5.2 The Bohr Model of Atom

Niels Bohr proposed the Bohr model of atom in 1913, which agreed well with experimental evident that time. He won the 1922 Nobel Prize "for his services in the investigation of the structure of atoms and of the radiation emanating from them". Now let's take a close look at the Bohr Model.

In the Bohr Model, electron is in a circular orbit around the nucleus. In this case, all centripetal force is provided by the electron-magnetic attraction between the electron itself and the nucleus. Suppose the distance between electron and nucleus is $r$, and the speed of electron is $v$. For a Hydrogen atom, we have

$$ \frac{mv^2}{r}=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}. $$

Here $m=9.109\times10^{-31}\ kg$ is the electron mass. $\varepsilon_0=8.854\times10^{12}\ C^2N^{-1}m^{-2}$ is the vacuum permittivity. $e=1.602\times10^{-19}\ C$ is the charge of electron. Or, we can write the expression as

$$ v=\sqrt{\frac{e^2}{4\pi\varepsilon_0m}\frac{1}{r}}. $$

The energy of the electron in this case can also be found simply by adding its kinetic energy and potential energy. Since we have the expression for its speed $v$, the kinetic part is pretty easy. The electric potential is also not hard to find. After simplification, the result would just be

$$ E=-\frac{e^2}{8\pi\varepsilon_0r}. $$

5.2.1 Angular Momentum

As a short detour, let's take a moment to analyze the angular momentum. Classically, we have angular momentum

$$ L=mvr. $$

Here Bohr made an assumption much like the one Planck did when solving the problem of Black-body radiation. Planck assumed that energy has smallest indivisible part which equals to $hf$. Bohr also assumed that the angular momentum also has such a smallest indivisible portion which is $\hbar=h/2\pi$. Therefore,

$$ L=mvr=n\hbar. $$

An alternative interpretation is that, we can treat the electron as a wavefunction circling around the orbit (wave-particle duality). Because the wavefunction has to be smooth, we require the circumference of the orbital to be an integer multiplication of wavelength:

$$ 2\pi r=n\lambda. $$

From the Section 2.2 we know the wavelength of a matter wave follows

$$ \lambda=\frac{h}{p}=\frac{h}{mv}. $$

Therefore, combining above two expression gives

$$ \begin{align} 2\pi r&=n\frac{h}{mv}\\ mvr&=n\frac{h}{2\pi}\\ &=n\hbar=L. \end{align} $$

The same result yeilds pretty naturally.

With above results, we can find the radius $r$ to be

$$ \bbox[silver]{r=\frac{(4\pi\varepsilon_0)\hbar^2}{me^2}n^2}. $$

We can notate constants in front as $a_0={(4\pi\varepsilon_0)\hbar^2}/{me^2}\approx0.0529\ nm$, which is known as the Bohr radius. With this expression for radius $r$, the energy can be written as

$$ \begin{gather} \bbox[silver]{E=-\frac{me^4}{2(4\pi\varepsilon_0)^2\hbar^2}\frac{1}{n^2}}\\ =-13.6\ eV\frac{1}{n^2}. \end{gather} $$