Very Small Physics 2

Let's work in the spirit of early quantum mechanics and model the atom by a massive stationary nucleus, with the electron kept in a circular orbit by a central force field and use Heisenberg's uncertainty relation to eliminate the electron's velocity from the formula. Here are the details -

If we differentiate the given potential, we find that the force on the electron is

$F=\frac{e^{2}}{4\pi\varepsilon_0 r^{2}}$

using Newton's second law we can equate this with mass times (the centripetal) acceleration of the electron to get

$\frac{e^{2}}{4\pi\varepsilon_0 r^{2}}=\frac{mv^{2}}{r} \dots(1)$ where m is the mass of the electron

Now $mv^{2}=\frac{p^{2}}{m}$ and invoking Heisenberg ( $rp\approx \hbar$ ) allows us to write (1) as

$mv^{2}\approx\frac{(\frac{\hbar}{r})^{2}}{m}$

Substituting this into equation (1) gives

$\frac{e^{2}}{4\pi\varepsilon_0 r^{2}}\approx\frac{\hbar^{2}}{mr^{3}}$

and easy algebra gives us our approximation

$\boxed{r\approx\frac{4\pi\varepsilon_0 \hbar^{2}}{me^{2}}}$

The run down

We need to break down the energy of an electron near a proton into it's kinetic and potential energy terms, and find a minimum. Classically that minimum would be the electron falling all the way into the proton, but quantum mechanics forbids that kind of spatial confinement, so there will hopefully be another minimum radius that the electron will tend to occupy. A classical Coulomb potential, with a minimum at $r=0$

Total energy

The total energy of our electron is the sum of its kinetic energy (bog-standard $E_k = \frac{p^2}{2m}$ ) and its potential energy, corresponding to the Coulomb potential.

$E_{tot} = \frac{p^2}{2m_e} - \frac{e^2}{4 \pi \epsilon_0 r}$

The problem here is that we'd like to minimize this energy with respect to the radius $r$ , but we can't really be sure what the relationship is between $p$ and $r$ . This is where the uncertainty principle comes in.

Uncertainty

This isn't a perfect calculation by any means, but it's a useful approximation. The trick here is to realize that the radius of the electron orbit must at the very least be larger than its minimum spread in position $\Delta x$ by the uncertainty principle. Similarly, the electron's momentum must be larger than the associated spread in momentum predicted by the principle. We've been instructed to use this approximation:

$\Delta x \Delta p \geq \hbar$

This can provide us with a relationship between $r$ and $p$ :

$r \geq \Delta x \\ p \geq \Delta p \geq \frac{\hbar}{\Delta x}$

So at a minimum the uncertainty principle limits the momentum of the electron according to $p \geq \frac{\hbar}{r}$ . The momentum could of course be must larger than this value, but to find the ground state (or lowest energy/momentum) we can use this approximation and change the $\geq$ into a $=$ .

Confinement Energy

The total energy of our electron is now:

$\begin{aligned} E_{tot} &= \frac{p^2}{2m_e} - \frac{e^2}{4 \pi \epsilon_0 r} \\ &= \frac{\hbar^2}{2m_2 r^2} - \frac{e^2}{4 \pi \epsilon_0 r} \end{aligned}$

We can now see that the kinetic and potential energies are coupled , and both depend on the electron-nucleus separation. This kinetic energy term is often termed a confinement energy , since it makes an energetic barrier stopping an electron from getting too close to the nucleus. The confinement energy approaches positive infinity much faster than the Coulombic energy approaches negative infinity as the radius approaches zero.

The Coulombic potential energy (blue), the confinement energy (red) and the total electron energy (black)

Minimize!

We can minimize this total energy with respect to radius to find the minimum energy:

$\begin{aligned} \frac{dE}{dr} &= \frac{-\hbar^2}{m_e r^3} + \frac{e^2}{4 \pi \epsilon_0 r^2} = 0\\ \frac{\hbar^2}{m_e r^3} &= \frac{e^2}{4 \pi \epsilon_0 r^2} \\ r &= \frac{4 \pi \epsilon_0 \hbar^2}{m_e e^2} \end{aligned}$

We could plug in the fundamental constants and yield an estimate for the radius of Hydrogen: $\SI{0.53}{\angstrom}$ .

Since the radius of the electron's orbit and its momentum/energy are coupled, we can calculate the energy corresponding to this ground state as well by plugging in the ground state radius:

$\begin{aligned} E_{tot} &= \frac{\hbar^2}{2m r^2} - \frac{e^2}{4 \pi \epsilon_0 r} \\ &= \frac{m_e e^4}{2 (4 \pi \hbar \epsilon_0)^2} - \frac{m_2 e^4}{(4 \pi \hbar \epsilon_0)^2} &= - \frac{m_e e^4}{2 (4 \pi \hbar \epsilon_0)^2} \end{aligned}$

Plugging in fundamental constants here yields a ground state energy for hydrogen of $\SI{13.6}{\electronvolt}$ .

I don't know much about atoms, electrons, the Heisenberg bound and all the stuff that this question has to do with, so I had to guess. Here's my thought process:

1) The correct answer must:

a) Be reasonable. For example saying that a car has an acceleration of 2000 $\frac {m}{s^2}$ cannot by any means be a correct solution to a problem.

b) Look the most like all the others.

2) Looking at the answers we can see that both B and C contain $m_e$ in their denominator whereas A has $m_p$ . Therefore we can quite safely conclude that A might not be the correct answer.

3) Now we have to choose between B and C. We can observe that B has "2" while C has "4 $\pi$ " in the numerator. Also we notice that the shape of the Hydrogen atom is circular so we can deduce that $\pi$ might be included in the equation somehow.

Therefore, from options B and C we choose $\boxed{C}$ which contains " $\pi$ ". And voila. This method of course is not 100% accurate and should only be used as a last resort whenever we face a problem which we cannot solve.

Yeah, It's Bohr radius