Very Small Physics 3

Treating the set-up as a classical two-body problem, the electrostatic attraction between the nucleus and the electron must provide the necessary centripetal acceleration, and so $\frac{Ze^2}{4\pi \varepsilon_0 r^2} \; = \; \frac{m_ev^2}{r} \hspace{2cm} \mathrm{or} \hspace{2cm} m_ev^2r \; = \; \frac{Ze^2}{4\pi \varepsilon_0}$ We are assuming that angular momentum is quantized, and that the electron is in the ground state. Thus we require $m_e vr \; = \; \hbar$ In both of these equations $r$ is the radius of the electron's orbit, and $v$ the electron's velocity. Thus $v \; = \; \frac{Ze^2}{4\pi\varepsilon_0\hbar}$ In atomic units, this means that $v = Z$ , and hence that $\gamma = \big(1 - \tfrac{Z^2}{c^2}\big)^{-\frac12}$ . Thus $\gamma \ge 1.05$ when $Z \; \ge \; c\frac{\sqrt{1.05^2 - 1}}{1.05} = 0.3049c$ With $c \approx 137$ we deduce that $Z \ge 41.773$ , and hence $Z \ge 42$ , making $\boxed{\mathrm{Molybdenum}}$ the answer.

The run down

In the second Very Small Physics problem, we treated an atom as a two-body problem and found the radius of the ground state of hydrogen. In this problem we'll do something similar, but allow the charge on the nucleus $Z$ to vary. Then we'll find the kinetic energy associated with the ground state electron, and find it's velocity. When the charge $Z$ gets higher than 42, the velocity of the ground state electron becomes greater than $30 \%$ of the speed of light and results in a Lorentz correction $\gamma > 1.05$ .

Total Energy

As we did in the Intermediate problem, we set up the total energy of an electron with a confinement energy. This energy was yielded by making an approximation for the coupling of momentum and position in a quantum mechanical electron according to the uncertainty principle: $rp \sim \hbar$ . This time we'll allow the charge in the nucleus to vary, and so there is a factor of $Z$ in the Coulombic term.

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

This simplifies greatly if we use atomic units , a great suggestion by @Mark Hennings :

$E_{tot} = \frac{1}{2 r^2} - \frac{1}{ r}$

We can now see that the kinetic and potential energies are coupled , and both depend on the electron-nucleus separation. The confinement energy approaches positive infinity much faster than the Coulombic energy approaches negative infinity as the radius goes to zero. The expected radius of an electron can be found by minimizing this energy:

$\begin{aligned} \frac{dE}{dr} &= \frac{-1}{r^3} + \frac{Z}{r^2} = 0\\ \frac{1}{r^3} &= \frac{Z}{r^2} \\ r &= \frac{1}{Z} \end{aligned}$

By using our uncertainty principle approximation, the momentum of the electron at this expected radius is given by:

$p = \frac{1}{r} = Z$

The "velocity" of the electron is just this momentum divided by the mass of the electron ( $m_e = 1$ in atomic units).

$v = \frac{\lvert p \rvert}{m_e} = Z$

In Hydrogen ( $Z=1$ ) the electron isn't even close to relativistic. A velocity of $v = 1$ is about 137th the speed of light in these units. We can calculate the Lorentz factor for this Hydrogen ground state electron without using any physical constants other than $c = 137$ in atomic units:

$\gamma = \frac{1}{sqrt{1-Z^2/c^2}} = 1.0000266$

This is a tiny relativistic correction.

To reach a Lorentz factor of $\gamma = 1.05$ , we need $v/c = 0.3$ , or $30 \%$ the speed of light. Based on the linear relation between velocity and charge $Z$ , we just need $0.3/0.0073 = 41.1$ times the charge. The element with an atomic number of $42$ is Molybdenum. In atomic units this calculation is even easier:

$\frac{Z}{c} = \frac{Z}{137} \geq 0.3$

So again the minimum charge is $Z = 42$ .

For elements larger than Molybdenum, relativistic effects begin to significantly change the orbits of the electrons. These effects are possibly most well known for the color of gold (element 79, Au). In Gold, not only are the ground state orbits relativistic, but so is the highest occupied orbital, $6s$ . Electrons in $6s$ are traveling so fast that their increased relativistic weight leads to a contracted orbital radius and a lower energy.

This narrows the energy gap between the two highest orbitals $5d$ and $6s$ . In lower $Z$ elements of the same period (like Silver), this gap ( $4d \rightarrow 5s$ ) is wide enough that only UV light can excite electrons up to this state. So silver absorbs UV but reflects the entire visible spectrum, resulting in a white coloring. In gold, the narrowed gap can be excited with blue light, and so gold absorbs blue but reflects the rest of the visible spectrum. We perceive this color as yellow.

I generally don't care for the physics problems because they are like easy math problems with hidden information. But I take a stab at them and can sometimes get the answer. The following solution took some lucky guesses.

I took a hint from an intermediate problem for this week. Plugging all the 'atomic units' into the solution formula gives a radius of 1. That's from having a +1 in the nucleus. What happens if we crank it up? Faster electron. How much faster?

The two body problem link gives $v=\sqrt{\frac{GM}{r}}$ . I figured the $M$ and $r$ can't change so $v$ is probably proportional to $\sqrt{G}$ . $G$ is usually a gravitational force, maybe here it can be substituted for the charge of the nucleus.

Solving the $\gamma$ formula gives $v \geq 41.77$ Molybdenum is #42. That can't be a coincidence.

Something is slightly wrong, though, because it looks like $v$ is actually proportion to $G$ . Or I made some other mistake. I don't really care. Back to the math problems, where everything is laid out.

I don't know much about physics, but Uranium, Lead and Helium are all elements people talk about. You never hear about Molybdenum. The only reason it could be one of the answer choices would be if it was the correct answer. Although not always effective, tapping into human psychology can help improve the odds of a random guess.