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

$m_{e}v_{e}a_{0}=n\hbar \,\!$ Next, we balance the force required to keep the $e^{-}$ in a circular orbit against the electric force:

$m_{e}{\frac {v_{e}^{2}}{a_{0}}}={\frac {Ze^{2}}{a_{0}^{2}}}\,\!$ This gives us the following solution for the Bohr radius:

$a_{0}={\frac {\hbar ^{2}}{m_{e}Ze^{2}}}\approx 0.52{\frac {\mathrm {\AA} }{Z}}\,\!$ A Rydberg is the energy required to ionize an H atom from the ground state. It is $\sim 13.6eV$ . We can estimate it as half of the energy we get integrating the electric force from $r=a_{0}\rightarrow \infty$ , (the other half is the kinetic energy of the orbiting electron):

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

Fine Structure comes from the interaction of the magnetic moment of the $e^{-}$ 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:

$\Delta E\sim \mu B\,\!$ We can estimate B by just applying a Lorentz boost to the electric field of the nucleus:

$B\sim {\frac {Ze}{a_{0}^{2}}}{\frac {v}{c}},\,\!$ with ${\frac {v}{c}}\sim {\frac {Ze^{2}}{\hbar c}}$ . For reference, ${\frac {v}{c}}$ for $Z=1$ evaluates to ${\frac {e^{2}}{\hbar c}}\equiv \alpha \approx {\frac {1}{137}}$ , which is the fine structure constant.

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

$\mu \sim B_{e}{\big |}_{r_{e}}r_{e}^{3}.\,\!$ We can estimate $r_{e}$ by setting the rest mass energy of the $e^{-}$ to the electrostatic potential energy of the electron:

{\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 $B_{e}$ , we can use the Maxwell Equations:

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

$\mu _{e}={\frac {e\hbar }{2m_{e}c}}\,\!$ (Bohr magneton for $e^{-}$ )

$\mu _{p}={\frac {Ze\hbar }{2m_{p}c}}\,\!$ (Bohr magneton for nucleus) Finally, we plug these in for our expression for the energy:

{\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 $Z$ dependence explicitly.

• 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 $\hbar$ .