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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?

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    $\begingroup$ 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. $\endgroup$ – Oscar Lanzi Jan 28 '18 at 11:36
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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 [1, p. 10-204]: $E_\mathrm{i}^\mathrm{IV}(\ce{Be}) = \pu{217.71865 eV}$.

Reference:

  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.; 2017; Vol. 97.
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    $\begingroup$ Thanks a lot! this solved my problem and organized my data. @andselisk $\endgroup$ – Mabadai Jan 29 '18 at 18:50
  • $\begingroup$ @Mabadai You are very welcome:) $\endgroup$ – andselisk Jan 29 '18 at 21:06
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    $\begingroup$ 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. $\endgroup$ – Gaurang Tandon Jan 31 '18 at 1:05

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