The electron interference pattern of the figure was made by shooting electrons w

Question

The electron interference pattern of the figure was made by shooting electrons with 50 keV of kinetic energy through two slits spaced 1.0 um apart. The fringes were recorded on a detector 1.0 m behind the slits.


What was the speed of the electrons? (The speed is large enough to justify using relativity, but for simplicity do this as a nonrelativistic calculation.)

Totally lost on how to go about this problem. Have been researching it for three days, and can not find an explanation on how to solve this problem.

Explanation / Answer

kinetic energy of the electron is K =50 keV                                                    =50*10^3*1.6*10^-19 J      since 1 eV =1.6*10^-19 J                                                    =80*10^-16 J we know the formula for kinetic energy K =1/2mv^2....................(1) where 'm' is the mass of the electron =9.1*10^-31 kg from equation (1) speed of the electron v =2K/m                                    =(2*80*10^-16/9.1*10^-31)                                    =13.2598*10^7 m/s __________________________________________________________________________ where the fringe spacing is not given let fringe spacing is =y from the interference condition wave length of the wave is =yd/L................(2) where given that d =1.0m =1*10^-6 m                           L =1 m the de'Brogle wave equation wave length =h / mv.........................(3) where h is plank's constant =6.625*10^-34 J .s and m ia mass of the electron=9.1*10^-31 kg by using equation (2) and (3) we can calculate the speed of the electron

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