Answer
$K = 2\times 10^{-19}~J$
Work Step by Step
On the graph, we can see that $~~U = -1\times 10^{-19}~J~~$ when $~~r = 0.3~nm$
We can find the kinetic energy:
$K + U = E_1$
$K = E_1 - U$
$K = (1\times 10^{-19}~J) - (-1\times 10^{-19}~J)$
$K = 2\times 10^{-19}~J$
