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.