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]
