Define electron Volt and convert it into Joule unit.
If an electron with charge $q=e=1.6 \times 10^{-19} \mathrm{C}$ is accelerated by a potential difference of 1 volt it would gain energy $=q \Delta \mathrm{V}$
$=1.6 \times 10^{-19} \times 1 \quad[\because \Delta \mathrm{V}=1 \mathrm{~V}]$
$=1.6 \times 10^{-19} \mathrm{~J}$
This unit of energy is defined as 1 electron volt or $1 \mathrm{eV}$.
Definition: "If an electron according through a potential difference of one volt the difference in potential (or kinetic) energy of an electron is known as one electron volt".
This unit are most commonly used in atomic, nuclear and particle physics.
Multiplier and submultiplier of $e \mathrm{~V}$ :
$1 \mathrm{meV}=1.6 \times 10^{-22} \mathrm{~J}$ $1 \mathrm{keV}=1.6 \times 10^{-16} \mathrm{~J}$ $1 \mathrm{MeV}=1.6 \times 10^{-13} \mathrm{~J}$ $1 \mathrm{GeV}=1.6 \times 10^{-10} \mathrm{~J}$ $1 \mathrm{TeV}=1.6 \times 10^{-7} \mathrm{~J}$
Two identical particles of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $v$. The distance of closest approach of these charges is
There exists a uniform electric field $E=4 \times 10^5 \,Vm ^{-1}$ directed along negative $x$-axis such that electric potential at origin is zero. Acharge of $-200 \,\mu C$ is placed at origin, and a charge of $+200 \,\mu C$ is placed at $(3 \,m , 0)$. The electrostatic potential energy of the system is ...........$J$
A square of side ‘$a$’ has charge $Q$ at its centre and charge ‘$q$’ at one of the corners. The work required to be done in moving the charge ‘$q$’ from the corner to the diagonally opposite corner is
${\rm{ }}1\,ne\,V{\rm{ }} = {\rm{ }}......\,J.$ (Fill in the gap)
Choose the $CORRECT$ option