Predicting spectral lines

Question

In the spectrum of some hydrogen-like ions the three lines are known, which belong to the same series and have the wavelengths 992, 1085, 1215 angstroms. What other lines can be predicted?

I don't think it's very useful to use $$\dfrac{1}{\lambda}=R_0 Z^2\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)$$ We know three lambdas and we need to find $Z, n, m$. It's true that n is the same for all three wavelengths, but that still leaves us with an underdetermined system.

Another approach would be to approximate the above formula for large $n$ and $m$. That gives us $$\dfrac{1}{\lambda}=\dfrac{2R_0Z}{n^3}$$ This way we can find $n$ and $Z$ and than go back in the first formula and predict more spectral lines.

Am I off the rail here? Any clarifications would much appreciated.

Answer 1

Given the throw-away style in which the question has been phrased, I strongly suspect that you're not expected to fully solve for the parameters $\{Z,n,m_1,m_2,m_3\}$ of the system.

Instead, you're expected to recognize that even without solving for that system at all, you can still use sets of lines to predict new ones. This is because every atomic transition has the form of a difference between two terms, $$ \frac1\lambda = E_1-E_2, $$ and if you know several transitions that belong to the same spectral series and therefore have one of those terms in common, \begin{align} \frac{1}{\lambda_1} & = E_1-E_0, \\ \frac{1}{\lambda_2} & = E_2-E_0, \\ \frac{1}{\lambda_3} & = E_3-E_0, \end{align} then you can calculate the energy differences between the levels $1$, $2$ and $3$, and use those to predict the wavelengths of the transitions between them.

---

That said, it's important to note that while it looks like the system is under-determined, as you've got five parameters $\{Z,n,m_1,m_2,m_3\}$ with only three equations, that would only be true if $\{n,m_1,m_2,m_3\}$ are allowed to take on real values, but that's not true - the question explicitly marks the system as hydrogen-like, which means that those indices need to be integers. This means, in turn, that there's only a very reduced set of parameters that's even vaguely consistent with the data you've been given.

To have a stab at that, I recommend looking at the lines that you've predicted above (basically, eliminate $n$ from the equations), and then see what choices of $Z$ allow you to produce transition energies which are consistent with differences between the squares of integers.

I'll leave it to you to work out, but the problem can indeed be fully solved given enough ingenuity.

Answer 2

We have: $\lambda_1, \lambda_2$ and $\lambda_3$, 1215, 1085, 992 nm respectively.

Lets use Rydberg formula: $\frac{1}{\lambda}=R_0 Z^2 (\frac{1}{n_0^2}-\frac{1}{m^2})$

Lets write formulas for $\lambda_1$ and $\lambda_2$:

1. $\frac{1}{\lambda_1}=R_0 Z^2 (\frac{1}{n_0^2}-\frac{1}{m_1^2})$
2. $\frac{1}{\lambda_2}=R_0 Z^2 (\frac{1}{n_0^2}-\frac{1}{m_2^2})$

Now subtract the first equation from the second one:

$\frac{1}{\lambda_2}-\frac{1}{\lambda_1}=R_0 Z^2 (\frac{1}{m_1^2}-\frac{1}{m_2^2})$

$\frac{\lambda_1-\lambda_2}{\lambda_1\lambda_2}=R_0 Z^2 (\frac{1}{m_1^2}-\frac{1}{m_2^2})$

So we have obtained the fourth possible line $\lambda_{2-1}$.

We have four energy states $n_0, m_1, m_2, m_3$, so besides the known lines $\lambda_{1-0}, \lambda_{2-0}, \lambda_{3-0}$, we can calculate $\lambda_{2-1}$ (what has been done), $\lambda_{3-1}$ and $\lambda_{3-2}$.