# Deducing the shell when transition occurs such that two photons are discharged

Find $$n$$ (quantum number) corresponding to the excited state of $$\ce{He+}$$ ion if on transition to ground state the ion emits two photons in succession with wavelengths $$\pu{108.5 nm}$$ and $$\pu{30.4 nm}$$.

Solution:

Using Rydberg's equation where $$n_1$$ is the shell from which the electron transitions to shell $$n = 1$$ (ground state). $$(1)$$. $$n_2$$ is the excited state $$n$$ from which the electron transitions to $$n_1$$. $$(2)$$

When finding $$n_1$$ the wavelength $$\pu{30.4 nm}$$ is used, while the wavelength $$\pu{108.5 nm}$$ is used for the other transition.

My Question: How are the wavelengths coupled with those transitions specifically?

My Guess: Since the absolute value of energy in a transition is inversely proportional to the wavelength of the photon and since the energy to remove electrons from outer shells is lower, the smaller wavelength corresponds to the transition closer to the nucleus (i.e : to the ground state). Is this the reason for this correlation?