Question

PHYS 3350 Quiz 4 March 5, 2018 ame (wrie it on the back, too) In the Electron Diffraction pattern shown, from graphite, the electron beam is accelerated through a potential of 10.0 keV (so the electrons have kinetic energy 10.0 keV) What is the de Broglie wavelength of the electrons? Assume the electrons are non- relativistic. The rest energy of an electron is mc 51 1ke

Answer 1

Kinetic energy of electrons KE = Vd KE = P^2/2m P rightarrow Linear momentum P = squareroot 2m KE P = squareroot 2m Vd De-Broglie wavelength lambda = h/P lambda = h/squareroot 2m Vd lambda = 6.62 times 10^-34/squareroot 2 times 9.1 times 10^-31 times 10 times 1.6 times 10^-19 lambda = 0.3879 times 10^-9 lambda = 3.879 times 10^-10 m lambda = 3.879 A degrees