## 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

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.

## Answers ( )

Answer:Ek=142.7keV

Explanation:First we have to compute the associated wavelength of the electrons by using the formula for the interference condition:

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:

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

hope this helps!!