Nuclei

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 13.

A teaspoon that weighs a billion tonnes

The nucleus is about $10^4$ times smaller than the atom in radius, so it fills about one trillionth ($10^{-12}$) of the atom's volume. Yet it carries more than 99.9% of the atom's mass. Pack nuclear matter together with no gaps and a teaspoon of it (about 5 cm³) would weigh around $10^{12}$ kg, roughly a billion tonnes. Neutron stars are made of matter that dense. This chapter explains what the nucleus is made of, what holds it together and why splitting it or fusing it releases enormous energy.

What is inside the nucleus

A nucleus contains protons (charge $+e$, mass 1.00727 u) and neutrons (no charge, mass 1.00866 u), together called nucleons. James Chadwick discovered the neutron in 1932. The three numbers that describe a nucleus are:

A nuclide is written with A as a superscript and Z as a subscript, for example ²³⁵₉₂U ($A = 235$, $Z = 92$, so $N = 143$). A neutron outside a nucleus is unstable: it decays into a proton, an electron and an antineutrino in roughly 15 minutes. Inside a nucleus it is stable.

The atomic mass unit is defined as one twelfth of the mass of a carbon-12 atom: $1 \text{ u} = 1.66054 \times 10^{-27} \text{ kg}$. Atoms and nuclei are weighed with a mass spectrometer, and it also shows that atoms of one element can have different masses.

Hydrogen has three isotopes with masses 1.0078 u, 2.0141 u and 3.0160 u: protium, deuterium and tritium. Tritium is radioactive, with a half-life of about 12 years.

Worked example: the atomic mass on the periodic table. Chlorine has two isotopes, mass 34.98 u (abundance 75.4%) and mass 36.98 u (abundance 24.6%). Find the average atomic mass.

The common wrong step: taking the plain mean of the two masses, $(34.98 + 36.98)/2 = 35.98 \text{ u}$. The abundances are not equal, so the average must be weighted.

How big is a nucleus?

Scattering of electrons and α-particles shows that a nucleus of mass number A has radius

$$R = R_0 A^{1/3}, \qquad R_0 = 1.2 \text{ fm} = 1.2 \times 10^{-15} \text{ m}$$

The volume of a sphere is $(4/3)\pi R^3$, so the volume of a nucleus is proportional to A. Each nucleon takes up the same amount of room, as if nucleons were tiny marbles packed together. Mass is also proportional to A, so the density does not depend on A at all. It works out at about $2.3 \times 10^{17}$ kg m⁻³, some $10^{14}$ times the density of water.

Worked example: radius and density. Find the radius of a ⁶⁴Cu nucleus, compare it with ²⁷Al, and check its density.

The common wrong step: taking the radius ratio equal to the mass-number ratio (64 : 27). Radius goes as the cube root of A. The density, by contrast, is the same for every nucleus.

Mass and energy: why a nucleus weighs less than its parts

Einstein showed that mass is a form of energy: $E = mc^2$. A mass of 1 g corresponds to $9 \times 10^{13} \text{ J}$, and a handy atomic conversion is $1 \text{ u} \equiv 931.5 \text{ MeV}$.

Now weigh a nucleus. It always has less mass than the separate protons and neutrons that make it. The difference is the mass defect:

$$\Delta M = [Z m_p + (A - Z) m_n] - M$$

where M is the nuclear mass. Because mass is energy, the missing mass is the energy that was given off when the nucleons came together, and it is the energy you must supply to pull the nucleus apart again. This is the binding energy:

$$E_b = \Delta M\, c^2 \quad (\text{with } \Delta M \text{ in u}, ; E_b = \Delta M \times 931.5 \text{ MeV})$$

The more useful quantity is the binding energy per nucleon, $E_{bn} = E_b/A$. It tells you how tightly an average nucleon is held, so the higher it is, the more stable the nucleus.

Worked example: binding energy of helium. The nuclear mass of ⁴₂He is 4.001506 u. Given $m_p = 1.00727 \text{ u}$ and $m_n = 1.00866 \text{ u}$, find $E_b$ and $E_{bn}$.

The common wrong step: mixing atomic and nuclear masses. Tables give atomic masses, which include the electrons. If you use the nuclear mass $M$ you must use $m_p$ for the protons. If you use the atomic mass of helium (4.002603 u), you must use the mass of hydrogen atoms (1.007825 u) in place of $m_p$. Then the electron masses cancel and you get the same 28.3 MeV.

The binding energy curve

Plot $E_{bn}$ against $A$ and you get one of the most important graphs in physics.

Why is the middle flat? The nuclear force is short ranged, so a nucleon bonds only to its nearest neighbours however large the nucleus is, and the binding energy grows in proportion to A. At the heavy end, the protons repel each other with a long-range electric force that does not saturate, which pulls $E_{bn}$ down.

This one graph explains both nuclear energy sources. A heavy nucleus such as $A = 240$ has $E_{bn} \approx 7.6$ MeV. If it splits into two nuclei with $A \approx 120$, each has $E_{bn} \approx 8.5$ MeV. The nucleons are now more tightly bound, and the gain, about $(8.5 - 7.6) \times 240 \approx 216$ MeV, is released. That is fission. Two very light nuclei ($A \leq 10$) that merge move up the steep left side of the curve and also release energy. That is fusion. The curve explains why we get energy by moving towards iron from either side.

Beyond the textbook. For the exam: the peak of the $E_{bn}$ curve is at $A \approx 56$, about 8.75 MeV, so iron is the most stable. In reality: the nuclide with the highest binding energy per nucleon is actually nickel-62, with iron-56 and iron-58 only a whisker behind. The three differ by less than 0.01 MeV per nucleon, so the "iron peak" is a region, not a single point. It also explains why stars cannot get energy by fusing elements beyond this region. Why it matters: when a question asks for the peak, answer $A \approx 56$ and about 8.75 MeV.

Pause and predict: which has the greater binding energy per nucleon, ⁵⁶Fe or ²³⁵U? Which has the greater total binding energy?

Check your answer: ⁵⁶Fe has the greater $E_{bn}$ (about 8.75 MeV against about 7.6 MeV). The total $E_b$ of ²³⁵U is greater (about 1790 MeV against about 490 MeV for iron), simply because it has many more nucleons. Do not mix up the total with the per nucleon value.

The nuclear force

What holds protons together when they repel each other? A third force, much stronger than electric repulsion at nuclear distances.

Because the protons repel with a long-range force while the nuclear force acts only on neighbours, heavy nuclei need extra neutrons to dilute the repulsion. The neutron-to-proton ratio is about 1 for light stable nuclei and rises to about 1.5 for heavy ones. Nuclei with the wrong ratio are unstable.

Radioactivity

In 1896 Henri Becquerel found by accident that uranium salts blacken a photographic plate wrapped in black paper. Unstable nuclei change into more stable ones by emitting radiation. There are three kinds.

A positron has the same mass as an electron and the opposite charge. The electron emitted in β-decay is not an orbital electron. It is created when a neutron turns into a proton. In every decay the sum of A and the sum of Z are conserved on the two sides, and this is how you find a missing daughter nucleus.

The decay law. Radioactive decay is random for any one nucleus but predictable for many. The number of undecayed nuclei falls as

$$N = N_0 e^{-\lambda t} \quad \text{which is} \quad N = N_0 \left(\dfrac{1}{2}\right)^{t/T_{1/2}}$$

The decay constant $\lambda$ (unit s⁻¹) is the probability per second that a given nucleus decays. The half-life $T_{1/2}$ is the time in which half the nuclei decay, and the mean life $\tau$ is the average lifetime, the time when $N$ has dropped to $1/e$ (about 37%) of its starting value. The activity $R = \lambda N$ is the number of decays per second, measured in becquerel (1 Bq = 1 decay per second). The older unit curie equals $3.7 \times 10^{10}$ Bq.

$$T_{1/2} = \dfrac{0.693}{\lambda} = 0.693\,\tau, \qquad \tau = \dfrac{1}{\lambda}$$

Worked example: how much is left? The half-life of a nuclide is 20 days. Starting with 640 mg, find what remains after 60 days and after 50 days, and the decay constant and mean life.

The common wrong step: assuming everything has decayed after two half-lives. After two half-lives 25% remains, and the sample never reaches exactly zero. Also, the mean life is longer than the half-life by the factor $1/0.693 \approx 1.44$.

Pause and predict: a sample of a nuclide with half-life 10 minutes has an activity of 800 Bq at some moment. What is its activity 30 minutes later?

Check your answer: 100 Bq. Thirty minutes is three half-lives, and activity $R = \lambda N$ falls by half each half-life: 800 → 400 → 200 → 100.

Nuclear energy

Q-value. For any nuclear reaction or decay, the energy released is the difference in rest mass between the starting and final nuclei:

$$Q = (\text{sum of initial masses} - \text{sum of final masses})\, c^2$$

A positive Q means energy is released and the reaction can happen by itself. A negative Q means energy must be supplied. Use atomic masses only if the number of electrons on the two sides is equal, so that they cancel.

Worked example: the Q-value of a fusion reaction. For ²₁H + ²₁H → ³₂He + n, the atomic masses are ²H = 2.014102 u, ³He = 3.016029 u and n = 1.008665 u. Find Q.

The common wrong step: writing final minus initial and getting a negative Q for a reaction that releases energy. Also keep to the electron balance: two H atoms (2 electrons) on the left and one He atom (2 electrons) on the right.

Nuclear fission is the splitting of a heavy nucleus. When a slow (thermal) neutron strikes ²³⁵U, the nucleus splits into two medium-mass fragments and releases 2 to 3 neutrons, for example

¹₀n + ²³⁵₉₂U → ²³⁶₉₂U∗ → ¹⁴⁴₅₆Ba + ⁸⁹₃₆Kr + 3 ¹₀n + about 200 MeV

The released neutrons can strike more ²³⁵U nuclei, giving a chain reaction. If it is uncontrolled you get an atom bomb. In a nuclear reactor it is controlled: a moderator (heavy water or graphite) slows the neutrons to thermal speeds, where ²³⁵U captures them most readily, and control rods (such as cadmium or boron) absorb neutrons to keep exactly one neutron per fission causing another fission. The fragments are radioactive and decay by β emission to stable products.

Worked example: energy from one kilogram of uranium. Each ²³⁵U fission releases about 200 MeV. Estimate the energy from 1 kg of pure ²³⁵U fully fissioned.

That is a few million times (about $3 \times 10^6$) what a kilogram of coal gives (about $3 \times 10^7 \text{ J}$). The common wrong step: using the atomic mass in kilograms for the molar mass, or forgetting to convert MeV to joules ($1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$).

Nuclear fusion is the joining of light nuclei, for example ²₁H + ²₁H → ³₂He + n + 3.27 MeV, or ²₁H + ²₁H → ³₁H + ¹₁H + 4.03 MeV. Two nuclei both carry positive charge, so they repel strongly and must overcome a Coulomb barrier of about 400 keV for two protons. Matching that energy to the thermal energy $(3/2)kT$ gives $T$ of about $3 \times 10^9$ K, which is why this is called thermonuclear fusion.

The Sun's core is only about $1.5 \times 10^7$ K, far below that. Fusion goes ahead because a small fraction of protons in the high-energy tail of the speed distribution is fast enough. The Sun's energy comes from the proton–proton cycle, whose net result is

4 ¹₁H + 2e⁻ → ⁴₂He + 2ν + 6γ + 26.7 MeV

The first step, two protons becoming a deuteron with the emission of a positron and a neutrino, depends on the weak interaction and is extremely slow. That slowness is why the Sun burns steadily for billions of years instead of exploding. In about 5 billion years the Sun's core hydrogen will be used up, helium will start to fuse into carbon at about $10^8$ K, and so on. Controlled fusion on Earth aims at a steady power supply at about $10^8$ K, where the fuel is a plasma that cannot touch the walls of any container.

Beyond the textbook. For the exam: Sun's fusion works because of the high-energy tail of the proton speed distribution. In reality: even the tail is not enough. Quantum tunnelling lets protons slip through the Coulomb barrier without climbing over it. Without tunnelling, the Sun's core would be too cold for fusion at any useful rate. Why it matters: the same idea of a wave leaking through a barrier appears in the α-decay of heavy nuclei.

Beyond the textbook. For the exam: mass changes matter only in nuclear reactions, and chemistry treats mass as conserved. In reality: chemical reactions also change mass, by about one part in a billion, which no balance can detect. In the Sun, 4 million tonnes of mass are turned into radiation every second, found by dividing its power output (about $3.8 \times 10^{26}$ W) by $c^2$. Why it matters: it shows how big a nuclear mass defect is, and why $E = mc^2$ only matters at nuclear scales.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Atoms found the nucleus through α scattering. This chapter opens it up.
• Next: Semiconductor Electronics moves from the nucleus back to the electrons in a solid.
• Related: Dual Nature of Radiation and Matter supplies the photon picture needed for γ-rays, and Electric Charges and Fields is the source of the proton–proton repulsion that the nuclear force must beat.