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?

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:

   ( 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:

where you have used the constant's Planck h.

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

the width of the central maximum is:

v 154.9= 0.9 length length s little over 154.9cm

less than one you can't create energy