 1

Wavelength = h / p

p = m v = sqrt [2 m KE] = sqrt [2 * 9.11 * 10^-31 * 0.613 * 10³ * 1.6 * 10^-19]

lambda = ( 6.63 * 10⁻³⁴ ) / (1.34 * 10^-23 ) = 4.9 * 10^-11 m

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b) Wavelength = h / p

p = m v = sqrt [2 m KE] = sqrt [2 * 1.675 * 10^-27 * 0.613 * 10³ * 1.6 * 10^-19]

lambda = ( 6.63 * 10⁻³⁴ ) / sqrt [2 * 1.675 * 10^-27 * 0.613 * 10³ * 1.6 * 10^-19]

= 1.16 * 10^-12 m

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c) Wavelength = h / p

p = m v = sqrt [2 m KE] = sqrt [2 * 1.675 * 10^-27 * 0.613 * 10³ * 1.6 * 10^-19]

lambda = 1.16 * 10^-12 m