Answer
$p = 4.45\times 10^{-12}~C\cdot m$
Work Step by Step
At $r = 0.020~m$, the kinetic energy of the electron is zero.
We can find the electric potential at that point:
$\Delta K = -q~\Delta V$
$\Delta K = -q~(V-0)$
$V = -\frac{\Delta K}{q}$
$V = -\frac{-100~eV}{-1e}$
$V = -100~V$
We can use Equation (24-30) to find the dipole moment:
$V = \frac{1}{4\pi~\epsilon_0}~\frac{p~cos~\theta}{r^2}$
$p = \frac{V~4\pi~\epsilon_0~r^2}{cos~\theta}$
$p = \frac{(-100~V)~(4\pi)(8.854\times 10^{-12}~C/V~m)~(0.020~m)^2}{cos~180^{\circ}}$
$p = 4.45\times 10^{-12}~C\cdot m$
