Question

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.

Answer 1

Answer:

  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