# 6.4 Bohr’s model of the hydrogen atom  (Page 5/37)

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## Electron orbits

To obtain the size ${r}_{n}$ of the electron’s n th orbit and the electron’s speed ${v}_{n}$ in it, we turn to Newtonian mechanics. As a charged particle, the electron experiences an electrostatic pull toward the positively charged nucleus in the center of its circular orbit. This electrostatic pull is the centripetal force that causes the electron to move in a circle around the nucleus. Therefore, the magnitude of centripetal force is identified with the magnitude of the electrostatic force:

$\frac{{m}_{e}{v}_{n}^{2}}{{r}_{n}}=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{{e}^{2}}{{r}_{n}^{2}}.$

Here, $e$ denotes the value of the elementary charge. The negative electron and positive proton have the same value of charge, $|q|=e.$ When [link] is combined with the first quantization condition given by [link] , we can solve for the speed, ${v}_{n},$ and for the radius, ${r}_{n}:$

${v}_{n}=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{{e}^{2}}{\hslash }\phantom{\rule{0.2em}{0ex}}\frac{1}{n}$
${r}_{n}=4\pi {\epsilon }_{0}\frac{{\hslash }^{2}}{{m}_{e}{e}^{2}}{n}^{2}.$

Note that these results tell us that the electron’s speed as well as the radius of its orbit depend only on the index n that enumerates the orbit because all other quantities in the preceding equations are fundamental constants. We see from [link] that the size of the orbit grows as the square of n . This means that the second orbit is four times as large as the first orbit, and the third orbit is nine times as large as the first orbit, and so on. We also see from [link] that the electron’s speed in the orbit decreases as the orbit size increases. The electron’s speed is largest in the first Bohr orbit, for $n=1,$ which is the orbit closest to the nucleus. The radius of the first Bohr orbit is called the Bohr radius of hydrogen    , denoted as ${a}_{0}.$ Its value is obtained by setting $n=1$ in [link] :

${a}_{0}=4\pi {\epsilon }_{0}\frac{{\hslash }^{2}}{{m}_{e}{e}^{2}}=5.29\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\text{m}=0.529\phantom{\rule{0.2em}{0ex}}\text{Å}.$

We can substitute ${a}_{0}$ in [link] to express the radius of the n th orbit in terms of ${a}_{0}:$

${r}_{n}={a}_{0}{n}^{2}.$

This result means that the electron orbits in hydrogen atom are quantized because the orbital radius takes on only specific values of ${a}_{0},4{a}_{0},9{a}_{0},16{a}_{0},...$ given by [link] , and no other values are allowed.

## Electron energies

The total energy ${E}_{n}$ of an electron in the n th orbit is the sum of its kinetic energy ${K}_{n}$ and its electrostatic potential energy ${U}_{n}.$ Utilizing [link] , we find that

${K}_{n}=\frac{1}{2}{m}_{e}{v}_{n}^{2}=\frac{1}{32{\pi }^{2}{\epsilon }_{0}^{2}}\phantom{\rule{0.2em}{0ex}}\frac{{m}_{e}{e}^{4}}{{\hslash }^{2}}\phantom{\rule{0.2em}{0ex}}\frac{1}{{n}^{2}}.$

Recall that the electrostatic potential energy of interaction between two charges ${q}_{1}$ and ${q}_{2}$ that are separated by a distance ${r}_{12}$ is $\left(1\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}4\pi {\epsilon }_{0}\right){q}_{1}{q}_{2}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}{r}_{12}.$ Here, ${q}_{1}=+e$ is the charge of the nucleus in the hydrogen atom (the charge of the proton), ${q}_{2}=\text{−}e$ is the charge of the electron and ${r}_{12}={r}_{n}$ is the radius of the n th orbit. Now we use [link] to find the potential energy of the electron:

${U}_{n}=-\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{{e}^{2}}{{r}_{n}}=-\frac{1}{16{\pi }^{2}{\epsilon }_{0}^{2}}\phantom{\rule{0.2em}{0ex}}\frac{{m}_{e}{e}^{4}}{{\hslash }^{2}}\phantom{\rule{0.2em}{0ex}}\frac{1}{{n}^{2}}.$

The total energy of the electron is the sum of [link] and [link] :

${E}_{n}={K}_{n}+{U}_{n}=-\frac{1}{32{\pi }^{2}{\epsilon }_{0}^{2}}\phantom{\rule{0.2em}{0ex}}\frac{{m}_{e}{e}^{4}}{{\hslash }^{2}}\phantom{\rule{0.2em}{0ex}}\frac{1}{{n}^{2}}.$

Note that the energy depends only on the index n because the remaining symbols in [link] are physical constants. The value of the constant factor in [link] is

${E}_{0}=\frac{1}{32{\pi }^{2}{\epsilon }_{0}^{2}}\phantom{\rule{0.2em}{0ex}}\frac{{m}_{e}{e}^{4}}{{\hslash }^{2}}=\frac{1}{8{\epsilon }_{0}^{2}}\phantom{\rule{0.2em}{0ex}}\frac{{m}_{e}{e}^{4}}{{h}^{2}}=2.17\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-18}\text{J}=13.6\phantom{\rule{0.2em}{0ex}}\text{eV}.$

It is convenient to express the electron’s energy in the n th orbit in terms of this energy, as

${E}_{n}=\text{−}{E}_{0}\frac{1}{{n}^{2}}.$

Now we can see that the electron energies in the hydrogen atom are quantized because they can have only discrete values of $\text{−}{E}_{0},\text{−}{E}_{0}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}4,\text{−}{E}_{0}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}9,\text{−}{E}_{0}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}16,...$ given by [link] , and no other energy values are allowed. This set of allowed electron energies is called the energy spectrum of hydrogen    ( [link] ). The index n that enumerates energy levels in Bohr’s model is called the energy quantum number    . We identify the energy of the electron inside the hydrogen atom with the energy of the hydrogen atom. Note that the smallest value of energy is obtained for $n=1,$ so the hydrogen atom cannot have energy smaller than that. This smallest value of the electron energy in the hydrogen atom is called the ground state energy of the hydrogen atom    and its value is

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