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?
 A: 
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?

The Rydberg formula for a hydrogenic atom reads:
$$
\frac{1}{\lambda} = Z^2 R_\infty \left({\frac{1}{n_I^2} - \frac{1}{n^2}}\right)\;,
$$
where $n_I$ is the initial un-filled orbital quantum number (the lower energy orbital), $n>n_I$ is the orbital from which the electron will "drop down" to fill the lower orbital, $Z$ is the atomic number, and $R_\infty$ is $1.09677\times 10^7$ inverse meters.
Inverting and converting units gives an expression for the wavelength in nanometers:
$$
\lambda = \frac{91.16}{Z^2}\frac{n_I^2 n^2}{n^2 - n_I^2}\tag{nano-meters}
$$
It is easier to search for the shortest wavelength first, since we can simplify a lot by sending $n$ to $\infty$. This leads to:
$$
\lambda_{shortest} = \frac{91.16}{Z^2}n_I^2\tag{nano-meters}\;,
$$
where we know that $\lambda_{shortest}=40.5$nm, so we can also write this as:
$$
0.444 = \frac{n_I^2}{Z^2}\;.
$$
Now we just need to try and find integers $n_I$ and $Z$ that could satisfy this equation. For example, $n_I=1$ will not work since it can only give a LHS that is either 1 (Z=1), or 0.25 (Z=2), or lower. Searching a parameter space of integers between, say, 1 and 10 means we have to do 100 different calculations. It is easiest to do so many calculations with the help of a computer. It turns out that $n_I=4$ and $Z=6$ provides one solution since
$$
4^2/6^2 = 0.444\;.
$$
Of course, if we only look at the shortest wavelength, there are many other possible solutions. (For example, $n_I=8$ and $Z=12$ is also a solution.) But we also know that the longest wavelength has to be $112.5$nm.
$$
\lambda_{longest} = \frac{91.16}{Z^2}\frac{n_I^2(n_I+1)^2}{2n_I + 1}\tag{nano-meters}\;,
$$
which is only consistent with the $n_I=4$ and $Z=6$ solution.
A: 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 don't want to say more, I don't know the answer and I won't work it out, but I agree that a transition to n=1 gives a non-integer z value even if you try n=2 to n=1.
A: $Z=6$ for the atomic number and $n=4$ for the lowest term of the series seem to work fine.
