Answer
$-4.25\times10^4 V $
Work Step by Step
The kinetic energy gained by the helium nucleus is the work done on it by the electric field. It loses PE and gains KE. Use equation 17–2b.
$$\Delta PE=q \Delta V=-\Delta KE$$
$$\Delta V=\frac{-\Delta KE}{q}=\frac{KE_i-KE_f}{q}$$
The charge of a helium nucleus is 2e.
$$=\frac{0-85.0\times10^3 eV}{2e}$$
$$=-4.25\times10^4 V $$
The final potential is lower than the initial potential to accelerate the positively-charged helium nucleus.
