# Classical Bohr Atom

## Order of Magnitude Interaction of Radiation and Matter

It turns out that you can tell a cute little story about the energy levels in hydrogen-like atoms using a simple Bohr model of the hydrogen atom and classical physics. You just need a couple small doses of quantum mechanics in the beginning to get you started on the right track.

### Energy Levels

• Electronic Transitions:
Sketch of Rydberg-${\displaystyle \alpha }$ and Lyman-${\displaystyle \alpha }$ for hydrogen

We’ll start with the Bohr atom, and follow a classical derivation/estimation path. The only quantum mechanics that we need to inject is that angular momentum comes in units of ${\displaystyle \hbar }$. So let’s begin by quantizing the angular momentum associated with the electron’s orbit around the nucleus. We’ll define the radius of the electron’s orbit as ${\displaystyle a_{0}}$, and the velocity that the electron travels at will be ${\displaystyle v_{e}}$. This gives us the following expression for the angular momentum for ${\displaystyle n}$ units of ${\displaystyle \hbar }$:

${\displaystyle m_{e}v_{e}a_{0}=n\hbar \,\!}$

If we balance the force required to keep the ${\displaystyle e^{-}}$ in a circular orbit with the electric force:

${\displaystyle {m_{e}v_{e}^{2} \over a_{0}}={Ze^{2} \over a_{0}^{2}}\,\!}$
${\displaystyle a_{0}={\hbar ^{2} \over m_{e}e^{2}Z}\approx 0.52{\mathrm {\AA} \over Z}\,\!}$

A Rydberg is the energy required to ionize an H atom from the ground state. It is ${\displaystyle \sim 13.6eV}$. We can estimate it by integrating the electric force from ${\displaystyle r=a_{0}}$ to ${\displaystyle r=\infty }$, but in reality, there is another factor of 2:

${\displaystyle {Rydberg={Ze^{2} \over 2a_{0}}={Z^{2}e^{4}m_{e} \over 2\hbar ^{2}}=13.6\cdot Z^{2}eV}\,\!}$
• Fine Structure:

Fine Structure comes from the interaction of the magnetic moment of the ${\displaystyle e^{-}}$ with the a ${\displaystyle {\vec {B}}}$ caused by the Lorentz-transformed Coulomb field of the proton (generated by the ${\displaystyle e^{-}}$’s motion). The energy of a dipole interaction is ${\displaystyle E={\vec {\mu }}\cdot {\vec {B}}}$, so we’d expect that:

${\displaystyle \Delta E\sim \mu B\,\!}$

To estimate B: ${\displaystyle B\sim {Ze \over a_{0}^{2}}{v \over c}}$, and ${\displaystyle {v \over c}\sim {e^{2} \over \hbar c}=\alpha ={\frac {1}{137}}}$ (${\displaystyle \alpha }$ is the fine structure constant):

${\displaystyle B\sim {Ze^{3} \over \hbar ca_{0}^{2}}\,\!}$

Estimating ${\displaystyle \mu }$: The ${\displaystyle {\vec {B}}}$ of a dipole goes as ${\displaystyle B_{di}\sim {\mu \over r^{3}}}$, so ${\displaystyle \mu \sim B_{e}{\big |}_{r_{e}}r_{e}^{3}}$, where ${\displaystyle r_{e}}$ is the classical ${\displaystyle e^{-}}$ radius. Estimating that the rest mass energy of the ${\displaystyle e^{-}}$ should be about the electrostatic potential energy left in the ${\displaystyle e^{-}}$, ${\displaystyle m_{e}c^{2}\sim {e \over r_{e}}}$:

${\displaystyle r_{e}\sim {e^{2} \over m_{e}c^{2}}\,\!}$

We can estimate ${\displaystyle B_{e}}$ by reverting to Maxwell’s equation:

${\displaystyle {\vec {\nabla }}\times {\vec {B}}={4\pi J \over c}\,\!}$
${\displaystyle {B_{e} \over 2\pi r_{e}}={4\pi \over c}{I \over 4\pi r_{e}^{2}}\,\!}$

Again, we estimate that ${\displaystyle I\sim {e \over t_{spin}}}$, and because the electron spin is quantized, ${\displaystyle \hbar =m_{e}r_{e}^{2}{2\pi \over t_{spin}}}$. After some algebra, we get that:

${\displaystyle \mu _{e}={e\hbar \over 2m_{e}c}\,\!}$

(Bohr magneton for ${\displaystyle e^{-}}$)

${\displaystyle \mu _{p}={Ze\hbar \over 2m_{p}c}\,\!}$

(Bohr magneton for nucleus) Getting back to the energy:

{\displaystyle {\begin{aligned}\Delta E&\sim {e\hbar \over 2m_{e}c}{Ze \over a_{0}^{2}}\alpha \\&\sim Z^{3}\alpha ^{2}\cdot Ryd\\\end{aligned}}\,\!}
• Hyperfine Structure:

Instead of using the Bohr magneton, we use the intrinsic magnetic moment (spin) of the nucleus:

${\displaystyle B{\big |}_{p}\sim {\mu _{p} \over a_{0}^{3}}\sim B_{fine}{m_{e} \over m_{p}}\,\!}$

Thus:

${\displaystyle \Delta E\sim \Delta E_{fine}{m_{e} \over m_{p}}\,\!}$

Note that Fine and Hyperfine are magnetic dipole transitions. ${\displaystyle e^{-}}$ level transitions are electric dipole transitions. Magnetic dipole transitions are generally weaker.

• Vibrational Transitions in Molecules:

Our general technique with vibrational transitions is to model them as harmonic oscillators. Thus, they should have the characteristic harmonic energy series:

${\displaystyle E_{n}=(n+{\frac {1}{2}})\hbar \omega _{0}\,\!}$

For a harmonic oscillator, ${\displaystyle \omega _{0}={\sqrt {k \over m}}}$. We estimate that since the force for a spring is ${\displaystyle k\cdot x}$, and that force should be about the Coulomb force on ${\displaystyle e^{-}}$’s. If we say that atoms stretch with respect to each other about a Bohr radius:

${\displaystyle ka_{0}\sim {e^{2} \over a_{0}^{2}}\,\!}$
${\displaystyle \Delta E{\big |}_{vib \atop trans}\sim Ryd\cdot {\sqrt {m_{e} \over A\cdot m_{p}}}\,\!}$

where A is the atomic mass # of our atoms.

• Rotational Transitions in Molecules:

The thing to remember is that angular momentum comes in units of ${\displaystyle \hbar }$.