Answer
a. $ 43.5 eV $.
b. $2.85\times10^{-8}m$.
Work Step by Step
a. The energy levels in a 1D box are given by the equation $E_n=\frac{n^2 h^2}{8mL^2}$.
The ground state has n=1, and the third excited level has n = 4.
$\Delta E = E_4-E_1=\frac{4^2 h^2}{8mL^2}-\frac{1^2 h^2}{8mL^2}=\frac{15 h^2}{8mL^2}$
We evaluate with L = 0.360 nm to find an energy difference of $ 43.6 eV $.
b. Now solve for the wavelength.
$$\lambda =\frac{hc}{E}=\frac{1.24\times10^{-6}eV\cdot m}{43.6eV}=2.85\times10^{-8}m$$
