Answer
$3.58\times 10^{-14}m$
Work Step by Step
We know that
$\frac{1}{2}mv^2=\frac{Kq_{\alpha}q}{r}$
This can be rearranged as:
$r=K\frac{qq_{\alpha}}{\frac{1}{2}mv^2}$
We plug in the known values to obtain:
$r=(9\times 10^9Nm^2/C^2)\frac{(79\times 10^{-19}C)(3.2\times 10^{-19}C)}{\frac{1}{2}(6.64\times 10^{-27}Kg)(1.75\times 10^7m/s)^2}$
$r=3.58\times 10^{-14}m$
