Question

In a television picture tube electron are accelerated from rest through potential difference of 10 kV in an electron gun.
(a) What is the muzzle velocity of the electrons emerging from the gun?
(b) If the gun is directed at a screen 35 cm away, how long does it take the electron to reach the screen?

Answer 1

Given that,

Initial velocity of the electrons, u = 0

Potential difference in an electron gun, V = 10 kV

To find,

(a) The muzzle velocity of the electrons emerging from the gun.

(b) If the gun is directed at a screen 35 cm away, how long does it take the electron to reach the screen?

Solution,

(a) Let v is the muzzled velocity of the electrons. Using conservation of energy such that :

$\dfrac{1}{2}mv^2=qV\\\\v=\sqrt{\dfrac{2qV}{m}}$

m is mass of an electron

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 10\times 10^3}{9.1\times 10^{-31}}} \\\\v=5.92\times 10^7\ m/s$

(b) If distance is 35 cm or 0.35 m, let t is the time taken by the electron to reach the screen. So,

$t=\dfrac{d}{v}\\\\t=\dfrac{0.35}{5.92\times 10^7}\\\\t=5.91\times 10^{-9}\ s$

Hence, this is the required solution.