Why does the first lithium stationary state have n = 9/4?


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> The ionization energy of $\ce{He+}$ is $19.6 \times 10^{-18}~\mathrm{J~atom^{-1}}$. What is the energy of the first stationary state ($n=1$) of $\ce{Li^{2+}}$?
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Since the question specifically states that $n=1$, I used the formula 


\begin{align}
E &= -2.178 \times 10^{-18} \cdot \frac{Z^2}{n^2}.\\
E &= -2.178 \times 10^{-18} \cdot \frac{3^2}{1}\\
E &= -1.96 \times 10^{-17}~\mathrm{J}
\end{align}


But the correct answer is supposed to be $-4.14 \times 10^{-17}~\mathrm{J}$ and many people are telling me to use $n=\frac{9}{4}$ which gives the right answer but I don't know why I am supposed to set $n$ at $\frac{9}{4}$. 




I think your calculation is correct. There is an error in this textbook. All you have to do is to switch between $\ce{Li^{2+}}$ and $\ce{He^{+}}$. And 'll prove my point: 


$\ce{Li^{2+}}$ and $\ce{He^{+}}$ are hydrogen-like atoms.


For these species, the ionization energy is given by the equation:
$$E\_\mathrm{i}=E\_\mathrm{i}(\ce{H})\times Z^2$$


For $\ce{Li^{2+}}$: 
\begin{align}
E\_\mathrm{i}(\ce{Li^{2+}})&=E\_\mathrm{i}(\ce{H})\times Z^2\\
19.6 \times 10^{-18}&=E\_\mathrm{i}(\ce{H})\times 3^2\\
E\_\mathrm{i}(\ce{H})&=2.18 \times 10^{-18} \,\mathrm {J/atom}\\
E\_\mathrm{i}(\ce{H})&=\frac{2.18 \times 10^{-18}}{1.6 \times 10^{-19}}\\
 &= 13.6 \,\mathrm{eV}
\end{align}


$13.6 \,\mathrm{eV}$ is well-known to be the first ionization energy of hydrogen. So, my assumption is true.
To solve the problem after correction:
 The ionization energy of $\ce{He^{+}}$ is given by the equation:
\begin{align}
E\_\mathrm{i}(\ce{He^{+}})&=\frac{E\_\mathrm{i}(\ce{Li^{2+})}\times Z^2(\ce{He^{+})}}{Z^2(\ce{Li^{2+})}}\\
E\_\mathrm{i}(\ce{He^{+}})&=8.72\times 10^{-18}\,\mathrm {J/atom}
\end{align}
The energy of the first stationary state:
$$E\_\mathrm{i,1}=E\_\infty-E\_1$$
But, $E\_\infty=0$, we find:
$$E\_1=-E\_\mathrm{i,1}=-8.72\times 10^{-18}\,\mathrm {J/atom}$$ 


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