Answer
$1.4\times 10^{5}~eV$
Work Step by Step
In part (a), we found that the speed is $v = 2.60\times 10^6~m/s$
We can find the kinetic energy:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(4.00)(1.661\times 10^{-27}~kg)(2.60\times 10^6~m/s)^2$
$K = 2.25\times 10^{-14}~J (\frac{J}{1.6\times 10^{-19}})$
$K = 1.4\times 10^{5}~eV$
