Radioactive Decay

A nuclide is written $^{A}_{Z}X$, where $Z$ counts the protons and $A$ counts the total nucleons. Isotopes of an element share the same $Z$ but contain different numbers of neutrons, so they have identical chemistry yet very different nuclear stability: carbon-12 is stable while carbon-14 is radioactive.

Careful measurement shows that a bound nucleus is always lighter than the separated protons and neutrons that make it up. This missing mass is the mass defect, given by:

$$\Delta m = (Z m_p + (A-Z)m_n) - m_{\text{nucleus}}$$

When individual nucleons combine to form a nucleus, the strong nuclear force pulls them into a lower potential-energy state and the released energy carries the lost mass away with it. The same amount of energy, the binding energy, would have to be supplied to tear the nucleus apart again.

The higher the binding energy per nucleon, the more stable the nucleus, because more energy is required per particle to break it up. Dividing the total binding energy by $A$ gives the binding energy per nucleon, the best single measure of stability. Plotted against $A$ it:

• rises steeply through the light nuclei
• peaks near $^{56}\text{Fe}$ at about $8.8\ \text{MeV}$
• then declines gently for heavier nuclei

This is why fusing light nuclei and splitting heavy ones both release energy. You must be able to compute $\Delta m$ from tabulated masses in $\text{u}$ and convert it to energy using $1\ \text{u} = 931.5\ \text{MeV}/c^2$.

Einstein's relation says that mass and energy are interchangeable, expressed as:

$$E = mc^2$$

Any energy released by a system carries mass away with it, and any energy absorbed adds mass. Because $c^2$ is about $9.00 \times 10^{16}\ \text{m}^2\,\text{s}^{-2}$, a minute change in mass represents an enormous energy. Converting just one gram of mass entirely into energy would release $9.0 \times 10^{13}\ \text{J}$, roughly the output of a large power station running for a day.

In nuclear physics the relation is applied as:

$$E = \Delta m c^2$$

The mass defect of a nucleus gives its binding energy, and the difference between the mass of a parent nucleus and the total mass of its decay products gives the energy released in the decay, which appears as kinetic energy of the emitted particles and the energy of any photons.

At the nuclear scale it is usually easier to avoid SI units: masses are quoted in unified atomic mass units with $1\ \text{u} = 931.5\ \text{MeV}/c^2$, and energies in $\text{MeV}$.

You must be able to apply the conversion in both directions, deciding when kilograms and joules are appropriate (macroscopic energy comparisons) and when $\text{u}$ and $\text{MeV}$ are quicker (single-nucleus calculations), and to keep full precision in mass values because the defect is a small difference between large numbers.

Every pair of protons in a nucleus repels electrostatically, so electromagnetism alone would blow a nucleus apart. Nuclei hold together because of the strong nuclear force, an attraction between nucleons that does not depend on charge: proton-proton, proton-neutron and neutron-neutron pairs all feel it equally.

Its defining feature is its range:

• The force is strongly attractive between separations of roughly $1\ \text{fm}$ and $3\ \text{fm}$
• It becomes strongly repulsive below about $0.5\ \text{fm}$ (which stops nucleons collapsing into one another)
• It falls to zero beyond about $3\ \text{fm}$

Contrast this with the electrostatic force, which weakens with distance but never switches off.

This short range explains the pattern of nuclear stability. Each nucleon is bound only to its nearest neighbours, while every proton repels every other proton across the entire nucleus, however far apart they sit. Larger nuclei therefore need a growing excess of neutrons, which contribute strong-force attraction without adding any repulsion; eventually even a neutron surplus is not enough, and all very heavy nuclides are unstable.

You should be able to describe these range characteristics precisely and use them to explain why the neutron-to-proton ratio of stable nuclei increases with $Z$, and why heavy nuclei tend to decay by emitting alpha particles.

An unstable nucleus can move to a lower-energy state in three main ways:

• Alpha decay ejects a tightly bound helium nucleus $^{4}_{2}\alpha$, lowering $A$ by 4 and $Z$ by 2; for example $^{241}_{95}\text{Am} \to {}^{237}_{93}\text{Np} + {}^{4}_{2}\alpha$.
• Beta decay emits a fast electron or positron together with an antineutrino or neutrino.
• Gamma emission occurs when a nucleus left in an excited state after a decay relaxes by emitting a high-energy photon; $A$ and $Z$ are unchanged.

Decay equations must balance: total nucleon number $A$ and total charge $Z$ are the same on both sides, which lets you deduce the daughter nuclide. For carbon-14:

$$^{14}_{6}\text{C} \to {}^{14}_{7}\text{N} + e^- + \bar{\nu}_e$$

While alpha decay is governed by the competition between the strong force and electrostatic repulsion, beta decay is mediated by the weak nuclear force. In $\beta^-$ decay a neutron transforms into a proton, $n \to p + e^- + \bar{\nu}_e$; in $\beta^+$ decay a proton transforms into a neutron, $p \to n + e^+ + \nu_e$.

These conversions allow an unbalanced nucleus to adjust its neutron-to-proton ratio and settle into a more stable configuration.

Alpha particles from a given nuclide carry discrete energies, but beta particles emerge with a continuous spread of energies, the evidence that a neutrino shares the released energy. You must be able to write and balance equations for all three decay types.

Radioactive decay is random and spontaneous: there is no way to predict when a particular nucleus will decay, and no physical or chemical condition (temperature, pressure, bonding) changes the probability.

What is predictable is the statistical behaviour of a large sample, summarised by the half-life $T_{1/2}$: the time for the number of active nuclei, and therefore the activity, to fall by half. After $n$ half-lives the fraction remaining is:

$$\left(\frac{1}{2}\right)^n$$

Iodine-131, with $T_{1/2} \approx 8\ \text{days}$, falls to one sixteenth of its initial activity in $32\ \text{days}$.

Activity $A$ is the number of decays per second occurring in the whole sample, measured in becquerels ($\text{Bq}$). A Geiger-Muller tube does not measure activity directly: it records a count rate, which is smaller because the detector intercepts only a fraction of the emitted radiation, and the reading also includes background radiation from rocks, cosmic rays and other everyday sources.

Always measure the background separately and subtract it before analysing data.

You must be able to extract a half-life from a decay curve or a table of corrected count rates using the repeated-halving method, and to explain why repeated measurements over several half-lives give a more reliable value than a single interval.

Each nucleus of a given nuclide has a fixed probability of decaying per unit time, the decay constant $\lambda$. A sample of $N$ nuclei therefore decays at a rate proportional to $N$, so the activity is $A = \lambda N$. Whenever the rate of change of a quantity is proportional to the quantity itself, the result is exponential decay:

$$N = N_0 e^{-\lambda t} \quad \text{and} \quad A = A_0 e^{-\lambda t}$$

Setting $N = N_0/2$ connects the two descriptions of decay speed:

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

A large decay constant therefore means a short half-life.

The exponential form handles times that are not whole numbers of half-lives. Carbon-14 has $T_{1/2} = 5730\ \text{years}$, so $\lambda = \frac{\ln 2}{5730} = 1.21 \times 10^{-4}\ \text{yr}^{-1}$; a bone sample retaining $30\%$ of its original carbon-14 has age $t = \frac{\ln(0.30)}{-\lambda} \approx 9.95 \times 10^{3}\ \text{years}$.

Taking natural logarithms gives $\ln A = \ln A_0 - \lambda t$, so a graph of $\ln A$ against $t$ is a straight line of gradient $-\lambda$, the standard experimental route to a decay constant.

You must be able to:

• convert between $\lambda$ and $T_{1/2}$
• rearrange the exponential equations with logarithms to solve for any unknown
• keep $\lambda$ and $t$ in the same time units throughout a calculation