Answer
$K = 28~MeV$
Work Step by Step
An alpha particle has 2 protons
A gold nucleus has 79 protons
To find the required energy of the alpha particle, we can equate the final potential energy with the initial kinetic energy:
$K = U$
$K = \frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{r}$
$K = \frac{1}{(4\pi)~(8.854\times 10^{-12}~F/m)}~\frac{(2)(1.6\times 10^{-19}~C)(79)(1.6\times 10^{-19}~C)}{(1.80\times 10^{-15}~m+6.23\times 10^{-15}~m)}$
$K = 4.527\times 10^{-12}~J$
$K = (4.527\times 10^{-12}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$K = 2.8\times 10^7~eV$
$K = 28~MeV$
