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?

  • 6
    $\begingroup$ You are right on target. Go ahead and use the Bohr model. You should be able to get the energy levels of the electron in the Bohr model from your text and you would use the most stable (ground) state. $\endgroup$ Commented Jan 28, 2018 at 11:36

1 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]: $E_\mathrm{i}^\mathrm{IV}(\ce{Be}) = \pu{217.71865 eV}$.


  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.
  • 2
    $\begingroup$ I was also taught the straight forward formula for energy of the $n$-th orbit in atom $Z$, as $E_n^Z=-13.6\cdot\frac{Z^2}{n^2}$, using the Bohr model. Posting just in case anyone else finds it useful. $\endgroup$ Commented Jan 31, 2018 at 1:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.