Classical Bohr Atom

Short Topical Videos

• Radiation and Matter, to Order of Magnitude (Aaron Parsons)
• Bohr Hydrogen Model 1: Radius (Quantum Chemistry)
• Bohr Hydrogen Model 2: Energy (Quantum Chemistry)
• Rotational Spectroscopy Example (Quantum Chemistry)

Reference Material

• Bohr Model (Wikipedia)
• Diatomic Molecule (Wikipedia)
• Rotational Levels (Tuckerman, NYU)

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-$\alpha$ and Lyman-$\alpha$ 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 $\hbar$. 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 $a_{0}$, and the velocity that the electron travels at will be $v_{e}$. This gives us the following expression for the angular momentum for $n$ units of $\hbar$:

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

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

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

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

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

• Fine Structure:

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

$\Delta E\sim \mu B\,\!$

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

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

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

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

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

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

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

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

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

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

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

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

• Hyperfine Structure:

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

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

Thus:

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

Note that Fine and Hyperfine are magnetic dipole transitions. $e^{-}$ 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:

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

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

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

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 $\hbar$.