Answer
$E_1 = 0.0205~eV$
Work Step by Step
Note that in the ground state, $n=1$
We can find the ground-state energy of a proton in the potential well:
$E_1 = (\frac{h^2}{8m~L^2})~n^2$
$E_1 = \frac{h^2}{8m~L^2}~(1)^2$
$E_1 = \frac{(6.626\times 10^{-34}~J~s)^2}{(8)(1.67\times 10^{-27}~kg)~(100\times 10^{-12}~m)^2}$
$E_1 = 3.28622\times 10^{-21}~J$
$E_1 = (3.28622\times 10^{-21}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E_1 = 0.0205~eV$
