Answer :

Given:

Potential difference between collector and emitter = 500V

Specific charge of electron (charge per unit mass **e/m**) = 1.76 × 10^{11} C

Kinetic energy of an electron is given by:

…(1)

Where,

M = mass of electron

v = velocity of electron

e = charge of electron

V = potential difference (accelerating potential)

(a) From equation (1), we can write

…(2)

By putting the values in equation (2) we can find electron velocity.

v = 1.327 × 10^{7} ms^{-1}

(b) Accelerating potential, V = 10MV = 10^{6}V

Let speed of electron be v_{1}

Again putting the values in equation (2),

v_{1} =

v_{1 =} 1.8 × 10^{9}ms^{-1}

This result is wrong as we understand that speed of light

(i.e. 3 × 10^{8} ms^{-1}) is the theoretical limit of the speed.

Such problems can be dealt using relativistic mechanics,

Relativistic mass is given by:

m =

Where,

m = relativistic mass

m_{0} = rest mass

v = velocity of particle

c = speed of light

At relativistic speeds, kinetic energy is given by,

KE = mc^{2}-m_{0}c^{2}

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