Answer
$\lambda = 17.2~nm$
Work Step by Step
We can find an expression for energy of an electron in this potential well:
$E_n = (\frac{h^2}{8m~L^2})~n^2$
$E_n = \frac{h^2}{8m~L^2}~n^2$
$E_n = \frac{(6.626\times 10^{-34}~J~s)^2}{(8)(9.109\times 10^{-31}~kg)~(250\times 10^{-12}~m)^2}~n^2$
$E_n = (9.640\times 10^{-19}~J)~n^2$
$E_n = (9.640\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})~n^2$
$E_n = (6.02~eV)~n^2$
Note that light with a shorter wavelength has more energy than light with a longer wavelength.
The second shortest wavelength will be emitted when the electron jumps between states with the second largest energy difference.
We can find the energy in the third excited state where $n=4$:
$E_4 = (6.02~eV)~(4)^2 = 96.3~eV$
We can find the energy in the first excited state where $n=2$:
$E_2 = (6.02~eV)~(2)^2 = 24.1~eV$
We can find the difference in energy:
$E_4-E_2 = (96.3~eV) - (24.1~eV) = 72.2~eV$
We can find the wavelength associated with this energy:
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(4.136\times 10^{-15}~eV~s)(3.0\times 10^8~m/s)}{72.2~eV}$
$\lambda = 17.2\times 10^{-9}~m$
$\lambda = 17.2~nm$
