Answer
a) $1.62\times 10^{-12}J$ or $10.1MeV$
b) Distance of closest approach will increase by a factor of 4.
Work Step by Step
(a) We know that
$E_f=K\frac{(Ze^{i\theta}e)q}{d}$
We plug in the known values to obtain:
$E_f=\frac{8.99\times 10^9(79)(2)(1.60\times 10^{-19})(1.60\times 10^{-19})}{22.5\times 10^{-15}}$
$E_f=1.62\times 10^{-12}J\times \frac{1MeV}{1.6\times 10^{-13}}$
$E_f=10.1MeV$
(b) If the speed is reduced by a factor of 2, then kinetic energy is reduced by a factor of 4 (since KE is proportional to $v^2$). We see from (a) that distance ($d$) and kinetic energy are inversely proportional, so the distance of closest approach will increase by a factor of 4.
