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

Question

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 $\ce{Li^{2+}}$?

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 $\frac{9}{4}$.

Answer 1

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: $$E_1=-E_\mathrm{i,1}=-8.72\times 10^{-18}\,\mathrm {J/atom}$$