Atoms

In the experiment itself (Geiger and Marsden, 1911), a narrow beam of alpha particles, helium nuclei carrying charge $+2e$, was fired at a very thin gold foil. Almost all of them went straight through, but about 1 in 8000 was turned back through more than 90°. The ones that bounce almost straight back are the ones aimed dead centre at a nucleus, a head-on approach, and they let us put an actual number on how small the nucleus has to be.

Follow one such alpha particle, mass $m_\alpha$, starting far away with speed $v$, so with kinetic energy $K = \tfrac{1}{2}m_\alpha v^2$. The target nucleus has charge $+Ze$, where $Z$ is its atomic number (79 for gold), so the two repel. As the alpha particle closes in, that repulsion slows it down, until at some separation $d$ it stops completely for an instant, and is then pushed straight back the way it came. That turning point is the distance of closest approach.

A heavy nucleus like gold’s is about 50 times heavier than an alpha particle, so treat it as staying put. Then the total mechanical energy (kinetic plus electrostatic potential) is conserved the whole way in, and all we need to do is compare it at the start and at the turning point.

At the start, the separation is so large it might as well be infinite, so the electrostatic potential energy is zero and the energy is all kinetic:

At the turning point, the alpha particle is momentarily at rest, so its kinetic energy is zero and all of the energy has become the potential energy $kq_1q_2/r$ of the two charges $q_1 = +2e$ and $q_2 = +Ze$ at separation $d$, where $k = 1/4\pi\varepsilon_0 \approx 8.988\times10^9\text{ N m}^2\text{C}^{-2}$:

Energy conservation sets the two equal:

Multiply both sides by $d$, then divide both sides by $K$:

or, writing out $K = \tfrac{1}{2}m_\alpha v^2$ (and dividing by $\tfrac{1}{2}m_\alpha v^2$ is the same as multiplying by $2/m_\alpha v^2$):

Reading it: a more energetic alpha particle gets closer ($d \propto 1/K$), and a more highly charged nucleus holds it further off ($d \propto Z$). An alpha particle that is not aimed exactly head-on gets deflected sideways before it can come this close, so $d$ is the nearest any alpha particle of that energy can reach. And since the alpha particle turns round without ever touching the nucleus, with Coulomb’s law holding all the way in, the nucleus must be smaller than $d$: the distance of closest approach is an upper limit on the size of the nucleus.

This model had a fatal flaw: by classical electromagnetism, an electron orbiting the nucleus is constantly accelerating (circular motion), and an accelerating charge must radiate electromagnetic energy continuously. That radiated energy would come out of the electron’s orbital energy, so the electron should spiral inward and crash into the nucleus in a tiny fraction of a second. Real atoms are stable, so classical physics was missing something.

Electrons can only occupy certain special circular orbits in which they do not radiate energy, called stationary states. These orbits are exactly the ones where the orbital angular momentum is an integer multiple of $h/2\pi$:

An atom emits or absorbs electromagnetic radiation only when an electron jumps from one stationary state to another, and the photon’s energy exactly equals the difference between the two states’ energies:

A jump to a lower energy state (smaller n) releases a photon (emission); a jump to a higher state (larger n) requires absorbing a photon of exactly that energy.

At separation $r$, the proton pulls the electron towards itself with a Coulomb force of magnitude $k\cdot e\cdot e/r^2 = ke^2/r^2$. To move in a circle of radius $r$ at speed $v$, the electron needs a net inward (centripetal) force of exactly $mv^2/r$. The Coulomb pull is the only force acting on it, so it has to supply all of that:

Multiply both sides by $r^2$ to clear the fractions:

On its own, this is pure classical physics and allows any radius: pick an $r$, and some speed satisfies it. It is the quantisation condition that singles out particular radii, and the next step brings the two together.

The key move is to spot that the left-hand side of the force balance, $mv^2r$, is just $mvr$ multiplied by one more factor of $v$, and $mvr$ is exactly the quantity the postulate fixes at $n\hbar$:

So the force balance becomes

and dividing both sides by $n\hbar$ gives the speed in the n-th orbit:

For the radius, go back to the quantisation condition and rearrange it for $r$ (divide both sides by $mv$), then put in the speed just found:

(dividing by $v_n = ke^2/n\hbar$ is the same as multiplying by its reciprocal, $n\hbar/ke^2$). Multiplying the two fractions together:

So the speed falls as $1/n$ (the electron in the second orbit moves at half the speed it has in the first), while the radius grows as $n^2$. The textbook writes the same result without the shorthands; putting $\hbar = h/2\pi$ and $k = 1/4\pi\varepsilon_0$ back in:

In $r_n = n^2\hbar^2/mke^2$, everything except $n^2$ is a fixed constant of nature, so it pays to evaluate that part once. It is the radius of the $n=1$ orbit, called the Bohr radius, $a_0$:

Using $\hbar = 1.0546\times10^{-34}\text{ J s}$, $m = 9.109\times10^{-31}\text{ kg}$, $k = 8.988\times10^{9}\text{ N m}^2\text{C}^{-2}$, and $e = 1.602\times10^{-19}\text{ C}$, work out the top and the bottom separately:

(The units do come out as metres: J² s² divided by kg × N m² C⁻² × C² reduces to m, using 1 J = 1 N m and 1 N = 1 kg m s⁻².) Every other orbit is just $n^2$ times this:

The same constants give the ground-state speed, $v_1 = ke^2/\hbar$:

Fast, but still under 1% of the speed of light (about $c/137$), which is why ordinary Newtonian mechanics was good enough in Step 1.

Kinetic energy. Divide the force balance $mv^2r = ke^2$ by $r$ to get $mv^2 = ke^2/r$, then take half of it:

Potential energy. Two point charges $q_1$ and $q_2$ at separation $r$ have electrostatic potential energy $kq_1q_2/r$, taking zero at infinite separation. Here $q_1 = +e$ (proton) and $q_2 = -e$ (electron):

Total energy. Add the two, writing $ke^2/r$ as $2ke^2/2r$ so both terms share a denominator:

Compare the three: $E = -K$ and $U = 2E$. The kinetic energy is positive, the potential energy is twice as big and negative, and the total comes out negative, exactly the pattern found for a satellite in the Gravitation chapter, because both are inverse-square attractions.

The energy still depends on $r$, but Step 2 says only the radii $r_n$ are allowed. Substitute $r_n = n^2\hbar^2/mke^2$:

(again, dividing by $r_n$ means multiplying by its reciprocal). The two factors of $ke^2$ multiply to give $k^2e^4$:

In the textbook’s notation, $\dfrac{k^2}{\hbar^2} = \dfrac{1/16\pi^2\varepsilon_0^2}{h^2/4\pi^2} = \dfrac{1}{4\varepsilon_0^2h^2}$, so this is the same as $E_n = -\dfrac{me^4}{8\varepsilon_0^2h^2n^2}$.

The fraction in front is just the size of the ground-state energy, and the quickest way to evaluate it is to notice that at $n=1$ the energy is $-ke^2/2a_0$, using $ke^2$ and $a_0$ already worked out in Step 3:

Dividing by $1.602\times10^{-19}$ J per eV:

and therefore, for every level:

Ultraviolet. Includes n=2→1, n=3→1, n=4→1, and so on.

The only series with visible lines (n=3→2 is red, n=4→2 is blue-green, and so on), which is why it was discovered first, historically.

Infrared. Includes n=4→3, n=5→3, n=6→3, and so on.