Question

Hi,

I am having trouble answering the following physics question.

The electron beam inside a television picture tube is 0.500mm in diameter and carries a current of 53.0 ?A. 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? Express your answer with the units A/m^2.

c) The electrons move with a velocity of 4.50

Answer 1

current i = 46*10^-6 A

charge of electron e = 1.6*10^-19 C

diameter of the television tube d = 0.5 mm

radius of the tube r = d/2 = 0.25 mm

= 0.25*10^-3 m

a)

charge q = ne

it = ne (since , current i = q/t ? q = it)

number of electrons strike the screen each second is

n = it/e

= (46*10^-6 A)(1 s) / (1.6*10^-19 C)

= 28.75*10¹³ electrons

= 287.5*10¹² electrons

= 2.875*10¹⁴ electrons

b)

area of the tube A = ?r²

current density J = i/A

= i / ?r²

= (46*10^-6 A) / (3.14)(0.25*10^-3 m )²

= 234.39 A/m²

c)

mass of the electron m = 9.11*10^-31 kg

velocity of the electron v = 4*10⁷ m/s

initial velocity of the electron v₀ = 0 m/s

distance S = 5.4*10^-3 m

from kinematic equation's ,

v² - v₀² = 2aS

v² - 0² = 2aS

acceleration a = v² / 2S

= (4*10⁷ m/s)² / (2)(5.4*10^-3 m)

= 1.48*10¹⁷ m/s²

electrostatic force F = qE ........... (1)

from , Newton's second law of motion ,

F = ma ............ (2)

compare eq (1) and (2) , we get

ma = qE

electric field E = ma/q

=(9.11*10^-31 kg)(1.48*10¹⁷ m/s²)/(1.6*10^-19 C)

= 8.43*10⁵ N/C

d)

number of electrons = n
current i = nq/t

n/t = i/q ................... (3)
total energy U = n(1/2)mv² ................. (4)
power = total energy / time

= [n(1/2)mv²]/t

= (n/t)(1/2)mv²

= (i/q)((1/2)mv²)

= (46*10^-6 A / 1.6*10^-19 C)(1/2)(9.11*10^-31 kg)(4*10⁷ m/s)²

= 0.209 W

= 2.095*10^-3 W

Answer 2

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 = I*6.241*10^18 = 53*10^-6 * 6.241*10^18 = 337*10^12 electrons per second

B)

The area of the beam is ?*r^2:

A = ?*r^2 = ?*(.0005/2)^2 = 196*10^-9 m^2

The current density is:

J = I / A = 53*10^-6 / 196*10^-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)*m*v^2 = (1/2)*(9.109*10^-31)*( 4.50*10^7 )^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 / (.006) = 827*10^3 V/m

D)

P = N*KE = 337*10^12 * 729*10^-18 = .246 W

As a check, this should also be V*I = 54*10^-6 * 4550 = .246 W