Answer to Question #109176 in Atomic and Nuclear Physics for Tanya Saini

Question

Under certain conditions, a beam of electrons impinging on a crystal surface will diffract
and a scattering pattern of the beam can be obtained. What is the approximate kinetic
energy of the electrons needed in order to see the pattern?
(Assume the lattice spacing of the crystal to be 0.4nm )

Answer 1

A beam of electrons impinging on a crystal surface will diffract and a scattering pattern of the beam can be obtained if de Broglie wavelength of electrons $\lambda=\frac{h}{p}$ will be the same value as the distance between the atoms in the lattice of the crystal $a=0.4 nm$ . That is

$a=\frac{h}{p}$ . The kinetic energy of the electron is

$E=\frac{p^2}{2m_e}=\frac {h^2}{2m_e a^2}=\frac{(6.6×10^{−34})^2 J^2s^2}{2\cdot 9.1\cdot 10^{-31} kg\cdot (0.4\cdot 10^{-9})^2 m^2}=\frac{4.36\cdot 10^{-67}}{2.91\cdot 10^{-49}}=2.3\cdot 10^{-18}J$

It is often convenient to express the energy of particles in off-system units - electron-volts. This energy corresponds to the energy of an electron which it acquires after passing through the area of the electric field with a potential difference of 1V. $1eV=1.6\cdot 10^{-19}J$. Thus $E=\frac{2.3\cdot 01^{-18}J}{1.6\cdot 10^{-19} J/eV}=14 eV$

Answer: The approximate kinetic energy of the electrons needed in order to see the pattern from the crystal is $2.3\cdot 10^{-18}J$ or $14 eV$.