Answer
The lower quantum number is $~~n = 12$
Work Step by Step
We can find an expression for energy of an electron in this potential well when $n = 5$:
$E_n = (\frac{h^2}{8m~L^2})~n^2$
$E_5 = (\frac{h^2}{8m~L^2})~(5)^2$
$E_5 = 25~(\frac{h^2}{8m~L^2})$
Let $n_h$ be the state with the higher quantum number.
Let $n_l$ be the state with the lower quantum number.
We can use the expression for the energy difference to find $n_h$ and $n_l$:
$E_{n_h} - E_{n_l} = E_5$
$(\frac{h^2}{8m~L^2})~n_h^2 - (\frac{h^2}{8m~L^2})~n_l^2 = 25~(\frac{h^2}{8m~L^2})$
$n_h^2 - n_l^2 = 25$
$(n_h - n_l)(n_h+n_l) = 25$
$(n_h - n_l)(n_h+n_l) = (1)(25)$
Then:
$n_h = 13$
$n_l = 12$
The lower quantum number is $~~n = 12$
