A beam of electrons, each with the same kinetic energy, illuminates a pair of slits separated by a distance of 51 nm. The beam forms bright and dark fringes on a screen located a distance 2.6 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 0.4 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.

K = 24.5 keV

Explanation:

The interference phenomenon is described by the equation

. d sin θ = m λ m = 1,2,3, ...

The pattern is observed on a screen at a distance L = 2.6 m

tan θ = y / L

As these experiments the angle is very small we can approximate the tangent

tan θ = sin θ / cos θ

For small angles

tan θ = sin θ

Let's replace

d y / L = m λ

λ = d y / m L

Let's reduce the units to the SI system

d = 51 nm = 51 10⁻⁹ m

y = 0.4 mm = 0.4 10⁻³ m

Let's calculate the wavelength

Let's use m = 1 for the first interference line

λ = 51 10⁻⁹ 0.4 10⁻³ / 2.6

λ = 7.846 10⁻¹² m

Let's look for kinetic energy

K = ½ m v²

p = mv

K = ½ m p² / m

K = p² / 2m

Let's use the wave-particle duality relationship

p = h / λ

K = h² / 2m λ²

Let's calculate

K = (6.63 10⁻³⁴) ² / (2 9.1 10⁻³¹ (7.846 10⁻¹²) ²)

K = 3,923 10⁻¹⁵-15 J

K = 3.923 10⁻¹⁵ J (1 eV / 1.6 10⁻¹⁹ J) = 2.452 10⁴ eV

K = 24.5 keV