# Determining Ion from Emission Spectrum

Consider an ion that has had all its electrons stripped from it except for one. This series produces spectral lines as described by the Bohr model and corresponds to electronic transitions that terminate in the same final state.

The longest wavelength produced is $112.5 nm$ and the shortest is $40.5nm$. What is the ion?

From this information I gathered that the lowest energy released is $~11.043 eV$ and the highest is $30.676eV$. Since the highest energy released would be from a very high state to the ground state $(\infty \rightarrow 1)$, then:

$$30.676 eV = 13.6eV \cdot \frac{z^2}{1^2}$$

However, this gives me $z=1.5$, which is an impossible result. Where did my logic go wrong?

• Is this supposed to only be one possible series? As such, is the assumption that $n=1$ for the lowest state valid? – BowlOfRed Mar 21 '18 at 20:43
• The shortest wavelength seems to be from an $n=2$ to $n=1$ transition. I do not see why there would be a longest wavelength. – Pieter Mar 21 '18 at 20:44
• Yes it's supposed to be only one possible series. How would the shortest wavelength be from $n=2$ to $n=1$? Wouldn't, for example, $n=3$ to $n=1$ release more energy? – Bryden C Mar 21 '18 at 20:47
• BowlOfRed is right, disregard my comment. The series limit is given. – Pieter Mar 21 '18 at 20:59

The answer is that the line at 40.5 nm may not terminate at n=1 - or the question is incorrect. because the n=2 to n=1 line for He$^+$ is at 30.4 nm - although this has an energy of about 40.8 eV if I remember correctly, which makes me wonder if 40.5nm should be 40.5 eV....
• I got that $n=5$ to $n=2$ for $z=2$ works fairly well for the 112.5 nm line, although that was through an educated guess. Therefore since they end at the same final state, the 40.5 nm would be $n=\infty$ to $n=2$, but that gives me $z=3.3$. Am I right to assume that there exist some z and some transition where z is the same? – Bryden C Mar 21 '18 at 22:18