The kinetic energy K given to the helium nucleus is equal to its potential energy, which is [tex]E=q \Delta V[/tex] where q=2e is the charge of the helium nucleus, and [tex]\Delta V[/tex] is the potential difference applied to it. Since we know the kinetic energy, we have [tex]E=K=85~keV=q \Delta V[/tex] and from this we can find the potential difference: [tex]\Delta V = \frac{K}{q}= \frac{85~keV}{2e}=42.5~kV [/tex]
