"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]

think its the third option

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures