Question

A monoenergetic beam of electrons is incident on a single slit of width 8.0 Angstroms . A diffraction pattern is formed on a screen 30 cm from the slit. If the distance between successive minima of the diffraction pattern is 3.2 cm, what is the energy of the incident electron? answer is 13 ev, how do you get it? I keep getting 209 eV

Answer 1

Use the diffraction formula

for successive minima, n = 1

=>

now use the De-Broglie's relation to get the momentum of the electrons

this lies within the Non-Relativistic Regime.

so, the energy of the incident electron will be:

= 207.05 eV.