Atoms and Nuclei

Rutherford's students fired alpha particles at a gold foil. About one in eight thousand came straight back. Why did that single observation destroy the plum-pudding model?

Because that model makes backscattering arithmetically impossible.

If the positive charge is smeared through a sphere of atomic size, the field an alpha particle meets is feeble. The maximum deflection from one atom works out at about $0.02^{\circ }$. A foil a micrometre thick is roughly $10^4$ atoms deep, and the deflections are random, so they accumulate as a random walk:

${\theta }_{total}\approx 0.02^{\circ }\times \sqrt{10^4}=2^{\circ }$

To turn a $5$ MeV alpha particle through $180^{\circ }$ you need a single violent encounter, and that requires the entire positive charge packed into a volume small enough for the alpha to get very close before being repelled. Setting the kinetic energy equal to the potential energy at closest approach:

$r_0=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{K}=\frac{2\times 79\times 1.44\text{MeV fm}}{5\text{MeV}}=45\text{fm}$

which is about ten thousand times smaller than the atom. The nucleus had to exist.

Small impact parameter means large deflection. The relation is

$b=\frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{K}\cot \frac{\theta }{2},N(\theta )\propto \frac{1}{{\sin }^4\left(\theta /2\right)}$

and the fourth-power law was confirmed over five orders of magnitude in count rate — which is why the nuclear model was accepted so quickly.

Illustration 1

Find the distance of closest approach for a $7.7$ MeV alpha particle fired head-on at a gold nucleus, and compare it with the nuclear radius.

$r_0=\frac{2\times 79\times 1.44}{7.7}=\frac{227.5}{7.7}=29.5$ fm

The gold nucleus has $R=1.2\times 197^{1/3}=1.2\times 5.82=7.0$ fm.

The alpha never touches it. At $29.5$ fm it is still four nuclear radii away, so the whole experiment probes the Coulomb field alone — which is precisely why the analysis is purely electrostatic.

1. Bohr's model, and what it actually explains

Bohr kept the nuclear atom and added one postulate: angular momentum is quantised in units of $\hslash$. Everything else follows from balancing the Coulomb attraction against the centripetal requirement:

$r_n=\frac{0.529n^2}{Z}\text{A˚},E_n=-\frac{13.6Z^2}{n^2}\text{eV},v_n=\frac{c}{137}\frac{Z}{n}$

Three consequences are worth holding separately. The total energy is negative and equals minus the kinetic energy, so removing an electron costs exactly its kinetic energy. The speed in the ground state of hydrogen is $c/137$, which is why non-relativistic mechanics is adequate but only just. And the model works for any one-electron system — helium ion, lithium twice-ionised — provided $Z$ is inserted.

de Broglie later supplied the missing reason for the quantisation: a standing wave must fit the orbit,

$2\pi r_n=n\lambda =\frac{nh}{mv}\Rightarrow mv{r}_n=\frac{nh}{2\pi }$

which is Bohr's postulate, derived rather than assumed.

Illustration 2

Find the radius, energy and speed of the $n=2$ state of the ${\text{Li}}^{2+}$ ion.

With $Z=3$ and $n=2$:

$r=\frac{0.529\times 4}{3}=0.705$ Å

$E=-\frac{13.6\times 9}{4}=-30.6$ eV

$v=\frac{c}{137}\times \frac{3}{2}=3.28\times 10^6$ m s${}^{-1}$

Higher $Z$ pulls the orbit in and deepens the well. The radius scales as $n^2/Z$ and the energy as $Z^2/n^2$, so a threefold charge makes this ion nine times more tightly bound at each level.

2. Spectra: series, line counts and the isotope shift

Transitions between levels give

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),R=1.097\times 10^7{\text{m}}^{-1}$

with $n_1=1,2,3$ naming the Lyman, Balmer and Paschen series. Only Balmer falls in the visible, which is why hydrogen's visible spectrum has just four lines.

A gas excited to level $n$ can emit

$\frac{n\left(n-1\right)}{2}\text{distinct lines}$

which is simply the number of ways of choosing two levels out of $n$.

The isotope shift is where Advanced goes beyond Main. The nucleus is not infinitely heavy, so the electron and nucleus both orbit their common centre of mass and the correct mass to use is the reduced mass:

$\mu =\frac{m_{e}M}{m_e+M},R_M=R_{\infty }\frac{\mu }{m_e}$

Deuterium's nucleus is twice as heavy as hydrogen's, so its Rydberg constant is larger by about one part in $3700$. Every deuterium line sits slightly to the short-wavelength side of the corresponding hydrogen line — and that tiny shift, about $0.18$ nm at the red Balmer line, is how deuterium was discovered.

Illustration 3

Hydrogen atoms are excited to $n=4$. Find the number of spectral lines emitted and the longest wavelength in the Balmer series.

Lines $=\frac{4\times 3}{2}=6$

Longest Balmer wavelength is the smallest energy gap, $n=3\to 2$:

$\Delta E=3.40-1.51=1.89$ eV, so $\lambda =\frac{1240}{1.89}=656$ nm

That is the red line every hydrogen discharge shows. The shortest Balmer wavelength is the series limit at $364.6$ nm, corresponding to $n=\infty \to 2$.

Illustration 4

The ground state energy of hydrogen is $-13.6$ eV. Find the energy needed to remove the electron from the $n=3$ state, and the wavelength of the photon that would do it.

$E_3=-\frac{13.6}{9}=-1.51$ eV, so the ionisation energy from that level is $1.51$ eV.

$\lambda =\frac{1240}{1.51}=821$ nm

Excited atoms ionise with infrared light. This is why a gas already glowing is far easier to ionise further, and it underlies how a discharge sustains itself once started.

3. The nucleus: size, density and binding

Scattering experiments give

$R=R_0{A}^{1/3},R_0=1.2\text{fm}$

Since volume goes as $A$, the density is the same for every nucleus — about $2.3\times 10^{17}$ kg m${}^{-3}$, or a hundred million tonnes per cubic centimetre. Nuclear matter is incompressible in a way ordinary matter is not.

The mass of a nucleus is always less than the sum of its parts, and the difference is the binding energy:

$\Delta m=Z{m}_p+\left(A-Z\right)m_n-M_{nucleus},BE=\Delta m{c}^2,1\text{u}=931.5\text{MeV}$

Illustration 5

Find the radius and density of a ${}^{56}$Fe nucleus, given $m=56\times 1.66\times 10^{-27}$ kg.

$R=1.2\times 56^{1/3}=1.2\times 3.826=4.59$ fm

$V=\frac{4}{3}\pi {\left(4.59\times 10^{-15}\right)}^3=4.05\times 10^{-43}$ m${}^3$

$\rho =\frac{9.30\times 10^{-26}}{4.05\times 10^{-43}}=2.3\times 10^{17}$ kg m${}^{-3}$

Repeating this for any nucleus gives the same number. The $A^{1/3}$ law exists precisely so that the density comes out constant, which tells you nucleons touch rather than overlap.

4. The binding-energy curve: fission and fusion

Plot binding energy per nucleon against mass number and the curve rises steeply to a broad maximum of about $8.8$ MeV near iron, then falls slowly.

Energy is released by any process that moves nucleons towards the peak. Light nuclei do so by fusing; heavy ones by splitting. Both release energy for the same reason, and neither would if the curve were flat.

The energy released in any nuclear reaction is the $Q$-value:

$Q=\left(\sum m_{initial}-\sum m_{final}\right)c^2$

positive for an exoergic reaction.

Illustration 6

Estimate the energy released when a ${}^{236}$U nucleus (BE per nucleon $7.6$ MeV) splits into two fragments of mass number $118$ each with BE per nucleon $8.5$ MeV.

Initial binding $=236\times 7.6=1794$ MeV

Final binding $=236\times 8.5=2006$ MeV

$Q=2006-1794=212$ MeV

About $200$ MeV per fission, against a few electronvolts for a chemical reaction — a factor of roughly a hundred million, which is the entire reason nuclear energy densities are what they are.

Illustration 7

Find the $Q$-value of the deuterium-tritium fusion reaction, given mass defects such that the products are lighter by $0.01876$ u.

$Q=0.01876\times 931.5=17.5$ MeV

Per nucleon this is $\frac{17.5}{5}=3.5$ MeV.

Fusion beats fission per nucleon by about four times, which is why it is worth the enormous difficulty of achieving it. The obstacle is the Coulomb barrier, not the energetics.

5. The nuclear force and the stability belt

The force holding a nucleus together has four properties that between them explain almost every trend in nuclear physics.

It is short ranged, effective only out to about $2$ fm — roughly one nucleon diameter. It is charge independent, acting equally between any pair of nucleons. It saturates, meaning each nucleon binds only to its immediate neighbours rather than to every other nucleon. And below about $0.7$ fm it turns strongly repulsive, which is what stops nuclei from collapsing.

Saturation is the key to two facts already met. If every nucleon bound to all $A-1$ others, binding energy would go as $A^2$ and the binding energy per nucleon would rise without limit. Instead it flattens at about $8$ MeV, exactly as a nearest-neighbour interaction requires. For the same reason the nucleons pack at fixed spacing, which is why the density is constant and the radius goes as $A^{1/3}$.

The Coulomb repulsion, by contrast, is long ranged and every proton pushes on every other, so it grows roughly as $Z^2$. Heavy nuclei therefore need extra neutrons to space the protons apart without adding to the repulsion, and the neutron-to-proton ratio climbs from $1$ in light nuclei to about $1.5$ in uranium. Beyond $Z=83$ the Coulomb term wins outright and no nucleus is stable at all.

Illustration 8

Compare the neutron-to-proton ratios of ${}^4$He, ${}^{56}$Fe and ${}^{238}$U, and explain the trend.

${}^4$He: $\frac{2}{2}=1.00$

${}^{56}$Fe: $\frac{30}{26}=1.15$

${}^{238}$U: $\frac{146}{92}=1.59$

Extra neutrons add attraction without adding repulsion. As $Z$ grows, the Coulomb energy rises faster than the nuclear binding, and only a growing neutron surplus keeps the balance — until at $Z>83$ even that fails.

Illustration 9

Explain why fission fragments are radioactive and emit both neutrons and beta particles.

Uranium sits at $N/Z=1.59$, but a fragment of mass number around $118$ is stable only near $N/Z\approx 1.3$.

The fragments are therefore neutron-rich the instant they form. They shed the excess in two ways: a few neutrons are emitted promptly, which is what sustains the chain reaction, and the remainder is corrected by a series of beta-minus decays converting neutrons into protons.

This is the origin of fission waste. The long-lived radioactivity of a reactor's spent fuel is a direct consequence of the neutron-to-proton ratio being wrong for the fragments' mass.

6. Radioactive decay

Decay is a random process with a fixed probability per unit time, which gives

$-\frac{dN}{dt}=\lambda N\Rightarrow N=N_0{e}^{-\lambda t},A=\lambda N$

$T_{1/2}=\frac{\ln 2}{\lambda }=0.693\tau ,\tau =\frac{1}{\lambda }$

The mean life $\tau$ is longer than the half-life, because the long-lived tail of the exponential pulls the average up. Activity is measured in becquerel (one decay per second) or curie ($3.7\times 10^{10}$ Bq).

Illustration 10

A sample has an activity of $8000$ Bq and a half-life of $6$ hours. Find the activity after $18$ hours, the decay constant and the number of nuclei present initially.

$18$ hours is three half-lives, so $A=\frac{8000}{8}=1000$ Bq.

$\lambda =\frac{0.693}{6\times 3600}=3.21\times 10^{-5}$ s${}^{-1}$

$N_0=\frac{A_0}{\lambda }=\frac{8000}{3.21\times 10^{-5}}=2.5\times 10^8$ nuclei

Note how few nuclei that is. A visible speck of matter contains $10^{20}$ atoms, so a strongly radioactive sample can be far too small to see.

Illustration 11

A wooden artefact shows a ${}^{14}$C activity of $9$ counts per minute per gram against $15$ for living wood. Find its age, with $T_{1/2}=5730$ years.

$\frac{A}{A_0}=\frac{9}{15}=0.6=e^{-\lambda t}$

$\lambda t=\ln \frac{1}{0.6}=0.511$

$t=\frac{0.511}{0.693}\times 5730=4224$ years

Carbon dating works only up to about ten half-lives, roughly $50000$ years, beyond which the remaining activity is lost in the background.

7. Successive decay and equilibrium

When a parent decays to a radioactive daughter, the daughter is being created and destroyed at once:

$\frac{d{N}_2}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$

Starting from pure parent, the daughter's population rises, peaks, and then falls, with the maximum at

$t_{max}=\frac{\ln \left({\lambda }_2/{\lambda }_1\right)}{{\lambda }_2-{\lambda }_1}$

at which instant ${\lambda }_1{N}_1={\lambda }_2{N}_2$ — the two activities are momentarily equal.

If the parent is very long-lived compared with the daughter, ${\lambda }_2\gg {\lambda }_1$, the daughter settles into secular equilibrium, where its population becomes constant and

${\lambda }_1{N}_1={\lambda }_2{N}_2\Rightarrow \frac{N_1}{N_2}=\frac{T_1}{T_2}$

This is why radium is found with uranium ores in a fixed ratio, and why radon accumulates in basements at a steady rate.

Illustration 12

Uranium-238 has a half-life of $4.5\times 10^9$ years and decays through a chain to radium-226 with half-life $1600$ years. Find the ratio of radium to uranium atoms in an old ore.

Secular equilibrium gives $\frac{N_{Ra}}{N_U}=\frac{T_{Ra}}{T_U}$:

$\frac{N_{Ra}}{N_U}=\frac{1600}{4.5\times 10^9}=3.6\times 10^{-7}$

About one radium atom per three million uranium atoms, which is exactly the ratio the Curies had to work through to isolate a visible quantity from tonnes of pitchblende.

8. Decay modes, and what each conserves

Alpha decay emits a helium nucleus: $A$ falls by $4$, $Z$ by $2$. The alpha energies are discrete, because the transition is between two definite nuclear levels.

Beta-minus decay converts a neutron into a proton: $Z$ rises by $1$, $A$ is unchanged. But the beta energies are continuous, spread from zero to a sharp maximum — which for two decades looked like a violation of energy conservation. Pauli's resolution was that a third, nearly undetectable particle shares the energy: the antineutrino.

Beta-plus decay and electron capture both reduce $Z$ by $1$. Gamma emission changes neither, being the de-excitation of a nucleus left in an excited state by a previous decay.

Illustration 13

A ${}^{238}$U nucleus decays through a chain to ${}^{206}$Pb. Find the number of alpha and beta-minus decays involved.

Mass number falls by $238-206=32$, and only alphas change $A$:

Alphas $=\frac{32}{4}=8$

Those alone would reduce $Z$ by $16$, from $92$ to $76$. The actual final $Z$ is $82$, so betas must raise it by $6$:

Beta-minus decays $=6$

Always count alphas from $A$ first, then betas from $Z$. Doing it the other way round leaves two unknowns in one equation.

Illustration 14

Why is the alpha spectrum discrete while the beta spectrum is continuous?

Alpha decay is a two-body final state: daughter plus alpha. Momentum conservation then fixes the energy split uniquely, so every alpha emerges with the same energy.

Beta decay is a three-body final state: daughter, electron and antineutrino. The energy can be shared in any proportion, so the electron's energy ranges continuously up to a maximum.

The shape of the spectrum was the evidence for the neutrino, twenty-five years before one was detected.

Summary

• Plum-pudding scattering gives about $2^{\circ }$ after $10^4$ atoms; backscattering requires a concentrated charge.
• Closest approach $r_0=\frac{2Z{e}^2}{4\pi {\epsilon }_{0}K}$; impact parameter $b\propto \cot \left(\theta /2\right)$; count rate $\propto \frac{1}{{\sin }^4\left(\theta /2\right)}$.
• Bohr: $r_n=\frac{0.529n^2}{Z}$ Å, $E_n=-\frac{13.6Z^2}{n^2}$ eV, $v_n=\frac{c}{137}\frac{Z}{n}$.
• de Broglie explains the quantisation: $2\pi r_n=n\lambda$ gives $mvr=n\hslash$.
• $\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$; level $n$ gives $\frac{n\left(n-1\right)}{2}$ lines.
• Reduced mass shifts every deuterium line short of hydrogen's by about one part in $3700$ — how deuterium was found.
• $R=1.2A^{1/3}$ fm, so nuclear density is constant at $2.3\times 10^{17}$ kg m${}^{-3}$.
• $1$ u $=931.5$ MeV; $BE=\Delta m{c}^2$; $Q=\left(\sum m_i-\sum m_f\right)c^2$.
• The binding-energy curve peaks near iron at $8.8$ MeV per nucleon; both fusion and fission move towards that peak.
• Fission releases about $200$ MeV; fusion about $3.5$ MeV per nucleon, roughly four times better.
• Nuclear force: short ranged, charge independent, saturating, with a repulsive core — which is why $BE/A$ flattens and density stays constant.
• Coulomb repulsion grows as $Z^2$, so $N/Z$ climbs from $1$ to about $1.5$; beyond $Z=83$ nothing is stable.
• $N=N_0{e}^{-\lambda t}$, $A=\lambda N$, $T_{1/2}=0.693\tau$ — the mean life is longer than the half-life.
• Successive decay peaks where ${\lambda }_1{N}_1={\lambda }_2{N}_2$; secular equilibrium gives $\frac{N_1}{N_2}=\frac{T_1}{T_2}$.
• Alpha spectra are discrete (two-body), beta spectra continuous (three-body) — which is how the neutrino was predicted.