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 { m keV} of kinetic energy through two slits spaced 1.0 { m mu m} apart. The fringes were recorded on a detector 1.0 { m m} behind the slits.

Part A:

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.)

Express your answer using two significant figures.

Part B:

The figure is greatly magnified. What was the actual spacing on the detector between adjacent bright fringes?

Express your answer using two significant figures.

Explanation / Answer

Part A 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 __________________________________________________________________________ Part B 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 (3) solve for wavelength                                               =6.625*10^-34 J .s / (9.1*10^-31 kg)(13.2598*10^7 m/s)                                                    = 0.0548*10-10m From the equation (2)                                              y= L/d                                                  =(0.0548*10-10m )(1m)/ (1*10^-6 m)                                                   = 0.0548*10-4m                                                   = 0.00548 mm                                                     or ˜5.5 m Part A 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 __________________________________________________________________________ Part B 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 (3) solve for wavelength                                               =6.625*10^-34 J .s / (9.1*10^-31 kg)(13.2598*10^7 m/s)                                                    = 0.0548*10-10m From the equation (2)                                              y= L/d                                                  =(0.0548*10-10m )(1m)/ (1*10^-6 m)                                                   = 0.0548*10-4m                                                   = 0.00548 mm                                                     or ˜5.5 m

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