Ionization energy of beryllium

Suppose I have $\ce{Be^3+}$. What would be its 4th ionization energy?


By trying to solve the issue I saw that its a "hydrogen-like" atom – means that beryllium left with only $1$ electron in his valance shell. Hence, I can use the Bohr atom model to solve the problem. 


Is my guess right? If so, what values I need to plug in the equation and why so?




In terms of Bohr model ionization potential $E\_\mathrm{i}$ is the work $A\_\mathrm{i}$ on eliminating an electron in vacuum from its current non-excited orbital level to infinity:


$$E\_\mathrm{i} = \frac{A\_\mathrm{i}}{e}$$
$$A\_\mathrm{i} = h\nu = \frac{hc}{\lambda}$$


Unknown wavelength $\lambda$ can be determined from the [Rydberg formula](https://en.wikipedia.org/wiki/Rydberg_formula):


$$\frac{1}{\lambda} = R\_\infty Z^2\left(\frac{1}{n\_1^2} - \frac{1}{n\_2^2}\right)$$


so that final equation for ionization energy looks like this:


$$E\_\mathrm{i} = \frac{hc}{e}R\_\infty Z^2\left(\frac{1}{n\_1^2} - \frac{1}{n\_2^2}\right)$$


Since we are determining 4th ionization energy of beryllium ($Z = 4$), $n\_1 = 1$ and $n\_2 = \infty$:


$$E\_\mathrm{i}^\mathrm{IV} = \frac{\pu{6.63e-34 m^2 kg s-1}\cdot\pu{3e8 m s-1}}{\pu{1.602e-19 C}}\cdot\pu{10973732 m-1}\cdot 4^2\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = \pu{217.86 eV}$$


This value is in a good agreement with the one listed in *CRC Handbook of Chemistry and Physics* [1, p. 10-204]: $E\_\mathrm{i}^\mathrm{IV}(\ce{Be}) = \pu{217.71865 eV}$.


### References


1. Haynes, W. M.; Lide, D. R.; Bruno, T. J. *CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data*, 97th ed.; CRC Press, **2016**. ISBN 978-1-4987-5429-3.


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