Question

A beam of electrons is incident on a single slit that has a width of 1.00 × 10–6 m and a diffraction pattern is observed on a screen located 15.0 m from the slit. The momentum of an individual electron in the beam is 5.91 × 10–24 kg ⋅ m/s. What is the width of the central maximum fringe?

Answer 1

Answer:

w = 5.04 mm

Explanation:

To find the width of the central maximum you use the following formula for the height of the dark fringes:

$y=(m+\frac{1}{2})\frac{\lambda D}{d}$    ( 1 )

y: height of the m-th dark fringe

λ: wavelength of the electron

D: distance to the screen = 15.0m

d: distance between slits = 1.00*10^{-6} m

twice the height of the first dark fringe (m = 1) gives you the width of the central peak.

You first calculate the wavelength of the electron by using the Broglie's relation:

$\lambda=\frac{h}{p_e}=\frac{6.62*10^{-34}Js}{5.91*10^{-24}kgm/s}\\\\\lambda=1.12*10^{-10}m$

where you have used the constant's Planck h.

Next, you replace the values of the parameters in the equation ( 1 ) with m=1:

$y=(1+\frac{1}{2})\frac{(1.12*10^{-10}m)(15.0m)}{1.00*10^{-6}m}=2.52*10^{-3}m\\\\y=2.52\ mm$

the width of the central maximum is:

$w=2y=5.04*10^{-3}m=5.04\ mm$