The electron gun in an old TV picture tube accelerates electrons between two par

Question

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.2 cm apart with a 21 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole the positive plate. Assume that the holes are small enough not to affect the electric field or potential. What is the electric field strength between the plates? With what speed does an electron exit the electron gun if its entry speed is close to zero?

Explanation / Answer

here,

B)

potential difference , V = 21 kV

let the speed gained be v

using work energy theorm

the work done potential = kinetic energy gained

V * e = 0.5 * m * (v^2 - 0^2)

21000 * 1.6 * 10^-19 = 0.5 * 9.1 * 10^-31 * v^2

solving for v

v = 8.59 * 10^7 m/s

the speed of electron when exit electron gun is 8.59 * 10^7 m/s

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