Notes · PhysicsUK · A-Levels

Nuclear physics

This chapter probes the nucleus itself: how Rutherford scattering revealed it, the properties and hazards of the radiations it emits, and the exponential law of radioactive decay. It measures the nucleus - its radius, density and stability - and, through mass-energy equivalence and binding energy, explains the enormous energy released in nuclear fission and fusion.

6 sections~17 min reading time3 competenciesLevel Standard 2 · Advanced 4

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Exam profile

AO1 · Describe Rutherford scattering, the properties of alpha, beta and gamma radiation, radioactive decay and nuclear fission and fusionAO2 · Calculate activity, half-life, nuclear radius and density, mass defect, binding energy and the energy released in reactionsAO3 · Analyse decay curves and log-linear plots, apply the inverse-square law for gamma and interpret the binding-energy-per-nucleon curve

Operators:describecalculatedetermineshow thatexplainestimate

basic level

This is A2 (full A-Level) content. It requires the properties of radiation, the exponential decay law, half-life, nuclear radius, binding energy and the outline of fission and fusion.

higher level

The full A-Level demands the derivation and use of the decay law, log-linear analysis, nuclear-radius and density calculations, and quantitative energy calculations from mass defect via E = mc^2.

Depth

Text

Contents · 6 sections

Nuclear physics

Rutherford scattering and the nuclear atom#

StandardAQA 7408 3.8.1.1DfE GCE Physics - Rutherford scattering

The Rutherford alpha-scattering experiment

Fig. 1Alpha particles strike a thin gold foil: most pass straight through, a few deflect, and a very few bounce back - evidence for a tiny, dense, positive nucleus.

In the Geiger-Marsden experiment, a beam of alpha particles was fired at a very thin gold foil and the scattering angles recorded. Most alpha particles passed almost straight through, a few were deflected through large angles, and a very small fraction - about one in eight thousand - bounced almost straight back. This was, in Rutherford's words, as astonishing as a shell bouncing off tissue paper.

The results overturned the earlier 'plum-pudding' model of diffuse positive charge. They could only be explained if the atom's positive charge and nearly all its mass were concentrated in a tiny central nucleus, with the rest of the atom mostly empty space. Most alphas miss the nucleus and pass through; the rare large deflections occur when an alpha approaches a nucleus closely and is strongly repelled by the concentrated positive charge.

The experiment also allows an estimate of the nuclear size. An alpha particle heading straight at a nucleus slows as it climbs the electric potential, stopping momentarily at the distance of closest approach, where all its kinetic energy has become electric potential energy:

E_k = \dfrac{2Ze^2}{4\pi\varepsilon_0 r}

. Solving for

r

gives an upper limit on the nuclear radius, of the order of

10^{-14}\,\text{m}

- about ten thousand times smaller than the atom.

Electron diffraction, met earlier through the de Broglie wavelength, gives a more precise nuclear radius: high-energy electrons (wavelength comparable to the nucleus) diffract off nuclei, and the first minimum of the diffraction pattern fixes the radius. Together these methods confirmed the nuclear model and measured the nucleus directly.

E_k = \dfrac{2Ze^2}{4\pi\varepsilon_0 r}

Closest approach

At closest approach the alpha's kinetic energy has all become electric potential energy.

Worked example

Distance of closest approach

An alpha particle of kinetic energy $4.0\,\text{MeV}$ is fired directly at a gold nucleus ( $Z = 79$ ). Estimate the closest approach. Take $\dfrac{1}{4\pi\varepsilon_0} = 8.99\times10^{9}\,\text{N m}^2\,\text{C}^{-2}$ , $e = 1.60\times10^{-19}\,\text{C}$ , $1\,\text{MeV} = 1.60\times10^{-13}\,\text{J}$ .

Energy in joules
$E_k = 4.0 \times 1.60\times10^{-13} = 6.4\times10^{-13}\,\text{J}$ .
Equate to electric PE and solve for r
$r = \dfrac{2Ze^2}{4\pi\varepsilon_0 E_k} = \dfrac{8.99\times10^{9} \times 2 \times 79 \times (1.60\times10^{-19})^2}{6.4\times10^{-13}}$ .
$r = \dfrac{3.64\times10^{-26}}{6.4\times10^{-13}} = 5.7\times10^{-14}\,\text{m}$

Result: The closest approach is about $5.7\times10^{-14}\,\text{m}$ , an upper limit for the nuclear radius.

Typical mistakes

Saying the large deflections are common - they are very rare; most alphas pass nearly undeflected.
Forgetting the factor $2Z$ for the alpha (charge $+2e$ ) and the nucleus (charge $+Ze$ ) in the closest-approach energy.

Active revision

An alpha particle of kinetic energy $5.0\,\text{MeV}$ is fired head-on at a gold nucleus ( $Z = 79$ ). Estimate the distance of closest approach.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Properties of alpha, beta and gamma radiation#

StandardAQA 7408 3.8.1.2DfE GCE Physics - properties of radiation

Inverse-square law for gamma intensity

Fig. 2The intensity of gamma radiation from a point source falls as the inverse square of distance: at double the distance the intensity is a quarter.

The three common radiations differ sharply in nature and penetrating power. An alpha particle is a helium nucleus (two protons and two neutrons): highly ionising, but stopped by a few centimetres of air or a sheet of paper. A beta-minus particle is a fast electron: less ionising, penetrating a few millimetres of aluminium. A gamma ray is a high-energy photon: weakly ionising but very penetrating, reduced only by thick lead or concrete. Their charges (alpha

+2e

, beta

-e

, gamma

0

) determine how they deflect in electric and magnetic fields.

The strong ionising power of alpha comes from its charge and mass: it interacts frequently, losing energy quickly over a short range. Beta ionises less and travels further; gamma, being uncharged, interacts rarely and travels furthest. This inverse relationship between ionising power and range governs both detection and hazard.

Because a gamma source emits uniformly in all directions, its intensity obeys an inverse-square law with distance,

I = \dfrac{k}{x^2}

: doubling the distance quarters the intensity. This is the basis of a required practical and of radiation protection - increasing distance is one of the three defences, alongside reducing exposure time and using shielding.

The biological hazard depends on the route of exposure. Outside the body, gamma and beta are the greater risk because alpha cannot penetrate skin; but if an alpha emitter is inhaled or ingested, it becomes the most dangerous because its intense ionisation is deposited directly in living tissue. Background radiation - from rocks, cosmic rays, radon gas and medical sources - must be subtracted from any measured count rate to give the true rate from a source.

I = \dfrac{k}{x^2}

Inverse-square law for gamma

Intensity falls with the square of the distance from a point source.

Worked example

Inverse-square prediction

At $8.0\,\text{cm}$ from a gamma source the corrected count rate is $900\,\text{s}^{-1}$ . Predict the corrected count rate at $24\,\text{cm}$ .

Use the inverse-square ratio
$I \propto 1/x^2$ , so $\dfrac{I_2}{I_1} = \left(\dfrac{x_1}{x_2}\right)^2 = \left(\dfrac{8.0}{24}\right)^2 = \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}$ .
Scale the count rate
$I_2 = 900/9$ .
$I_2 = \dfrac{900}{9} = 100\,\text{s}^{-1}$

Result: The count rate falls to $100\,\text{s}^{-1}$ - tripling the distance cuts the intensity to a ninth.

Typical mistakes

Forgetting to subtract the background count rate before analysing a source.
Assuming alpha is always the safest - it is the most dangerous if the source is taken inside the body.

Active revision

A gamma detector reads a corrected count rate of $640\,\text{s}^{-1}$ at $10\,\text{cm}$ from a source. Predict the corrected count rate at $40\,\text{cm}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Radioactive decay, activity and half-life#

AdvancedAQA 7408 3.8.1.3DfE GCE Physics - radioactive decay

Exponential radioactive decay

Fig. 3The number of undecayed nuclei (and the activity) falls exponentially; each half-life halves the number remaining.

Radioactive decay is random and spontaneous: it is impossible to predict when a given nucleus will decay, and the decay is unaffected by temperature, pressure or chemical state. But with vast numbers of nuclei, the behaviour is statistically precise. The probability of decay per unit time is the decay constant

\lambda

, and the activity - the number of decays per second, in becquerels - is proportional to the number of undecayed nuclei present:

A = \lambda N

Because each decay reduces

N

, the rate of decay falls as the sample decays, giving exponential decay. Since

\dfrac{\text{d}N}{\text{d}t} = -\lambda N

, the solution is

N = N_0\,e^{-\lambda t}

; the activity follows the same law,

A = A_0\,e^{-\lambda t}

. This is identical in form to the capacitor-discharge law, and the same log-linear analysis applies.

The half-life

T_{1/2}

is the time for half the nuclei (or the activity) to decay, and it is constant for a given isotope. Setting

N = \tfrac{1}{2}N_0

in the decay law gives the link

T_{1/2} = \dfrac{\ln 2}{\lambda}

. Half-lives range from fractions of a second to billions of years, which is what makes different isotopes suitable for medical tracers (short) or geological dating (long).

Two applications appear regularly. Radiocarbon dating uses the known half-life of carbon-14 (about

5730\,\text{years}

): a living organism maintains a fixed proportion of carbon-14, which then decays after death, so the residual activity dates the sample. Taking natural logarithms,

\ln N = \ln N_0 - \lambda t

, gives a straight line whose gradient is

-\lambda

- the standard way to find a decay constant from measured data.

A = \lambda N

Activity

Decays per second, proportional to the number of nuclei.

N = N_0\,e^{-\lambda t}

Exponential decay law

The number of undecayed nuclei falls exponentially.

T_{1/2} = \dfrac{\ln 2}{\lambda}

Half-life

Constant for a given isotope; the time for half to decay.

Worked example

Decay constant and activity

A source has a half-life of $6.0\,\text{hours}$ and an initial activity of $8.0\times10^{5}\,\text{Bq}$ . Find the decay constant and the activity after $18\,\text{hours}$ .

Decay constant
$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{6.0 \times 3600} = 3.2\times10^{-5}\,\text{s}^{-1}$ .
Number of half-lives
$18\,\text{hours}$ is exactly three half-lives, so the activity halves three times: $8.0\times10^{5} \to 4.0 \to 2.0 \to 1.0\times10^{5}\,\text{Bq}$ .
$A = A_0\left(\tfrac{1}{2}\right)^{3} = 8.0\times10^{5} \times \tfrac{1}{8} = 1.0\times10^{5}\,\text{Bq}$

Result: The decay constant is $3.2\times10^{-5}\,\text{s}^{-1}$ and the activity after $18\,\text{hours}$ is $1.0\times10^{5}\,\text{Bq}$ .

Typical mistakes

Confusing the decay constant $\lambda$ (probability per unit time) with the half-life (a time).
Using half-lives only for whole numbers of half-lives instead of the exponential law for arbitrary times.

Active revision

A radioactive source of half-life $8.0\,\text{days}$ has an initial activity of $6.4\times10^{5}\,\text{Bq}$ . Find the decay constant and the activity after $20\,\text{days}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Nuclear stability, radius and density#

AdvancedAQA 7408 3.8.1.3AQA 7408 3.8.1.4DfE GCE Physics - nuclear stability and radius

Testing R against the cube root of A

Fig. 4Nuclear radius plotted against

A^{1/3}

gives a straight line through the origin, confirming

R = r_0 A^{1/3}

; the gradient is

r_0 \approx 1.2\,\text{fm}

Whether a nucleus is stable depends on the balance of protons and neutrons. Plotting neutron number

N

against proton number

Z

for stable nuclei gives a band of stability: light stable nuclei lie near

N = Z

, but heavier ones need proportionally more neutrons (whose strong-force attraction offsets the growing proton repulsion), so the band curves above the line. Nuclei above the band are neutron-rich and decay by beta-minus emission; those below are proton-rich and decay by beta-plus emission or electron capture; the heaviest decay by alpha emission.

The size of a nucleus is found from electron-diffraction and scattering experiments to follow

R = r_0 A^{1/3}

, where

A

is the nucleon number and

r_0 \approx 1.2\,\text{fm}

. The cube-root dependence is the signature of constant density: since volume

\propto R^3 \propto A

, each nucleon occupies the same volume regardless of the nucleus. A graph of

R

against

A^{1/3}

is therefore a straight line through the origin of gradient

r_0

The constant density is a striking result. Combining

R = r_0 A^{1/3}

with a nuclear mass of about

A

atomic mass units gives a density

\rho = \dfrac{m}{V} = \dfrac{A u}{\tfrac{4}{3}\pi (r_0 A^{1/3})^3}

; the

A

cancels, leaving a value of about

2.3\times10^{17}\,\text{kg m}^{-3}

for every nucleus. This is some fourteen orders of magnitude denser than everyday matter and is matched only in neutron stars.

The constancy of nuclear density confirms that nuclear matter is essentially incompressible and that nucleons are packed together at a fixed spacing, like drops in a liquid - the 'liquid-drop' picture that underlies the understanding of fission. It is also why the nucleus is so much smaller than the atom: nearly all the atom's volume is the electron cloud.

R = r_0 A^{1/3}

Nuclear radius

The cube-root dependence signals constant nuclear density.

\rho = \dfrac{A u}{\tfrac{4}{3}\pi (r_0 A^{1/3})^3}

Nuclear density

The nucleon number cancels, giving a value independent of $A$ .

Worked example

Nuclear density is constant

Find the radius of a $^{64}\text{Cu}$ nucleus and its density. Take $r_0 = 1.2\,\text{fm}$ and the nucleon mass $u = 1.66\times10^{-27}\,\text{kg}$ .

Radius
$R = r_0 A^{1/3} = 1.2\times10^{-15} \times 64^{1/3} = 1.2\times10^{-15} \times 4.0 = 4.8\times10^{-15}\,\text{m}$ .
Mass and volume
$m = 64 \times 1.66\times10^{-27} = 1.06\times10^{-25}\,\text{kg}$ ; $V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi(4.8\times10^{-15})^3 = 4.63\times10^{-43}\,\text{m}^3$ .
Density
$\rho = m/V$ .
$\rho = \dfrac{1.06\times10^{-25}}{4.63\times10^{-43}} = 2.3\times10^{17}\,\text{kg m}^{-3}$

Result: The radius is $4.8\,\text{fm}$ and the density $2.3\times10^{17}\,\text{kg m}^{-3}$ - the same for every nucleus.

Typical mistakes

Forgetting the cube root in $R = r_0 A^{1/3}$ , or cubing $r_0$ incorrectly in the density calculation.
Assuming stable heavy nuclei have equal numbers of protons and neutrons - they need an excess of neutrons.

Active revision

Calculate the radius of a $^{125}\text{Te}$ nucleus and show that the nuclear density is about $2\times10^{17}\,\text{kg m}^{-3}$ . Take $r_0 = 1.2\,\text{fm}$ and $u = 1.66\times10^{-27}\,\text{kg}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Mass-energy equivalence and binding energy#

AdvancedAQA 7408 3.8.1.5AQA 7408 3.8.1.6DfE GCE Physics - mass and energy

Binding energy per nucleon

Fig. 5Binding energy per nucleon rises steeply, peaks near iron (

A \approx 56

) at about

8.8\,\text{MeV}

, then falls slowly. Fusion (light nuclei) and fission (heavy nuclei) both move towards the peak, releasing energy.

Einstein's mass-energy equivalence,

E = mc^2

, states that mass and energy are two aspects of one quantity: a change in energy is accompanied by a change in mass,

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

. On the nuclear scale masses are quoted in atomic mass units (

1\,\text{u} = 1.66\times10^{-27}\,\text{kg}

) and energies in MeV, with the convenient conversion

1\,\text{u} \equiv 931.5\,\text{MeV}

The mass of any nucleus is always less than the total mass of its separate protons and neutrons; the difference is the mass defect

\Delta m

. This missing mass is the energy released when the nucleons came together to form the nucleus, and it equals the binding energy - the energy that must be supplied to pull the nucleus completely apart:

E_{\text{binding}} = \Delta m\,c^2

. A larger binding energy means a more tightly bound, more stable nucleus.

To compare nuclei of different sizes fairly, we use the binding energy per nucleon,

E_{\text{binding}}/A

. Plotted against nucleon number, it rises steeply for light nuclei, reaches a broad maximum of about

8.8\,\text{MeV}

near iron (

A \approx 56

), then declines slowly for heavy nuclei. Iron and nickel are therefore the most stable nuclei, and this single curve governs the direction of nuclear energy release.

Energy is released whenever a reaction moves nuclei towards the peak of the curve, increasing the binding energy per nucleon. Splitting a very heavy nucleus (fission) into two medium ones climbs the curve on the right; joining two very light nuclei (fusion) climbs it on the left. In both cases the products are more tightly bound than the reactants, so the mass defect is converted into released energy via

E = mc^2

E = mc^2, \qquad \Delta E = \Delta m\,c^2

Mass-energy equivalence

Energy released equals mass defect times $c^2$ .

1\,\text{u} \equiv 931.5\,\text{MeV}

Mass-energy conversion

A convenient nuclear-scale conversion.

Worked example

Binding energy per nucleon of helium-4

A $^{4}\text{He}$ nucleus has mass $4.00150\,\text{u}$ . Its constituents have masses: proton $1.00728\,\text{u}$ , neutron $1.00867\,\text{u}$ . Find the binding energy per nucleon. Take $1\,\text{u} \equiv 931.5\,\text{MeV}$ .

Total mass of separate nucleons
$2 \times 1.00728 + 2 \times 1.00867 = 2.01456 + 2.01734 = 4.03190\,\text{u}$ .
Mass defect
$\Delta m = 4.03190 - 4.00150 = 0.03040\,\text{u}$ .
Binding energy
$E = 0.03040 \times 931.5 = 28.3\,\text{MeV}$ .
Per nucleon
Divide by $A = 4$ .
$\dfrac{E}{A} = \dfrac{28.3}{4} = 7.1\,\text{MeV per nucleon}$

Result: The binding energy per nucleon of helium-4 is $7.1\,\text{MeV}$ , consistent with the curve.

Typical mistakes

Using the atomic mass instead of the nuclear mass, or forgetting to include every nucleon when finding the mass defect.
Confusing binding energy (total) with binding energy per nucleon (divided by $A$ ) when comparing stability.

Active revision

The measured mass of a $^{4}\text{He}$ nucleus is $4.00150\,\text{u}$ . Using proton mass $1.00728\,\text{u}$ and neutron mass $1.00867\,\text{u}$ , find the binding energy per nucleon.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

Nuclear fission and fusion#

AdvancedAQA 7408 3.8.1.7AQA 7408 3.8.1.8DfE GCE Physics - fission and fusion

In induced fission a heavy nucleus such as uranium-235 absorbs a slow (thermal) neutron, becomes unstable and splits into two medium-sized daughter nuclei plus two or three fast neutrons, releasing about

200\,\text{MeV}

per fission. Because the products lie nearer the peak of the binding-energy curve, they are more tightly bound, and the increase in binding energy appears as kinetic energy of the fragments. The emitted neutrons can trigger further fissions - a chain reaction.

A thermal reactor controls this chain reaction. A moderator (water or graphite) slows the fast neutrons to thermal speeds so they are readily absorbed by uranium-235; control rods (boron or cadmium) absorb excess neutrons to keep the reaction just critical (one further fission per fission on average); and a coolant carries the heat to raise steam and drive turbines. Removing the control rods speeds the reaction; inserting them fully shuts it down.

Nuclear fusion joins two light nuclei into a heavier one, as in the Sun where hydrogen nuclei fuse to helium. Per nucleon it releases even more energy than fission, because the left-hand side of the binding-energy curve rises so steeply. But fusion requires extreme temperatures (millions of kelvin) so that the nuclei have enough kinetic energy to overcome their mutual electrostatic repulsion and approach within range of the strong force - a formidable engineering challenge on Earth.

Both processes are quantified through mass-energy equivalence: find the total mass before and after, take the mass defect, and multiply by

c^2

(or use

1\,\text{u} \equiv 931.5\,\text{MeV}

). Fission produces long-lived radioactive waste that must be stored safely; fusion produces far less, using abundant fuel, which is why it is pursued as a future energy source despite the difficulty of containing a plasma at fusion temperatures.

^{235}_{92}\text{U} + \,^{1}_{0}\text{n} \rightarrow X + Y + 2\,\text{or}\,3\,^{1}_{0}\text{n}

Induced fission

A neutron splits uranium-235 into two fragments plus further neutrons.

E = \Delta m \times 931.5\,\text{MeV}

Energy released

From the mass defect of the reaction.

Worked example

Energy from deuterium-tritium fusion

In $^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n}$ , the masses are: deuterium $2.01355\,\text{u}$ , tritium $3.01550\,\text{u}$ , helium-4 $4.00150\,\text{u}$ , neutron $1.00867\,\text{u}$ . Find the energy released. Take $1\,\text{u} \equiv 931.5\,\text{MeV}$ and $1\,\text{MeV} = 1.60\times10^{-13}\,\text{J}$ .

Mass before and after
Before: $2.01355 + 3.01550 = 5.02905\,\text{u}$ . After: $4.00150 + 1.00867 = 5.01017\,\text{u}$ .
Mass defect
$\Delta m = 5.02905 - 5.01017 = 0.01888\,\text{u}$ .
Energy released
$E = 0.01888 \times 931.5 = 17.6\,\text{MeV}$ ; in joules $17.6 \times 1.60\times10^{-13}$ .
$E = 17.6\,\text{MeV} = 2.8\times10^{-12}\,\text{J}$

Result: The reaction releases $17.6\,\text{MeV}$ ( $2.8\times10^{-12}\,\text{J}$ ) per fusion - the reaction powering experimental fusion reactors.

Typical mistakes

Confusing the roles of the moderator (slows neutrons) and the control rods (absorb neutrons).
Forgetting to include every particle (including the neutrons) when finding the mass defect of a reaction.

Active revision

In the fusion reaction $^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n}$ , the mass defect is $0.0189\,\text{u}$ . Calculate the energy released in MeV and in joules.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Nuclear physics

Key points

Distance of closest approach

Key points

Inverse-square prediction

Key points

Decay constant and activity

Key points

Nuclear density is constant

Key points

Binding energy per nucleon of helium-4

Key points

Energy from deuterium-tritium fusion

Sources