An electron with a velocity of 107 m/s enters a set of infinite parallel plates

Question

An electron with a velocity of 107 m/s enters a set of infinite parallel plates a and b through a tiny hole in one of the plates, a, as shown. The plates are separated by a distance d = 0.1 m. Calculate the velocity with which the electron exits the second hole (in plate b) if Va - Vb - 100 V Calculate the velocity with which the electron exits the second hole (in plate b) if Va - Vb = -100 V Hint: Potential energy of a 1 eV electron (an electron accelerated through a potential of 1V is 1.6 times 10-19 joules. Thus add (or subtract) the energy that is gained in acceleration resulting from the electric potential from the initial electron energy and then convert the final energy of the electron into its new velocity.

Explanation / Answer

V0=107 m/s, distance between plates, d = 0.1 m, charge on electron,q=-e

initial kinetic energy of electron,KE0=0.5meV02, me=mass of electron = 9.1*10-31 kg

KE0=4.55*10-17 J

1)Va-Vb = Vab=100 V , PE=Vabq = -100 eV = -1.6*10-19*100 = -1.6*10-17 J

using energy conservation, KE1 = PE + KE0 = -1.6*10-17 + 4.55*10-17 = 2.95*10-17 J = 0.5meV12

V1=(2.95*10-17/(0.5*9.1*10-31)) = 8.05 * 106 m/s Ans.

2)If now Va-Vb = Vab=-100 V =>  PE'=Vabq = 100 eV = 1.6*10-19*100 = 1.6*10-17 J

using energy conservation, KE2 = PE' + KE0 = 1.6*10-17 + 4.55*10-17 = 6.15*10-17 J = 0.5meV22

V2=(6.15*10-17/(0.5*9.1*10-31)) = 1.16 * 107 m/s Ans.

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