Answer
Choice B.
Work Step by Step
According to the given expression, the spacing between adjacent energy levels is proportional to $\frac{1}{L^2}$.
Therefore, if L increases to 1.1 L, the energy spacing $\Delta E$ becomes smaller (decreases) by a factor of $(1.1)^2$.
We know that the energy spacing is related to the wavelength of the emitted photon:$\Delta E = \frac{hc}{\lambda}$.
If $\Delta E$ decreases by a factor of $(1.1)^2$, then the wavelength increases by a factor of $(1.1)^2$.
$$(550nm)(1.1)^2\approx 670 nm$$
