As a physicist, a way to think about the question of “relevance” is that of scales. In nature, there are time scales, length scales and energy scales. What therefore matters is not the minute details, but the order of magnitude. The difference in these scales generally requires different models and, in a nutshell, is also what makes physics hard!

In our problem, let’s compare the energy scales of thermal noise with the energy supplied (i.e electron volts). Whatever sources you look up for the energy due to thermal noise, you will find that the thermal energy $E_T$ at temperature $T$ is roughly \(E_T \sim k_B T.\)

Given that your operating temperature is $T=180^\circ C$, an order of magnitude estimation gives \(T\sim 10^2 K.\)

The Boltzmann constant $k_B$ has units of $\frac{\rm Energy}{\rm Temperature}$, and can be written in terms of $eV/K$ as

i.e the thermal noise is roughly \(E_T\sim 10^{-2} eV.\)

What this means is that any fluctuations you see will be on the 2nd decimal place to your right which is beyond the range that the FH power supply can work up to!

The Bohr model, that’s the problem! Specifically, the Bohr model is only valid for hydrogen-like atoms and our Hg atom is the last thing that is close to being hydrogen-like. However, most chemists uses hydrogen-like orbitals to do chemistry; the “s,p,d,f” orbitals we talk about are all solutions to the hydrogen atom shoe horned onto heavier atoms, so what gives? Start with the hydrogen atom which has one electron and a proton, the Bohr model tells us that the ground state corresponds to \(n=1\). The index “n” is what we call a quantum number and describes a quantum state. What’s interesting is that the solutions of the Schrödinger equation (for the hydrogen atom) produces new quantum numbers \(\ell\) and \(m\) from conservation of angular momentum. The states with \(\ell=0\) are what we call the “s” orbitals and the other funny letters are for \(\ell \geq 1\). So in the atomic configuration of H, we find

which is meant to be read as a state which has one electron in the so-called “s” shell in the \(n=1\) state. So if there are states labelled by different quantum numbers, then shouldn’t they have different energies contradicting the formula \(E_{n}= -13.6 {\rm eV}/n^2\)? It turns out that for the hydrogen atom, we’ve got a degeneracy! the energy levels only depend on the quantum number \(n\).

The story is different in the case of the atomic configuration of Hg in which you’ll find

The configuration tell us that that the outermost shell \(n=6\) contains two electrons in the “s” shell and is described succinctly as \(6s^2\). For non hydrogen-like atoms, the energy degeneracy is lifted due to the simple fact that the energetics depend on the electron-electron interaction! Coulomb repulsion forces electrons to optimize between the attractive force of the nucleus and the repulsive force from their neighbors making them occupy these other orbitals between \(n=6\) and \(n=7\) shells. These orbitals in between, too, are an approximation. Atomic physicists use something called term symbols to describe the states which you can find if you search energy level diagrams for your atom, but the main takeaway here is that “missing” different energy levels that were always present in the background for the hydrogen atom but washed away due to degeneracy arise from the coulomb repulsion between many electrons. This many-body interaction is precisely what makes Quantum mechanics so hard!

From class, you know two “peak” shaped functions (a trough is just an inverted peak): the Gaussian and Lorentzian, so you need to

1. Isolate the troughs (peaks)
2. Fit one of these curves over each trough (peak) accounting for the current uncertainty. The fit produces the mean and an associated width which will be used to find the location\(\pm\)error.

You are probably asking yourself which curve do I use to fit? This is a great question, and I will let you justify the fit based on your data. Recall that the key difference between a Lorentzian and a Gaussian lies in their tails. So, take some time to figure out which one is appropriate for your data, you only need one representative peak (or trough) to decide between a Lorentzian and a Gaussian, and after deciding on it, you need to fit your choice either on top of all the visible peaks or on top of all visible troughs.

Either way, whatever fit you decide is appropriate for your data, you need to estimate the vertical ($y$) errors to use the \(\chi^2\) test. I invite you to come up with a strategy to estimate these errors ; )

Here are some things to keep in mind in your adventure: recall that the Gaussian is defined as

where \(A\) is the amplitude (overall scaling), \(\mu\) is the mean (center) and \(\sigma\) is the standard deviation of the Gaussian.

The Lorentzian is defined as

where \(A\) is the amplitude, \(\mu\) the center and \(\Gamma\) is what’s called “Full width-half maximum”.

Both \(\sigma\) and \(\Gamma\) are the characteristic widths of your fits. Remember that in experiments, a measured value is always associated with an error. Therefore, your center (location of the trough or peak) is given by \(\mu \pm \sigma\) if you’re using a Gaussian distribution and \(\mu\pm \Gamma\) if you are using a Lorentzian.

Aside: To see why \(\Gamma\) is the natural equivalent of \(\sigma\), observe that for the standard Gaussian distribution that is normalized \(A=1\) (called the Normal distribution), the area under the curve between \(x=\mu\pm \sigma\) is

meaning that most (\(\approx 68\%\)) of the data from a random variable that follows a Normal distribution with mean \(\mu\) lies within the range \((\mu-\sigma,\mu+\sigma)\).

You might wonder what would happen if you do the same with the normalized (\(A=1\)) Lorentzian distribution and you find (exercise to you!)

i.e this area is \(70 \%\) of the whole! This is why the \(\Gamma\) of a Lorentzian distribution is the natural equivalent of \(\sigma\) even though it does not have a well defined notation of a variance!