I am tasked with the following question:
Find the wavelengths of the first four members of the Brackett series in $\mathrm{He}^{+}$ – the set of spectral lines corresponding to transitions from $n \gt 4$ to $n = 4$.
In order to find the wavelengths it was my understanding that the relevant formula to use is
$$\frac{1}{\lambda}=R_{\infty}\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\tag{1}$$
Rearranging,
$$\lambda=\frac{1}{R_{\infty}}\left(\frac{16n^2}{n^2-16}\right)$$
and taking the Rydberg constant $R_\infty$ to be $1.0974\times 10^7\text{m}^{-1}$
Then $n=8\rightarrow 4\implies \lambda\approx 1940\,\text{nm}$
and $n=7\rightarrow 4\implies \lambda\approx 2165\,\text{nm}$
and $n=6\rightarrow 4\implies \lambda\approx 2624\,\text{nm}$
and $n=5\rightarrow 4\implies \lambda\approx 4050\,\text{nm}$
The problem is that the answer is:
The wavelengths are given by $$\frac{1}{\lambda}=Rhc\left(\frac{1}{4^2}-\frac{1}{n^2}\right)$$ with $$Rhc=\frac{1}{91.1\,\text{nm}}$$ So the wavelengths are $1012\,\text{nm}$ ($n=5$ to $n=4$), $656\,\text{nm}$ ($n=6$ to $n=4$) and $541\,\text{nm}$
($n=7$ to $n=4$)
The formula that has been given in the answer I understood to be the change in energy $\Delta E$ when a transition takes place.
I am almost certain that $(1)$ is the correct formula to use (but it is apparently giving me the wrong results) unless there is something else I am missing.
Where am I going wrong?
Best Answer
The Rydberg formula has to be adjusted to that the nuclear charge $Z=2$. For hydrogen like ions (single electron + nucleus), the electron energies are proportional to $Z^2$, so you should multiply $R_{\infty}$ by four.