Question

Find the speed of an electron with kinetic energy $1eV $.
(A) $50.92 \times {10^5}\dfrac{m}{s} $
(B) $589.92 \times {10^5}\dfrac{m}{s} $
(C) $5.92 \times {10^5}\dfrac{m}{s} $
(D) $599.92 \times {10^5}\dfrac{m}{s} $

Answer 1

Hint Kinetic energy is the energy gained by a body when it is moving with a non-zero velocity. When we are dealing with small particles the unit joule and kilojoule seem too large thus there is a smaller unit for energy which is electron volt $eV $.

Formula Used:
 $K.E. = \dfrac{1}{2}m{v^2} $
Where $K.E. $ is the kinetic energy, $m $is the mass of the body, $v $ is the velocity with which the body is travelling.

Complete Step-by-step answer
It is given in the question that the kinetic energy of the electron is $1eV $ .
From this, we came to know the mass of the body i.e. electron is $9.1 \times {10^{ - 31}}Kg $.
We know that
 $K.E. = \dfrac{1}{2}m{v^2} $
Where $K.E. $ is the kinetic energy, $m $is the mass of the body, $v $ is the velocity with which the body is travelling.
And $1eV = 1.6 \times {10^{ - 19}}J $
 $1eV $is defined as the energy gained by an electron when it is accelerated through a potential difference of $1Volt $.
Hence Kinetic energy of the electron is $1.6 \times {10^{ - 19}}J $
Inserting the known values in the formula
 $ \Rightarrow K.E. = \dfrac{1}{2}m{v^2} $
 $ \Rightarrow 1.6 \times {10^{ - 19}} = \dfrac{1}{2} \times 9.1 \times {10^{ - 31}} \times {v^2} $
 $ \Rightarrow v = {(\sqrt {\dfrac{{9.1 \times {{10}^{ - 31}}}}{{2 \times 1.6 \times {{10}^{ - 19}}}}} )^{ - 1}} $
 $ \Rightarrow v = \sqrt {\dfrac{{2 \times 1.6 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}} $
On solving this further we get,
 $v = 592999.4533288\dfrac{m}{s} $
 $ \Rightarrow v \approx 5.92 \times {10^5}\dfrac{m}{s} $

Hence the correct answer is (C) $5.92 \times {10^5}\dfrac{m}{s} $.

Additional information
We know that the force acting on a charge $q $ placed in an electric field of magnitude $E $is $qE $ and
Energy is equal to $\int {F.dx} $where $F $ is the force
So from this, the energy of the charge $q $ placed in an electric field of magnitude $E $ will be
\[\int {qE.dx} \]
As $\int {E.dx} $is the potential difference $V $
So the energy becomes $qV $.

Note
The energy of a body is always conserved; it can’t be destroyed or created but it can always be converted from one form to another here also while defining $1eV $we used this as potential energy is being converted into kinetic energy.