Question

"what potential difference is needed to give a helium nucleus (q=3.2×10^−19c) 140 kev of kinetic energy?"

Answer 1

whenever a charge particle is moving under the potential difference at that time the kinetic energy gained by the charged particle is given as

            $K.E=\frac{1}{2} mv^{2} =qv'$

where v' is the potential difference .

as per the question the charged particle is the helium nucleus whose charge q=$3.2*10^{-19} coulomb$

the K.E is given as 140 kev i.e 140$*10^{3} *1.602*10^{-19} joule$

hence the potential difference V'$=\frac{K.E}{q}$

                                                       =140$*10^{3} *$$*1.602*10^{-19} *\frac{1}{3.2*10^{-19} }$

                                                      =70.0875$*10^{3} volt$ [ans]