Answer
$\lambda = 68.7~nm$
Work Step by Step
We can find an expression for energy of an electron in this potential well:
$E_n = (\frac{h^2}{8m~L^2})~n^2$
$E_n = \frac{h^2}{8m~L^2}~n^2$
$E_n = \frac{(6.626\times 10^{-34}~J~s)^2}{(8)(9.109\times 10^{-31}~kg)~(250\times 10^{-12}~m)^2}~n^2$
$E_n = (9.640\times 10^{-19}~J)~n^2$
$E_n = (9.640\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})~n^2$
$E_n = (6.02~eV)~n^2$
We can find the energy difference between states when light of wavelength 29.4 nm is emitted:
$E = \frac{hc}{\lambda}$
$E = \frac{(4.136\times 10^{-15}~eV~s)(3.0\times 10^8~m/s)}{29.4\times 10^{-9}~m}$
$E = 42.2~eV$
$E = (6.02~eV)(7.0)$
$E = (6.02~eV)(16-9.0)$
Note that this is the energy difference between state $n=4$ and $n=3$. This light was emitted when the electron moved from state $n=4$ to state $n=3$
Note that light with a shorter wavelength has more energy than light with a longer wavelength.
The longest wavelength will be emitted when the electron jumps between states with the smallest energy difference.
We can find the energy in the first excited state where $n=2$:
$E_2 = (6.02~eV)~(2)^2 = 24.08~eV$
We can find the energy in the ground state where $n=1$:
$E_1 = (6.02~eV)~(1)^2 = 6.02~eV$
We can find the difference in energy:
$E_2-E_1 = (24.08~eV) - (6.02~eV) = 18.06~eV$
We can find the wavelength associated with this energy:
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(4.136\times 10^{-15}~eV~s)(3.0\times 10^8~m/s)}{18.06~eV}$
$\lambda = 68.7\times 10^{-9}~m$
$\lambda = 68.7~nm$
