A beam of electrons, each with the same kinetic energy, illuminates a pair of slits separated by a distance of 65 nm. The beam forms bright and dark fringes on a screen located a distance 2 m beyond the two slits. The arrangement is otherwise identical to that used in the optical two-slit interference experiment. The bright fringes are found to be separated by a distance of 0.2 mm. What is the kinetic energy of the electrons in the beam? Planck’s constant is 6.63 × 10−34 J · s. Answer in units of keV.

where m is the order of the fringe, lambda is the wavelength of the electrons, d is the distance between the slits and D IS the distance to the screen. By taking m=1, we have that the distance between the first fringes is 0.2mm. Hence, from the middle of the screen we have that y=0.1mm=0.1*10^{-3}m.

By replacing we have:

[tex]\lambda=\frac{dy}{mD}=\frac{(0.1*10^{-3}m)(65*10^{-9}m)}{(1)(2m)}=3.25*10^{-12}m[/tex]

[tex]\lambda=\frac{dy}{mD}=\frac{(2m)(65*10^{-9}m)}{(1)(0.1*10^{-3}m)}=1.3*10^{-3}m[/tex]

Now, by using the Broglie's relation we get:

[tex]p=\frac{h}{\lambda}=\frac{6.63*10^{-34}Js}{3.25*10^{-12}m}=2.04*10^{-22}kg\frac{m}{s}\\\\E_k=\frac{1}{2}mv^2=\frac{p^2}{2m}=\frac{(2.04*10^{-22}kg\frac{m}{s})^2}{2(9.1*10^{-31}kg)}=2.28*10^{-14}J=142.7keV[/tex]

hope this helps!!