-2
$\begingroup$

My textbook, Silberberg & Amateis, Chemistry: The Molecular Nature of Matter and Change (9th ed.), gives the following equation for calculating the energy level of an atom: $$E = \pu{-2.18E-18 J}\left(\frac{Z^2}{n^2}\right)$$ Where $Z$ is the charge of the nucleus.

They never explained what the constant term that is multiplying the (Z/n)^2 term means, or where it comes from. Where does this come from? It looks similar to the Rydberg equation, but I still can't see its relation to it exactly.

$\endgroup$
2
  • 1
    $\begingroup$ Well, if you figure out how to compute the energy associated with a particular wavelength of light, you can check for yourself if it is Rydbergs equation. $\endgroup$
    – Buck Thorn
    Mar 3, 2023 at 18:40
  • 3
    $\begingroup$ Rydberg formula is but an empirical observation, it doesn't explain anything either. The real explanation comes from quantum mechanics. $\endgroup$ Mar 3, 2023 at 19:02

1 Answer 1

3
$\begingroup$

Essentially, the equation conveys how the energy of a one-electron atom varies with factors such as the nuclear charge and electronic orbit number.

$-2.18\times10^{-18}\ \mathrm J$ is an empirically obtained factor, but you can also derive it from first principles using Bohr's atomic model.

$$2.18\times10^{-18}\ \mathrm J=\frac{(k_\mathrm ee^2)^2m_\mathrm e}{2\hbar^2}$$

Where $k_\mathrm e$ is Coulomb's constant, $e$ is the electronic charge, $m_\mathrm e$ is the mass of an electron, and $\hbar$ is the reduced Planck constant.

You can find the derivation on Wikipedia: https://en.wikipedia.org/wiki/Bohr_model

Intuitively, you can think of $-2.18\times10^{-18}\ \mathrm J$ as the energy of a hydrogen atom system with an electron in its first orbit. Here $Z=1$ and $n=1$.

$\endgroup$

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.