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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?

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3 Answers 3

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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.

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    $\begingroup$ Thanks, that now gives the correct results. Any idea why the author is writing $Rhc$ instead of $Z^2\,R_{\infty}$? $\endgroup$
    – BLAZE
    May 6, 2017 at 22:16
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    $\begingroup$ @BLAZE Wild guess: it was supposed to be $R_\text{He}$ and was mis-transcribed somewhere in the editing process. Note that $Rhc$, for the usual $hc = 1240\rm\,eV\,fm$, doesn't have units of inverse length, so your text's solution must contain at least one error. $\endgroup$
    – rob
    May 7, 2017 at 8:49
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The general formula to calculate the wavelength of a hydrogen like species is $$\frac1\lambda=R_H\times z²\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)$$ where z is the atomic number of the species which in your case is equal to 2 for $He^+$ and $n_i$ and $n_f$ are the respective energy levels in the transition and $R_H$ is the Rydberg Constant for hydrogen atom.

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  • $\begingroup$ What is $R_H$ equal to here? $\endgroup$
    – BLAZE
    May 6, 2017 at 22:45
  • $\begingroup$ It is the Rydberg Constant =1.0974×$10^7$ per metre $\endgroup$
    – Kira
    May 6, 2017 at 22:48
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    $\begingroup$ Okay, this hasn't really told me anything that I didn't already know, but do you know why the author writes $Rhc$ instead of $z^2\,R_H$? Possible error? $\endgroup$
    – BLAZE
    May 6, 2017 at 22:52
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    $\begingroup$ That has to be a typo but you can guess why those particular quantities appeared in the typo by looking at the derivation of the value of $R_H$ which is equal to $\frac {E_1}{hc}$ where $E_1$ is the energy of first orbit of hydrogen atom,h the Planck's contact and c the speed of light in vacuum $\endgroup$
    – Kira
    May 6, 2017 at 23:01
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    $\begingroup$ Well done Kira, I managed to finally contact my lecturer to ask him about this error, and it turns out it was exactly as you pointed out :-) $\endgroup$
    – BLAZE
    May 8, 2017 at 0:20
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Okay so just a wild guess at why your book says $Rhc$ instead of $R$ as you mentioned in the comments. It might be $R_Hc$ where $R_H$ is the Rydberg's constant and $c$ is the speed of light.

I believe it could be because it's dealing with the frequency on the left hand side? Because we know that $\frac{c}{\lambda}$ = $\nu$ (where the symbols have their usual meanings).

Your book might want to say that

$$\nu= R_H\cdot c\left(\frac{1}{{n_i}^2}−\frac{1}{{n_f}^2}\right)Z^2$$

The symbols for frequency and wavenumber are very similar. The first is $\nu$ and the other is $\bar\nu$

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  • $\begingroup$ Or it could be as @rob pointed out. $\endgroup$ May 7, 2017 at 10:18

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