Answer
$E_1 = 9.41~eV$
Work Step by Step
In the ground state, $n=1$
We can find the ground-state energy of an electron in a potential well:
$E_1 = (\frac{h^2}{8m~L^2})~n^2$
$E_1 = \frac{h^2}{8m~L^2}$
$E_1 = \frac{(6.626\times 10^{-34}~J~s)^2}{(8)(9.109\times 10^{-31}~kg)~(200\times 10^{-12}~m)^2}$
$E_1 = 1.51\times 10^{-18}~J$
$E_1 = (1.506\times 10^{-18}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E_1 = 9.41~eV$
