Question

What potential difference is needed to give a helium nucleus (q=2e) 85.0 kev of kinetic energy?

Answer 1

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