Question

A modified oscilloscope is used to perform an electron interference experiment. Electrons are incident on a pair of narrow slits 0.052 0 µm apart. The bright bands in the interference pattern are separated by 0.400 mm on a screen 21.5 cm from the slits. Determine the potential difference through which the electrons were accelerated to give this pattern.

Answer 1

bright band will be formed at, y = m * lambda * D / d

seperation between two bands will be lambda*D / d

     0.400 x 10^-3 = lambda * 0.215 / (0.052 x 10^-6)

lambda = 9.67 x 10^-11 m = 967 x 10^-9 m = 967 nm

E =hc / lambda = 1240 eV-nm / 967nm = 1.28 eV

energy gain = qV

1.28 eV = 1e deltaV

deltaV = 1.28 volt