Atoms and Nuclei

Splitting uranium releases energy. Joining hydrogen releases energy. Both cannot be right — can they?

They can, and for the same reason.

Binding energy per nucleon rises steeply for light nuclei, peaks near iron, and falls slowly for heavy ones. Any nucleus releases energy by moving toward that peak — and there are exactly two ways to move toward it. Heavy nuclei split. Light nuclei join.

Two topics share this chapter, each resting on one sentence:

• Bohr added exactly one new assumption: angular momentum is quantised in units of $h/2\pi$. Every radius, level and spectral line in hydrogen follows from that plus Newton and Coulomb.
• One graph explains all of nuclear energy. Fission and fusion are two routes to the same peak.

Running through both: mass and energy are the same quantity. A nucleus weighs measurably less than its parts, and the missing mass is the energy holding it together.

1. Rutherford's Experiment

Geiger and Marsden fired alpha particles at thin gold foil and recorded where they came out.

Rutherford called it the most incredible event of his life: like firing a fifteen-inch shell at tissue paper and having it bounce back.

The nucleus is $\sim 10^{-15}$ m across and the atom $\sim 10^{-10}$ m — a ratio of $10^5$ in radius, $10^{15}$ in volume. Scale the nucleus to a marble and the atom is a kilometre wide.

$r_0=\frac{2kZ{e}^2}{K}\text{(distance of closest approach)}$

from energy conservation: a head-on alpha stops when all its kinetic energy has become electrostatic potential energy.

Illustration 1

A 5.5 MeV alpha particle is fired head-on at a gold nucleus ($Z=79$). Find the distance of closest approach, and compare it with the radius of that nucleus.

At the turning point every joule of kinetic energy has become potential energy:

$r_0=\frac{2kZ{e}^2}{K}=\frac{2(9\times 10^9)(79)(1.6\times 10^{-19}{)}^2}{(5.5)(1.6\times 10^{-13})}=4.1\times 10^{-14}\text{m}=41\text{fm}$

Gold has $A=197$, so its nuclear radius is $1.2(197{)}^{1/3}=7.0$ fm.

The alpha stops almost six times further out than the nuclear surface, so it never enters the range of the nuclear force at all. Rutherford's analysis assumed a pure Coulomb interaction throughout, and this is exactly why that assumption was safe. Fire alphas hard enough to close the gap and the scattering does start to depart from his formula — which is how nuclear radii were first measured.

2. Why the Classical Atom Fails

An orbiting electron is accelerating, and accelerating charges radiate. Two consequences, both fatal:

• Collapse. A hydrogen atom should spiral into its nucleus in about $10^{-8}$ s. Matter is observed to be stable.
• Continuous spectra. As the orbit shrank the frequency would drift, smearing the emission. Atoms emit sharp discrete lines instead, characteristic of each element.

Nothing classical explains either.

3. Bohr's Postulates

Everything computable in the model comes from the second postulate combined with ordinary mechanics.

4. Radii, Velocities and Energies

Set the Coulomb force equal to the centripetal force, impose $mvr=nh/2\pi$, and all three follow at once:

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

$K=-E_n,U=2E_n=-2K$

• 0.529 Å is the Bohr radius, and it sets the scale of every atom.
• $c/137$ in the hydrogen ground state — that 137 is the reciprocal of the fine structure constant.
• Energy is negative because the electron is bound, and its magnitude falls as $1/n^2$, so levels crowd together as $n$ rises.
• Ionisation energy of hydrogen is 13.6 eV.

The virial relations $K=-E$ and $U=2E$ are the same ones seen in gravitational orbits, and for the same reason: both are inverse-square attractions.

Illustration 2

A hydrogen electron sits in $n=4$. Find its radius, speed, kinetic energy and potential energy.

$r_4=0.529(16)=8.46\text{A˚},v_4=\frac{3\times 10^8}{137\times 4}=5.5\times 10^5{\text{m s}}^{-1}$

$E_4=-\frac{13.6}{16}=-0.85\text{eV}\Rightarrow K=+0.85\text{eV},U=-1.70\text{eV}$

Sixteen times the ground-state radius, a quarter the speed, and a sixteenth of the binding. The electron is barely held — which is why highly excited atoms are so easily ionised.

5. The Hydrogen Spectrum

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

Only Balmer is visible, which is why it was found first and why it dominates stellar spectra. Within a series the longest wavelength comes from the smallest jump and the series limit from $n_2=\infty$.

Trap. Bohr's model works only for single-electron systems — H, He⁺, Li²⁺. It makes no provision for electron-electron repulsion, so it fails outright for helium. It also explains neither fine structure, nor line intensities, nor the Zeeman effect, and offers no reason why the quantisation should hold. It is a bridge, not a final theory.

Illustration 3

Find the longest and shortest wavelengths in the Lyman series.

Longest is the smallest jump, $n=2\to 1$:

$\Delta E=13.6\left(1-\frac{1}{4}\right)=10.2\text{eV}\Rightarrow \lambda =\frac{1240}{10.2}=121.6\text{nm}$

Shortest is the series limit, $n=\infty \to 1$, which takes the full 13.6 eV:

${\lambda }_{min}=\frac{1240}{13.6}=91.2\text{nm}$

The series limit is the ionisation wavelength — the shortest photon the series can emit is exactly the longest one that can ionise the atom from the ground state. Every series limit works this way.

Illustration 4

A transition in He⁺ ($Z=2$) has exactly the same wavelength as hydrogen's Lyman-alpha line. Identify it.

Hydrogen Lyman-alpha is the $2\to 1$ jump:

$\frac{1}{\lambda }=R\left(1-\frac{1}{4}\right)=\frac{3R}{4}$

For He⁺ the factor $Z^2=4$ multiplies everything, so the requirement is

$4R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{3R}{4}\Rightarrow \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{3}{16}$

Taking $n_1=2$ and $n_2=4$ gives $\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$, as required. So He⁺ $4\to 2$ matches hydrogen $2\to 1$ exactly.

The pattern generalises: doubling both quantum numbers in He⁺ reproduces any hydrogen line, because $Z^2=4$ cancels the factor of 4 that doubling $n$ introduces. Coincidences of this kind are why spectroscopists took years to disentangle helium lines from hydrogen ones.

A different counting question comes up just as often. Excite a gas so its atoms reach level $n$; each atom can cascade down by any route, so the number of distinct wavelengths the sample emits is the number of available pairs of levels:

$\text{number of lines}=\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{n(n-1)}{2}$

Illustration 5

Hydrogen atoms are excited to $n=5$. How many distinct spectral lines appear, and how many of them are visible?

$\frac{5(4)}{2}=10\text{lines}$

The visible ones are the Balmer lines, which end on $n=2$ — the jumps $5\to 2$, $4\to 2$ and $3\to 2$. So three of the ten are visible. Four end on $n=1$ and are ultraviolet; the other three end on $n=3$ or $n=4$ and are infrared.

Group the jumps by the level they land on and the total confirms itself: $4+3+2+1=10$.

6. The Nucleus

$Z$ protons, $N$ neutrons, $A=Z+N$. Same $Z$ are isotopes, same $A$ isobars, same $N$ isotones.

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

Volume goes as $R^3\propto A$, and mass goes as $A$ — so every nucleus has the same density, about $2.3\times 10^{17}$ kg m⁻³. A teaspoon would weigh a billion tonnes.

Constant density means the nuclear force saturates: each nucleon binds only its immediate neighbours, exactly like molecules in a liquid drop. The force is short-ranged, charge-independent, attractive at typical separations, and strongly repulsive below about 0.8 fm — which is what stops the nucleus collapsing.

Illustration 6

Compare the radii and densities of ${}^{27}$Al and ${}^{64}$Cu.

$\frac{R_{Al}}{R_{Cu}}={\left(\frac{27}{64}\right)}^{1/3}=\frac{3}{4}$

$\frac{{\rho }_{Al}}{{\rho }_{Cu}}=\frac{A_{Al}/A_{Al}}{A_{Cu}/A_{Cu}}=1$

The copper nucleus is a third larger in radius and 2.4 times heavier — and exactly as dense. Cube roots are worth spotting: 27 and 64 were chosen so the ratio comes out clean, and exam setters do the same.

7. Mass Defect and Binding Energy

$\Delta m=\left[Z{m}_p+N{m}_n\right]-M_{nucleus},B.E.=\Delta m{c}^2,1\text{u}\equiv 931.5\text{MeV}$

The missing mass is the binding energy — what you must supply to pull the nucleus apart into free nucleons.

Trap. This is not a bookkeeping trick. The mass really is lower, and a mass spectrometer measures it to many significant figures. Watch, too, whether a question gives you atomic or nuclear masses: atomic masses include the electrons, and using one where the other is meant is the standard error here.

Illustration 7

Find the energy released in ${}^{238}\text{U}\to {}^{234}\text{Th}+\alpha$, given atomic masses 238.05079, 234.04363 and 4.00260 u.

$\Delta m=238.05079-(234.04363+4.00260)=0.00456\text{u}$

$Q=(0.00456)(931.5)\Rightarrow Q=4.25\text{MeV}$

Atomic masses are safe to use here without correction, because 92 electrons appear on the left and $90+2=92$ on the right — they cancel exactly. That cancellation works for alpha decay and fails for beta decay, which is why the question always tells you which masses it is giving.

Illustration 8

Find the binding energy per nucleon of ${}^{16}$O, given a nuclear mass of 15.99053 u, with $m_p=1.00728$ u and $m_n=1.00867$ u.

Eight protons and eight neutrons:

$\Delta m=8(1.00728)+8(1.00867)-15.99053=16.12760-15.99053=0.13707\text{u}$

$B.E.=(0.13707)(931.5)=127.7\text{MeV}\Rightarrow \frac{B.E.}{A}=\frac{127.7}{16}=7.98\text{MeV}$

Just under 8 MeV, which puts oxygen on the rising part of the curve, still short of the iron peak at 8.8. Nearly every stable nucleus lands between 7.5 and 8.8 MeV per nucleon, so a value far outside that window means an arithmetic slip rather than a discovery — a useful check to run before writing an answer down.

8. The Binding Energy Curve

Iron is the most tightly bound nucleus, which is why it is the end point of stellar fusion and unusually abundant in the universe. Nuclei near iron can move neither way — which is exactly why energy production stops there and why massive stars collapse once an iron core forms.

9. Fission and Fusion

${}^{235}\text{U}+n\to {}^{141}\text{Ba}+{}^{92}\text{Kr}+3n+200\text{MeV}$

The extra neutrons make a chain reaction possible; control rods decide how many go on to cause further fissions. About 200 MeV per fission against a few eV per chemical reaction — a factor of $10^8$, which is why nuclear fuel is so concentrated.

Why a reactor needs a moderator

Trap. Uranium-235 is fissioned efficiently by slow neutrons, not fast ones, which sounds backwards. A slow neutron lingers near the nucleus and is far more likely to be captured — and capture is what triggers fission.

But fission emits fast neutrons, so something must slow them without absorbing them. That is the moderator — heavy water, graphite or ordinary water, chosen because light nuclei take away more energy per collision, exactly as a billiard ball loses most of its speed hitting another ball rather than a wall.

Control rods of cadmium or boron absorb neutrons, held so that each fission triggers precisely one more.

$4{}^1\text{H}\to {}^4\text{He}+2e^{+}+26.7\text{MeV}$

Fusion releases far more per nucleon — roughly 6 MeV against 0.9 — but is enormously harder. The obstacle is the Coulomb barrier: two positive nuclei must be forced within range of the nuclear force, needing $\sim 10^7$ K. Stars manage it by gravitational compression, which is why fusion happens naturally only in objects of stellar mass.

Illustration 9

How much ${}^{235}$U does a 1000 MW power station consume per day at 33 per cent efficiency?

Thermal output is three times the electrical, so 3000 MW:

$E_{day}=(3\times 10^9)(86400)=2.59\times 10^{14}\text{J}$

$\text{per fission }200\text{MeV}=3.2\times 10^{-11}\text{J}\Rightarrow n=8.1\times 10^{24}\text{fissions}$

$m=\frac{8.1\times 10^{24}}{6.02\times 10^{23}}\times 235\text{g}\Rightarrow m\approx 3.2\text{kg per day}$

Three kilograms a day. A coal station of the same output burns around ten thousand tonnes in the same time — the $10^8$ ratio in energy per reaction, showing up as a $10^6$ ratio in fuel mass.

Illustration 10

Four hydrogen atoms fuse to one helium-4 atom, releasing 26.7 MeV. The Sun radiates $3.8\times 10^{26}$ W. Find the rate at which it consumes hydrogen.

Energy released per kilogram of hydrogen consumed:

$\frac{26.7\text{MeV}}{4\text{u}}=\frac{(26.7)(1.6\times 10^{-13})\text{J}}{(4)(1.66\times 10^{-27})\text{kg}}=6.4\times 10^{14}{\text{J kg}}^{-1}$

$\frac{dm}{dt}=\frac{3.8\times 10^{26}}{6.4\times 10^{14}}\Rightarrow \approx 6\times 10^{11}{\text{kg s}}^{-1}$

Six hundred million tonnes of hydrogen every second. Only about 0.7 per cent of that mass actually disappears as energy — the rest becomes helium — so the Sun loses roughly four million tonnes of mass per second.

It has enough hydrogen left for another five billion years, and that is the point of the calculation. A chemical fuel of the same mass would have burned out in a few thousand.

10. Radioactivity

Gamma emission changes neither, because it only carries away energy as the nucleus drops from an excited state — exactly as an atom does with visible light.

The neutrino, and a near-abandonment of energy conservation

Alpha and gamma emissions come out at sharply defined energies, as expected when a nucleus drops between two definite states. Beta particles do not — they emerge with a continuous spread from zero to a maximum.

That looked like a violation of energy conservation, and Bohr was prepared to abandon the conservation law at nuclear scale rather than accept an unseen particle. Pauli proposed the alternative in 1930: a neutral, almost massless particle carrying away the balance and sharing it randomly with the electron.

That is the neutrino, and it was not detected until 1956 — twenty-six years later. Most neutrinos pass through the entire Earth undeflected. The continuous beta spectrum is direct evidence for it, and a good example of a conservation law being trusted over an apparent observation.

$N=N_0{e}^{-\lambda t},T_{1/2}=\frac{0.693}{\lambda },\tau =\frac{1}{\lambda }=1.44T_{1/2},A=\lambda N$

The decay law is statistical, not deterministic. It says nothing about when a particular nucleus decays, only what fraction of a large population will have. Activity $A=\lambda N$ decays on the same exponential with the same half-life.

Illustration 11

A radioactive sample registers 8000 disintegrations per second, and 30 minutes later 1000 per second. Find the half-life, the decay constant and the number of nuclei present at the start.

Activity tracks $N$, and 1000 is 8000 divided by $2^3$, so three half-lives have passed in 30 minutes:

$T_{1/2}=10\text{min},\lambda =\frac{0.693}{600}=1.16\times 10^{-3}{\text{s}}^{-1}$

Then $A=\lambda N$ gives the population directly:

$N_0=\frac{A_0}{\lambda }=\frac{8000}{1.16\times 10^{-3}}\Rightarrow N_0\approx 6.9\times 10^6$

The last step carries a unit trap. Activity is quoted per second, so $\lambda$ has to be per second as well — reaching for $0.693/10$ with the half-life in minutes undercounts the nuclei by a factor of sixty.

Illustration 12

Living wood gives 16 counts per minute per gram of carbon; a sample from an excavation gives 12. Carbon-14 has a half-life of 5730 years. Find the sample's age.

Activity tracks $N$, so $N/N_0=12/16=0.75$:

$0.75=e^{-\lambda t},\lambda =\frac{0.693}{5730}\Rightarrow t=\frac{\ln (1/0.75)}{\lambda }=\frac{0.288}{0.693}\times 5730$

$t\approx 2380\text{years}$

Not a whole number of half-lives, so counting halvings will not work — take logarithms. That is the only difference between this and the textbook one-eighth-remaining question, and it is the version that actually gets asked.

Summary

• Rutherford: the atom is mostly empty, with a tiny massive nucleus $10^5$ times smaller in radius.
• The classical atom collapses in $10^{-8}$ s and would give continuous spectra. Neither is observed.
• Bohr added one assumption — $L=nh/2\pi$ — and everything computable follows.
• $r_n=0.529n^2/Z$ Å, $E_n=-13.6Z^2/n^2$ eV, with $K=-E$ and $U=2E$.
• Ionisation energy 13.6 eV; ground-state speed $c/137$.
• Only Balmer is visible; the series limit is the ionisation wavelength.
• Excitation to level $n$ gives $n(n-1)/2$ distinct lines.
• Bohr fails for anything with two or more electrons, and explains no fine structure.
• $R=1.2A^{1/3}$ fm, so every nucleus has the same density — the nuclear force saturates.
• $\Delta m{c}^2$ is the binding energy, with 1 u $\equiv$ 931.5 MeV. Check whether masses given are atomic or nuclear.
• Binding energy per nucleon peaks at $\sim 8.8$ MeV near $A=56$ — the end point of stellar fusion.
• Fission and fusion both climb toward that peak, which is why both release energy.
• ~200 MeV per fission; a 1000 MW station burns about 3 kg of ${}^{235}$U a day.
• Reactors need a moderator because slow neutrons fission uranium far better than fast ones.
• Alpha changes $A$ by 4 and $Z$ by 2; beta changes $Z$ alone; gamma changes neither.
• The continuous beta spectrum is direct evidence for the neutrino — proposed 1930, detected 1956.
• $N=N_0{e}^{-\lambda t}$; mean life is 1.44 times the half-life, never shorter. For non-integer half-lives, take logarithms.