Question -
Answer -
(a)Potential difference acrossthe evacuated tube, V = 500 V
Specific charge of an electron, e/m =1.76 × 1011 C kg−1
The speed of each emittedelectron is given by the relation for kinetic energy as:
Therefore, the speed of each emitted electronis 
(b)Potential of the anode, V =10 MV = 10 × 106 V
Thespeed of each electron is given as:
This result is wrong because nothing canmove faster than light. In the above formula, the expression (mv2/2)for energy can only be used in the non-relativistic limit, i.e., for v << c.
For very high speedproblems, relativistic equations must be considered for solving them. In therelativistic limit, the total energy is given as:
E = mc2
Where,
m =Relativistic mass
m0 = Mass of the particle at rest
Kinetic energy is givenas:
K = mc2 − m0c2