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Jan 27 2007

Derive the Bohr radius in a pinch

We all know the Bohr radius is about half an angstrom. But what if you want to derive it exactly in terms of fundamental constants, and you're stuck in the mountains without your textbook? Check this out: all we need is the Uncertainty Principle. (This discussion is inspired by the Feynman Lectures.)

To begin, we say that in a hydrogen atom, the electron is around a proton somewhere. We don't expect it to always be in the nucleus (which would make its momentum indeterminate anyway) but instead to have a spread that we define as

\Delta x = a.
(1)

Heisenberg's principle says that the product of the spread in position and momentum is at least \hbar , so

\Delta p \sim \hbar/a
(2)

With a bit of sleight of hand, we say that the actual values of r and p are approximately equal to these uncertainties

r \sim a, \quad p \sim \hbar/a.
(3)

Write down the energy, which is just the sum of the kinetic term and potential energy due to
Coulomb attraction

E=\frac{\hbar}{2ma^2} - \frac{e^2}{4 \pi \epsilon_0} \frac{1}{a}
(4)

(in SI units), and minimize the energy

0 = \frac{dE}{da} = \frac{-\hbar^2}{ma^3}+\frac{e^2}{4 \pi \epsilon_0 a^2}.
(5)

Solving for a , we find that

a=\frac{4 \pi \epsilon_0 \hbar^2}{m e^2} \approx \unit[0.5 \times 10^{-10}]{m}.
(6)

It's rigged! This is a stupid calculation that could have gone very wrong. Instead, all of the approximations, which could have been off by a factor of 10, work out to the right values. Amazing.