Derive the radius of the Bohr orbit

Problem

Starting from Bohr’s quantisation rule $m v r = n\hbar$ and from the equality between the Coulomb force and the centripetal force, show that the radius of the $n$-th orbit in the hydrogen atom is $r_n = n^2 \hbar^2/(m e^2 k)$.

Solution

Step 1 — Coulomb = centripetal. The electrostatic attraction provides the centripetal force:

$\frac{k e^2}{r^2} = \frac{m v^2}{r} \;\Rightarrow\; k e^2 = m v^2 r$

Step 2 — Quantisation of angular momentum. From Bohr’s rule I obtain the velocity:

$m v r = n\hbar \;\Rightarrow\; v = \frac{n\hbar}{m r}$

Step 3 — Substitute $v$ into the dynamical relation.

$k e^2 = m r \left(\frac{n\hbar}{m r}\right)^2 = m r \cdot \frac{n^2\hbar^2}{m^2 r^2} = \frac{n^2\hbar^2}{m r}$

Step 4 — Isolate $r$.

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

For $n=1$ one obtains the Bohr radius $a_0 = \hbar^2/(m e^2 k) \approx 5{,}29\cdot 10^{-11}\;\text{m}$, the length scale of the atom. The radii grow as $n^2$: the excited orbits are much larger.

Links

Topics: Fisica quantistica Concepts: Modello di Bohr Skills: Impostazione simbolica Objects: Atomo di idrogeno