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

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

where v' is the potential difference .

as per the question the charged particle is the helium nucleus whose charge q=

the K.E is given as 140 kev i.e 140

hence the potential difference V'

                                                       =140

                                                      =70.0875 [ans]

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation: