Problem 20.15

20.15.
The potential is given by
$U = \frac{kq_{1}q_{2}}{r}$
$U = \left(\frac{1}{n^{2}}\right) \frac{-(9 \times 10^{9})(1.6 \times 10^{-19})^{2}}{.529 \times 10^{-10}}$
So, for n=1,
$U = -4.37 \times 10^{-18} J ~=~ -27.3 eV$

b)
For n=2, we find that
$U = \frac{1}{2^{2}} (-27.3 eV)$
U = -6.81 eV

c)
If $n\rightarrow \infty$, then the energy goes to zero.