# Difference between revisions of "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 ${\displaystyle a_{0}}$ (the Bohr radius), and the velocity that the electron travels at will be ${\displaystyle v_{e}}$. This gives us the following expression for the electron’s angular momentum for ${\displaystyle n}$ units of ${\displaystyle \hbar }$:

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

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

${\displaystyle m_{e}{\frac {v_{e}^{2}}{a_{0}}}={\frac {Ze^{2}}{a_{0}^{2}}}\,\!}$

This gives us the following solution for the Bohr radius:

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

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 as half of the energy we get integrating the electric force from ${\displaystyle r=a_{0}\rightarrow \infty }$, (the other half is the kinetic energy of the orbiting electron):

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

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

We can estimate B by just applying a Lorentz boost to the electric field of the nucleus:

${\displaystyle B\sim {\frac {Ze}{a_{0}^{2}}}{\frac {v}{c}},\,\!}$

with ${\displaystyle {\frac {v}{c}}\sim {\frac {Ze^{2}}{\hbar c}}}$. For reference, ${\displaystyle {\frac {v}{c}}}$ for ${\displaystyle Z=1}$ evaluates to ${\displaystyle {\frac {e^{2}}{\hbar c}}\equiv \alpha \approx {\frac {1}{137}}}$, which is the fine structure constant.

Next we need to estimate the magnetic dipole of an electron ${\displaystyle \mu _{e}}$. The ${\displaystyle {\vec {B}}}$ of a magnetic dipole goes as ${\displaystyle B_{di}\sim {\frac {\mu }{r^{3}}}}$, so we can estimate the magnetic dipole of an electron as something that produces the ${\displaystyle {\vec {B}}}$ of a spinning electron at ${\displaystyle r_{e}}$, the classical electron radius:

${\displaystyle \mu \sim B_{e}{\big |}_{r_{e}}r_{e}^{3}.\,\!}$

We can estimate ${\displaystyle r_{e}}$ by setting the rest mass energy of the ${\displaystyle e^{-}}$ to the electrostatic potential energy of the electron:

{\displaystyle {\begin{aligned}m_{e}c^{2}&\sim {\frac {e^{2}}{r_{e}}}\\r_{e}&\sim {\frac {e^{2}}{m_{e}c^{2}}}\\\end{aligned}}\,\!}

To estimate ${\displaystyle B_{e}}$, we can use Maxwell’s equation:

{\displaystyle {\begin{aligned}{\vec {\nabla }}\times {\vec {B}}&={\frac {4\pi }{c}}J\\{\frac {B_{e}}{2\pi r_{e}}}&={\frac {4\pi }{c}}{\frac {I}{4\pi r_{e}^{2}}}\\\end{aligned}}\,\!}

We can estimate the current, ${\displaystyle I}$ from the spin timescale of the electron, ${\displaystyle I\sim {\frac {e}{t_{spin}}}}$, and because the electron spin is quantized, ${\displaystyle \hbar =m_{e}r_{e}^{2}{\frac {2\pi }{t_{spin}}}}$. After some algebra, we get that:

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

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

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

(Bohr magneton for nucleus) Finally, we plug these in for our expression for the energy:

{\displaystyle {\begin{aligned}\Delta E&\sim {\frac {e\hbar }{2m_{e}c}}{\frac {Ze}{a_{0}^{2}}}Z\alpha \\&\sim Z^{4}\alpha \cdot {\rm {Ryd}}\\\end{aligned}}\,\!}

where we take the Rydberg to be 13.6 eV and have factored out its ${\displaystyle Z}$ dependence explicitly.

• 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 ${\displaystyle \hbar }$.