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 $\pu{30.4 nm}$.


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?


You've got yourself all twisted around with 1's and 2's.

The problem states that "the ion emits two photons in succession with wavelengths 108.5 nm and 30.4 nm." So the first photon emitted has a wavelength of 108.5 nm and the second photon has a wavelength of 30.4 nm.

Furthermore the problem stated that "on transition to ground state." So after the second photon was emitted the helium ion is in its ground state which means that the single electron is in the 1s subshell.

The gist here is that you have to work backwards. Since you know for the 30.4 nm photon was from a transition that ends up in the 1s subshell, then did the electron start in the 2s subshell, or the 3s subshell and so on. So once you figure out the two subshells for the 30.4 nm photon, then you repeat the procedure for the 108.5 nm photon.

The relative wavelengths of the two photons isn't the important clue here. In other words you can't always say that the shortest wavelength will always be to the ground state for some electronic transition.

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