# Is there a quick yet semi rigorous way to derive the Bohr radius of hydrogen?

I will be taking an oral exam, where I have to do some "airport physics", fast and easy magnitude estimations.

Currently I try to come up with a good way to find the Bohr radius of the hydrogen atom for this exam. Usually one would solve the differential equation by splitting the Hamiltonian in radial and angular momentum parts. But all of this needs a lot of calculations and is boring.

For the harmonic oscillator I found that one can easily recover the characteristic length by using the virial theorem and Schrödinger uncertainty relation. Is there a similarly nifty reasoning for the hydrogen atom? I think this particular method of using the virial theorem does only work for $$\propto x^2$$ potentials, I wasn't able to generalize it. Because of this I'm looking for a different method.

Just guessing the length by dimensional analysis does not count though, this is cheating.

• Comments are not for extended discussion; this conversation has been moved to chat. – David Z Jun 2 at 4:35

Have a look at The Feynman Lectures on Physics, Volume 3 (chapter 2, section 2-4). There Feynman determines $$a$$ by minimizing the energy of the atom. The total energy is roughly $$E = \frac{\hbar^2}{2ma^2} - \frac{e^2}{a}$$ Above you need to understand why equating the momentum $$p$$ with $$\Delta\,p\approx \frac{\hbar}{a}$$ is reasonable. Also, as defined in Volume I, $$e^2$$ here is the charge of an electron squared, divided by $$4\pi\varepsilon_0$$.

To minimize $$E$$ one differentiates wrt $$a$$ and set the result equal to $$0$$ and solve for $$a$$. $$\frac{dE}{da} = -\frac{\hbar^2}{ma^3} + \frac{e^2}{a^2}$$ Setting $$\frac{dE}{da} = 0$$ gives $$a = \frac{\hbar^2}{me^2} \approx 0.5\times10^{-10}$$m.

So, the "quick yet semi rigorous way to derive the Bohr radius hydrogen'' you're looking for is an application of the uncertainty principle.

From the Virial theorem, $$\langle T\rangle = -\frac{1}{2}\langle V\rangle \rightarrow \langle E \rangle = \langle T\rangle + \langle V\rangle = \frac{1}{2}\langle V \rangle$$. Since the potential is given by $$V(r) = -\frac{e^2}{4\pi \epsilon_0 r}$$, this tells you that in an energy eigenstate,

$$\frac{e^2}{8\pi\epsilon_0} \langle \frac{1}{r}\rangle = \frac{13.6 eV}{n^2}$$

or

$$\langle \frac{1}{r}\rangle = \frac{13.6 \text{ eV}}{n^2} \left(\frac{8\pi\epsilon_0}{e^2}\right)$$

Inverting this$$^\dagger$$ and setting $$n=1$$ yields the Bohr radius

$$r = \frac{e^2}{8\pi\epsilon_0} \frac{1}{13.6\text{ eV}}\approx 5.3 \times 10^{-11}\text{ m}$$

$$^\dagger$$The Bohr radius provides the characteristic length scale for the problem. If you wish, you could recall that ground state wavefunction is of the form $$\psi(r,\theta,\phi)=c e^{-r/a}$$, and evaluate the expected value of $$\frac{1}{r}$$; if you do so, you will find $$\langle \frac{1}{r}\rangle = \frac{1}{a}$$, justifying the inversion above.

Alternatively, you could go back to the Bohr model, in which one has that the centripetal force is provided by the electrostatic force

$$\frac{mv^2}{r} = \frac{e^2}{4\pi\epsilon_0 r^2} \implies r = \frac{e^2}{4\pi \epsilon_0 mv^2}$$

and that the angular momentum is an integer multiple of $$\hbar$$: $$L = mvr = n\hbar \implies v^2 = \frac{n^2 \hbar^2}{m^2r^2}$$

which together yield that $$r = \frac{e^2}{4\pi\epsilon_0 m}\frac{m^2 r^2}{n^2 \hbar^2} \implies r = \frac{4\pi \epsilon_0n^2\hbar^2}{m e^2}$$

setting $$n=1$$ yields the Bohr radius

$$r = \frac{4\pi \epsilon_0\hbar^2}{m e^2} \approx 5.3 \times 10^{-11}\text{ m}$$

• I don't think it's true in general that $\langle 1/r \rangle = 1/\langle r \rangle$. – Javier May 31 at 20:15
• @Javier I agree - If we start from the ansatz that $\psi \propto e^{-r/a}$, however, then $\langle 1/r\rangle = 1/a$. – J. Murray May 31 at 20:35

Assume from the Lyman series that $$E_1 = -13.6$$eV. Assume the electron is trapped in an infinite box, $$E_1 = \frac{\pi^2 \hbar^2}{2m a_o^2} = 13.6 \, \mathrm{eV}$$ then $$a^2_o \approx 2.74 \times 10^{-20}\mathrm{m}^2$$ which means $$a_o \approx 1.6 \times 10^{-10}$$m.