Ionization energy of beryllium

Question

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?

Answer 1

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:

$$\frac{1}{\lambda} = R_\infty Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

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.