Question

The kinetic energy of an electron, which is accelerated in the potential difference of 100 V, is
(A) $1.6 \times {10^{ - 10}}J$
(B) $1.6 \times {10^8}J$
(C) $1.6 \times {10^{ - 17}}J$
(D) $1.6 \times {10^{ - 18}}J$

Answer 1

Hint Consider a condition where the particle is experiencing a force when it is moving and derive the formula relating kinetic energy and potential difference. Relate force with acceleration from which you can get velocity and on substituting it in kinetic energy formula we get the final equation with potential difference.

Complete step-by-step solution
Suppose a charged particle having a charge Q and mass in the electric field E experiences an electric force F when in motion with a velocity v. Let a be the acceleration produced.
We know that force experienced by the particle is,
 $F = QE$
Force is also related as,
 $F = ma$
From the two equations of force, acceleration of the particle is known
 $a = \dfrac{{QE}}{m}$
Now, we know that acceleration is change in velocity at time t.
 $a = \dfrac{v}{t} \Rightarrow v = at = \dfrac{{QEt}}{m} = \sqrt {\dfrac{{2QE\Delta V}}{m}} $
Kinetic energy K is given by,
 $K = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}m \times \dfrac{{2Q\Delta V}}{m}$
Here velocity is substituted to get the final equation relating charge and potential difference with kinetic energy.
 $\therefore K = Q\Delta V$
From the given data,
Q= charge of an electron, $1.6 \times {10^{ - 19}}C$
 $\Delta V = 100V$
Substitute the given data in the final equation of kinetic energy.
 $
  K = 1.6 \times {10^{ - 19}} \times 100 \\
  K = 1.6 \times {10^{ - 17}}J \\
 $

Hence, the kinetic energy of the electron is $1.6 \times {10^{ - 17}}J$ when accelerated in the potential difference of 100V.

So, the correct option is C.

Note When the particle moves from one point to another it can be considered as work done in terms of kinetic energy using eV electron volt as the unit of energy.