The four equations of the Bohr model

Once the quantisation rule for angular momentum is accepted, the Bohr model becomes an entirely solvable problem in mechanics. The electron circles the nucleus in a circular orbit, held in its path by the electric attraction of the proton. To find which orbits are allowed — that is, the radius $r$ and the velocity $v$ at each level — only four equations are needed.

Principle — The 4 equations of the hydrogen atom

For the electron in a quantised orbit of the hydrogen atom:

1. Quantisation rule: $m\,v\,r = n\,\hbar$
2. Circular motion: $|a| = v^2/r$
3. Newton’s second law: $\pq{F} = m\pq{a}$
4. Coulomb’s law: $|F| = k\,e^2/r^2$

These are 4 equations in 4 unknowns ($r$, $v$, $F$, $a$).

Where does quantum mechanics come in?

Of the four equations, only the first is “new”. The other three — circular motion, Newton and Coulomb — are pure classical physics, the same ones we would use for a satellite or for a charge in a field. Quantum mechanics enters only through the quantisation rule, as a single ingredient added to an otherwise classical world. This is the signature of Bohr’s “old quantum mechanics”: an almost intact classical physics, with a single quantised bolt.

Solving the system. From the quantisation rule (1) we obtain the velocity, $v = n\hbar/(mr)$. Newton’s second law (3) applied to circular motion requires that the Coulomb force (4) act as the centripetal force $mv^2/r$ (2):

$\frac{k\,e^2}{r^2} = m\cdot\frac{v^2}{r} = m\cdot\frac{n^2\hbar^2}{m^2 r^3}$

Solving for $r$ gives the radius of the $n$-th orbit:

$\ev{r_n = \frac{n^2\,\hbar^2}{m\,e^2\,k}}$

The radius grows as $n^2$: the allowed orbits are not evenly spaced, but move increasingly further from the nucleus. For $n=1$ we obtain the fundamental scale of the atom, the Bohr radius:

$r_1 = \frac{\hbar^2}{m\,e^2\,k} \approx 5{,}29\cdot 10^{-11}\ \text{m} = 0{,}53\ \text{Å}$

about half an ångström: this is the “size” of the hydrogen atom, and it fixes the order of magnitude of all ordinary matter. The corresponding velocity is $v_n = e^2 k/(n\hbar)$, and decreases as $1/n$: electrons in outer orbits are slower. For the ground level we write $v_1 = \alpha\,c$, where

$\alpha = \frac{e^2 k}{\hbar c} \approx \frac{1}{137}$

is the fine-structure constant, one of the most mysterious numbers in physics: dimensionless, universal, with a value that nobody yet knows how to explain “why” it is exactly that one.

An electron at 1% of the speed of light

The constant $\alpha \approx 1/137$ measures the size of atomic relativistic effects: since $v_1/c = \alpha \approx 1/137$, the electron in the hydrogen ground state travels at about $1\%$ of the speed of light — slow enough that non-relativistic mechanics can be used as an excellent first approximation.

Links

Topics: Quantum physics Concepts: Bohr model · Angular momentum · Coulomb’s law · Centripetal force · Uniform circular motion · Newton’s second law Objects: Hydrogen atom

Related exercises: Conical pendulum · True or false on circular motion and inclined plane · Problem — Two charges in circular motion