whenever a charge particle is moving under the potential difference at that time the kinetic energy gained by the charged particle is given as
[tex]K.E=\frac{1}{2} mv^{2} =qv'[/tex]
where v' is the potential difference .
as per the question the charged particle is the helium nucleus whose charge q=[tex]3.2*10^{-19} coulomb[/tex]
the K.E is given as 140 kev i.e 140[tex]*10^{3} *1.602*10^{-19} joule[/tex]
hence the potential difference V'[tex]=\frac{K.E}{q}[/tex]
=140[tex]*10^{3} *[/tex][tex]*1.602*10^{-19} *\frac{1}{3.2*10^{-19} }[/tex]
=70.0875[tex]*10^{3} volt[/tex] [ans]
