Question

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.

Answer 1

Ek=142.7keV

Step-by-step explanation:

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

`y=(m\lambda D )/(d)`

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:

`\lambda=(dy)/(mD)=((0.1*10^(-3)m)(65*10^(-9)m))/((1)(2m))=3.25*10^(-12)m`

`\lambda=(dy)/(mD)=((2m)(65*10^(-9)m))/((1)(0.1*10^(-3)m))=1.3*10^(-3)m`

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

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

hope this helps!!