## Chapter 2

**On p. 20,** I wrote, "in a hydrogen atom, characteristic frequencies are on the scale of $10^{15}$ oscillations or cycles per second." The place to start in understanding this number is the typical energy scale of the hydrogen atom, which is a Rydberg, $13.6\,{\rm eV}$. The way I start to convert this to frequency is to remember that $\hbar c = 1973 \, {\rm eV} {\rm A}$, where by ${\rm A}$ I mean an Angstrom, i.e. $10^{-10}\,{\rm m}$. Now we solve $E = h\nu = 2\pi\hbar\nu$ for $\nu$ to get

$$\nu = {E \over 2\pi\hbar} = {Ec \over 2\pi\hbar c} = {13.6\,{\rm eV} \times 3 \times 10^{18}\,{\rm A}/s \over 2 \times 3.1415927 \times 1973\,{\rm eV} {\rm A}} = 3.3 \times 10^{15}\,{\rm Hz} \,,$$

where in the last step I remembered that $1\,{\rm Hz}$ is $1/{\rm s}$. Now, if one is going to quote an order of magnitude for the frequency $\nu$ I just calculated, the proper thing is to round up and say $10^{16}\,{\rm Hz}$. That's because $3.3$ is closer to $10$ on a logarithmic scale than to $1$. I instead quoted $10^{15}\,{\rm Hz}$ because the frequency just calculated is based on the largest energy scale you can usefully talk about in connection with bound states of an electron and a proton: the difference between the ground state and the ionization threshold. Transitions from one level to another are characterized by an energy that is some order unity multiple of a Rydberg. This multiple is $3/4$ for the $2p \to 1s$ transtion, leading to a frequency $2.5 \times 10^{15}\,{\rm Hz}$. For $3s \to 2p$ the frequency is $4.6 \times 10^{14}\,{\rm Hz}$. So if you want a single order of magnitude figure to characterize the frequencies relevant for the hydrogen atom, $10^{15}\,{\rm Hz}$ is the best choice.

Later, **on p. 26,** I note that "the frequency of an electron in its lowest energy state is about $3 \times 10^{15}\,{\rm Hz}$. That's directly from the calculation described above.

**On p. 30,** I wrote, "The minimum frequency of light that it takes to kick electrons out of sodium is $5.5 \times 10^{14}$ oscillations per second, which means green light." I go on to derive a value of Planck's constant. Let's see how that calculation works. First, if we accept that the work function for sodium is $2.28\, {\rm eV}$, then we convert to a frequency like this:

$$\nu = {E \over h} = {E \over 2\pi} {c \over \hbar c} = {2.28 \, {\rm eV} \over 2 \times 3.1415927} \times {3 \times 10^8\,{\rm m/s} \over 1973\,{\rm eV}\,{\rm A}} {10^{10} \, {\rm A} \over 1\,{\rm m}} = 5.52 \times 10^{14}\,{\rm Hz} \,.$$

(If I use $2.3\,{\rm eV}$ then I would get $5.56 \times 10^{14}\,{\rm Hz}$. So to get the work function and the frequency that I quoted to two significant figures, you have to start by calculating with three.)

Turning things around, if we grant that the frequency is $5.5 \times 10^{14}$ oscillations per second and the work function is $2.28\,{\rm eV}$ (both of which are measurable quantities in principle), then we can derive a value for Planck's constant using

$$h = {E \over \nu} = {2.28\,{\rm eV} \over 5.5 \times 10^{14}\,{\rm Hz}} = 4.1 \times 10^{-15} \, {\rm eV} \, {\rm s} \,.$$