# 6.2 The bohr model  (Page 3/9)

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Since Bohr’s model involved only a single electron, it could also be applied to the single electron ions He + , Li 2+ , Be 3+ , and so forth, which differ from hydrogen only in their nuclear charges, and so one-electron atoms and ions are collectively referred to as hydrogen-like atoms. The energy expression for hydrogen-like atoms is a generalization of the hydrogen atom energy, in which Z is the nuclear charge (+1 for hydrogen, +2 for He, +3 for Li, and so on) and k has a value of 2.179 $\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$ 10 –18 J.

${E}_{n}=\text{−}\frac{k{Z}^{2}}{{n}^{2}}$

The sizes of the circular orbits for hydrogen-like atoms are given in terms of their radii by the following expression, in which ${\text{α}}_{0}$ is a constant called the Bohr radius, with a value of 5.292 $×$ 10 −11 m:

$r=\phantom{\rule{0.2em}{0ex}}\frac{{n}^{2}}{Z}\phantom{\rule{0.2em}{0ex}}{a}_{0}$

The equation also shows us that as the electron’s energy increases (as n increases), the electron is found at greater distances from the nucleus. This is implied by the inverse dependence on r in the Coulomb potential, since, as the electron moves away from the nucleus, the electrostatic attraction between it and the nucleus decreases, and it is held less tightly in the atom. Note that as n gets larger and the orbits get larger, their energies get closer to zero, and so the limits $n\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty$ $n\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty ,$ and $r\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty$ $r\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty$ imply that E = 0 corresponds to the ionization limit where the electron is completely removed from the nucleus. Thus, for hydrogen in the ground state n = 1, the ionization energy would be:

$\text{Δ}E={E}_{n\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\infty }-{E}_{1}=0\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}k=k$

With three extremely puzzling paradoxes now solved (blackbody radiation, the photoelectric effect, and the hydrogen atom), and all involving Planck’s constant in a fundamental manner, it became clear to most physicists at that time that the classical theories that worked so well in the macroscopic world were fundamentally flawed and could not be extended down into the microscopic domain of atoms and molecules. Unfortunately, despite Bohr’s remarkable achievement in deriving a theoretical expression for the Rydberg constant, he was unable to extend his theory to the next simplest atom, He, which only has two electrons. Bohr’s model was severely flawed, since it was still based on the classical mechanics notion of precise orbits, a concept that was later found to be untenable in the microscopic domain, when a proper model of quantum mechanics was developed to supersede classical mechanics. Quantum numbers and energy levels in a hydrogen atom. The more negative the calculated value, the lower the energy.

## Calculating the energy of an electron in a bohr orbit

Early researchers were very excited when they were able to predict the energy of an electron at a particular distance from the nucleus in a hydrogen atom. If a spark promotes the electron in a hydrogen atom into an orbit with n = 3, what is the calculated energy, in joules, of the electron?

## Solution

The energy of the electron is given by this equation:

$E=\phantom{\rule{0.2em}{0ex}}\frac{-k{Z}^{2}}{{n}^{2}}$

The atomic number, Z , of hydrogen is 1; k = 2.179 $\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$ 10 –18 J; and the electron is characterized by an n value of 3. Thus,

$E=\phantom{\rule{0.2em}{0ex}}\frac{-\left(2.179\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-18}\phantom{\rule{0.2em}{0ex}}\text{J}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\left(1\right)}^{2}}{{\left(3\right)}^{2}}\phantom{\rule{0.2em}{0ex}}=-2.421\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\phantom{\rule{0.2em}{0ex}}\text{J}$

## Check your learning

The electron in [link] is promoted even further to an orbit with n = 6. What is its new energy?

## Answer:

−6.053 $×$ 10 –20 J

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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