The electron beam inside a television picture tube is 0.500 mm in diameter and carries a current of 54.0uA. This electron beam impinges on the inside of the picture tube screen.
A) How many electrons strike the screen each second?
B) What is the current density in the electron beam?
C) The electrons move with a velocity of 4.00×10^7 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.50 mm?
D) Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam?
Thank you so much for you help in advance! sorry its got four parts! :-
1 Answer
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A)
An Ampere is 1 Coulomb of charge passing a point in one second. A Coulomb is 6.241*10^18 electrons. At the stated current, the number of electrons impinging is:
N = I6.24110^18 = 5410^-6 6.24110^18 = 33710^12 electrons per second
B)
The area of the beam is π*r^2:
A = πr^2 = π(.0005/2)^2 = 196*10^-9 m^2
The current density is:
J = I / A = 5410^-6 / 19610^-9 = 275 A/m^2
C)
The mass of an electron is 9.109*10^-31 kg. The Kinetic Energy of each electron is:
KE = (1/2)mv^2 = (1/2)(9.10910^-31)( 4.0010^7 )^2 = 729*10^-18 J
According to Google, this is 4550 eV, so it takes 4550 V to accelerate each electron to the stated velocity. The electric field strength is then:
E = V/d = 4550 / (.0055) = 827*10^3 V/m
D)
P = NKE = 33710^12 72910^-18 = .246 W
As a check, this should also be V*I = 54*10^-6 * 4550 = .246 W
