Question

The electrons in a cathode ray tube are emitted by a hot filament cathode and are accelerated by an electric field applied between the cathode and two orthogonal pairs of parallel plate electrodes. The beam passes through the gap between the electrodes and can be deflected up and down, or left and right if the parallel plates are appropriately biased. As the electrons pass through the electrodes, their kinetic energy increases. Assume the electrons are emitted from the cathode with a thermal velocity corresponding to a temperature of 3000 °C, are accelerated through a potential of 5000 V over a distance of 10 mm and then travel horizontally over a distance of 200 mm before they are collected on the phosphor-coated surface of the display screen. (a) Write the equation of motion for the electrons. Assume the electric field acts only in the x-direction between the cathode and the plates. (b) Calculate the velocity of the electrons after they have been electrostatically accelerated. (c) Calculate the downward deflection of the electrons over the 200 mm gap caused by the gravitational force acting in the negative y-direction. Use Vthermal = (8kT/pi*m)1/2 where m is the mass of the electron.

Answer 1

Equation of motion of electron in cathode ray tube is obtained from the equation related to distance traveled S in a time t with initial velocity u and acceleration a , which is given below

S = u t + (1/2) a t² ......................(1)

In the case of motion of electron, we assume electron moves along the +x-axis direction.

Hence S = x(t) in eqn.(1), where x(t) is distance traveled in t seconds.

u = initial velocity = [ ( 8 k T ) / ( m ) ]^1/2 ..................(2)

where k = Boltzman constant = 1.38 10^-23 J/K

T = Absolute temperature = 3273 K

m = mass of electron = 9.1 10^-31 kg

By substituting respective values in eqn.(2), we get , initial velocity u = 3.56 10⁵ m/s

acceleration of electron in electrostatic field = qE/m ................(3)

where q = charge of electron = 1.602 10^-19 C

E = Electric field = V/d = 5000 V / 10 mm = 5 10⁵ m/s

m = mass of electron = 9.1 10^-31 kg

acceleration of electron = 8.8 10¹⁶ m/s²

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Part-(a)

Equation of motion is obtained from eqn.(1) as

x(t) = u t + (1/2) a t²

where initial speed u and acceleration are calculated as above.

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(b) velocity v of electron after electrostatic acceleration is obtained from

v² = u² + 2 a S

where u is initial speed, a is acceleration and S is distance traveled in electrostatic field which is 10 mm

v² = (3.56 10⁵ )² + 2 8.8 10¹⁶ 0.01

we get v from above eqn. as , v = 4.2 10⁷ m/s

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(c) downward deflection h due to gravitational field , h = (1/2)gt²

where t is time of travel through 200 mm that is obtained using x-component velocity

t = ( 200 10^-3 ) / ( 4.2 10⁷ ) = 4.76 10^-9 s

h = (1/2) 9.8 [ 4.76 10^-9 ]² = 1.1 10^-18 m