PROTON is accelerated through 1V, its kinetic energy \[K=qV\] For a proton, \[q=e=
1
.6\times {{10}^{-19}}\,C\] and \[V=1\text{ }
VOLT
\]
THUS
, \[K=(1.6\times {{10}^{-19}}\times 1)\,J=1\,eV\] (as \[1\,eV=1.6\times {{10}^{-19}}\,J\])