Nuclear physics: A-level Physics AQA revision notes

3.8.1

Radioactivity

Rutherford scattering 3.8.1.1

Notes

Rutherford, Geiger and Marsden fired a collimated beam of alpha particles at a thin gold foil in a vacuum, and used a movable fluorescent screen to observe the angles through which particles were scattered.
Most alpha particles passed through the foil with little or no deflection, showing that most of an atom's volume is essentially empty space.
A small fraction — roughly 1 in 8000 — scattered through large angles, some almost straight back the way they came. This was the genuinely unexpected result: the earlier 'plum pudding' model, with positive charge spread diffusely through the atom, could not produce any large-angle scattering at all.
Only a very small, dense, positively charged region could repel a positive alpha particle strongly enough at close range to reverse its direction — the direct evidence for a nuclear model of the atom, with almost all its mass and all its positive charge concentrated in a tiny central nucleus, surrounded by mostly empty space occupied by electrons.
The rare large-angle events carry the entire evidential weight of the experiment: without them, the scattering pattern alone could not distinguish a nuclear model from a diffuse-charge model.

α, β and γ radiation 3.8.1.2

Definitions

Ionising radiation: radiation energetic enough to remove electrons from atoms or molecules it passes through, creating ions.
Alpha (α) particle: a helium nucleus, $_{2}^{4} He$ , charge $+ 2 e$ .
Beta-minus (β⁻) particle: a fast electron emitted from the nucleus, charge $- e$ .
Gamma (γ) ray: a high-energy electromagnetic photon, with no charge and no rest mass.

Notes

α radiation is intensely ionising but has a very short range — a few centimetres in air — and is stopped completely by a sheet of paper or the dead outer layer of skin.
β⁻ radiation is less densely ionising than α and has a longer range, but is stopped by a few millimetres of aluminium.
γ radiation is only weakly (indirectly) ionising and is the most penetrating: its intensity is progressively reduced, never sharply cut off, by increasing thicknesses of a dense absorber such as lead — 'stopping' γ radiation really means reducing it to an acceptably low intensity, not eliminating it at a sharp thickness threshold the way paper stops α or aluminium stops β.
Penetrating power and ionising power trade off against each other across the three types: α is the most ionising and least penetrating, γ the least ionising and most penetrating, with β intermediate on both counts.
This trade-off is exactly why the same radiation type poses very different hazards depending on whether its source is outside or inside the body: an external γ source can irradiate organs deep inside the body, while an internal α source (inhaled or ingested) deposits all its energy over a very short range in the immediately surrounding tissue, causing intense local damage disproportionate to how far the radiation could otherwise travel.
Industrial thickness monitoring exploits differing penetration directly, matching a source and absorber to the material being checked — β radiation through paper or thin aluminium for thin sheet, γ radiation through thicker steel for heavier stock — so that small changes in transmitted count rate reveal small changes in sheet thickness.
'Danger' cannot be ranked from penetrating power alone: it depends on the absorbed dose actually received, on the type of radiation (each deposits energy differently per unit path length), and on which tissue is exposed.

γ radiation and the inverse-square law 3.8.1.2

Definitions

Corrected (net) count rate: the measured count rate with background radiation subtracted, $C_{net} = C_{measured} - C_{background}$ .

Key results

Inverse-square law for γ radiation from a point source: corrected count rate $C \propto \frac{1}{r ^{2}}$ , where $r$ is measured from the source to the effective position of the detector.
Plotting corrected count rate against $1/ r^{2}$ gives a straight line through the origin if the inverse-square law holds.

Method

Required practical 12: use a Geiger–Müller (G–M) tube and counter to measure count rate from a sealed γ source at several source–detector distances $r$ .
Measure background count rate separately, over a long time interval with the source removed or shielded, and subtract it from every reading to give the corrected (net) count rate at each distance.
Use long counting times or repeated counts at each distance to average out the random fluctuation inherent in radioactive decay, keeping detector settings identical throughout.
Plot corrected count rate against $1/ r^{2}$ : a straight line through the origin confirms the inverse-square law.

In practiceWith a background of 30 counts per minute, the counter reads 430 per minute at 10 cm. Predict the reading at 20 cm.

$C = 430 - 30 = 400 min^{- 1} at r = 10 cm$ subtract background first
$C \propto \frac{1}{r ^{2}} ⟹ C_{20} = 400 \times (\frac{10}{20})^{2} = 100 min^{- 1}$
$reading at 20 cm = 100 + 30 = 130 min^{- 1}$ background included again

Notes

Background radiation is always present and must be measured and subtracted, not ignored — it comes from cosmic rays, radon gas, naturally radioactive rocks and building materials, and radionuclides naturally present within living things.
The inverse-square law only holds cleanly if several practical conditions are met: the source and detector are small compared with $r$ , their relative alignment is kept fixed, absorption and scattering of γ rays by the intervening air is negligible, and count rates are kept low enough to avoid the detector's dead-time (its brief non-responsive period after each detection) causing it to undercount.
Measured count rate is not the same as source activity: it also depends on the detector's own efficiency and on the solid angle (geometry) it subtends at the source — exactly why this practical tracks how count rate changes with distance rather than inferring activity from a single reading.

Radioactive decay and half-life 3.8.1.3

Definitions

Decay constant, $λ$ : the probability of an individual nucleus decaying per unit time — a constant for a given isotope.
Half-life, $T_{1/2}$ : the time for the number of undecayed nuclei (or activity) to fall to half its previous value.
Activity: the number of nuclear decays per unit time, measured in becquerels ( $1 Bq = 1$ decay per second).

Key results

Activity: $A = λ N$ .
Exponential decay: $N = N_{0} e^{- λ t}$ , $A = A_{0} e^{- λ t}$ .
Half-life: $T_{1/2} = \frac{ln 2}{λ}$ .
Straight-line (log-linear) test: $ln N = ln N_{0} - λ t$ — plotting $ln N$ (or $ln A$ ) against $t$ gives a straight line of gradient $- λ$ , confirming exponential decay and giving $λ$ (and so $T_{1/2}$ ) directly from the gradient, without needing to fit a curve.

Notes

Decay is spontaneous and random for an individual nucleus — its decay probability per unit time is constant and completely unaffected by its past ('memoryless'); it is impossible to predict when any one nucleus will decay.
A large population of identical nuclei nonetheless decays with a precisely predictable statistical pattern. The dice analogy makes this concrete: rolling many dice and removing every six after each throw leaves, on average, $5/6$ of the survivors after each roll — an exponential-looking decline built entirely from unpredictable individual events, exactly mirroring how a genuinely random per-nucleus decay probability produces smooth, predictable exponential decay for a large sample.
The smooth exponential curves describe expected values only; real measured count rates fluctuate randomly about that expected curve, particularly for small samples or short counting intervals.
$A = λ N$ shows activity tracking the number of undecayed nuclei directly: activity is highest when a sample is freshly prepared and falls exponentially in step with $N$ , never linearly.
Radioactive dating (e.g. carbon-14 dating) uses an isotope's known half-life to estimate elapsed time from the fraction of the original isotope remaining, but this requires the system to have been closed (no isotope added or lost other than by decay) and the initial ratio of isotopes to be known or reasonably assumed.

Nuclear instability 3.8.1.4

Definitions

Nucleon number (mass number), $A = Z + N$ : the total number of protons and neutrons in a nucleus.
Band of stability: the narrow region of the neutron–proton ( $N$ – $Z$ ) chart within which stable nuclei are found; nuclei outside it are unstable and decay toward it.
Electron neutrino $ν_{e}$ / electron antineutrino $\overset{ν}{ˉ}_{e}$ : near-massless, chargeless particles emitted alongside β⁺/β⁻ decay respectively, carrying away some of the energy released.

Key results

α decay: $_{Z}^{A} X \to_{Z - 2}^{A - 4} Y +_{2}^{4} He$ , with $Δ Z = - 2$ , $Δ N = - 2$ , $Δ A = - 4$ .
β⁻ decay: $_{Z}^{A} X \to_{Z + 1}^{A} Y +_{- 1}^{0} e + \overset{ν}{ˉ}_{e}$ , with $Δ Z = + 1$ , $Δ N = - 1$ , $Δ A = 0$ .
β⁺ decay: $_{Z}^{A} X \to_{Z - 1}^{A} Y +_{+ 1}^{0} e + ν_{e}$ , with $Δ Z = - 1$ , $Δ N = + 1$ , $Δ A = 0$ .
Electron capture: $_{Z}^{A} X +_{- 1}^{0} e \to_{Z - 1}^{A} Y + ν_{e}$ , with $Δ Z = - 1$ , $Δ N = + 1$ , $Δ A = 0$ .
γ emission: photon energy $h f = E_{2} - E_{1}$ , the difference between an excited and a lower nuclear energy state; $A$ , $Z$ and $N$ all unchanged.

Notes

Plotting nuclei by neutron number $N$ against proton number $Z$ shows stable nuclei clustered along a narrow band: light stable nuclei lie close to the $N = Z$ line, while heavier stable nuclei need increasingly more neutrons than protons ( $N > Z$ ) to stay stable, since neutrons add to the strong nuclear force holding the nucleus together without adding to the mutual electrostatic repulsion between protons.
A neutron-rich nucleus (above the band) decays by β⁻ emission, converting a neutron into a proton and moving it toward the band; a proton-rich nucleus (below the band) decays by β⁺ emission or electron capture, converting a proton into a neutron and moving it toward the band from the other side.
Electron capture is an alternative to β⁺ decay available to some proton-rich nuclides: an inner orbital electron is captured directly by the nucleus, producing the same change in $Z$ and $N$ as β⁺ decay but emitting only a neutrino, with no positron.
γ emission alone changes neither $A$ nor $Z$ — it simply lets a nucleus left in an excited state by another decay lose its excess energy as a photon, without becoming a different nuclide.
Technetium-99m (a metastable excited state of technetium-99) decays almost entirely by γ emission with no accompanying particle emission, which is exactly what makes it valuable for medical imaging: it gives a diagnostic γ signal while minimising the more damaging particle dose delivered to the patient.

Nuclear radius 3.8.1.5

Definitions

Distance of closest approach, $r_{m i n}$ : the minimum separation a charged particle (e.g. an α particle) reaches when directed head-on at a nucleus, at which point all its initial kinetic energy has converted to electric potential energy.

Key results

Closest-approach method: equating initial kinetic energy to electric potential energy at closest approach, $E_{k} = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{r _{m i n}}$ , gives $r_{m i n}$ — an upper estimate of nuclear radius.
Electron diffraction method: high-energy electrons (de Broglie wavelength $λ = h / p$ , comparable to nuclear dimensions) diffract around nuclei; treating the nucleus as a circular obstacle, the angle $θ$ of the first diffraction minimum gives the nuclear radius $R$ — at fixed wavelength, a larger nucleus produces a smaller first-minimum angle.
Nuclear radius: $R = R_{0} A^{1/3}$ , with $R_{0} \approx 1.2 \times 1 0^{- 15} m$ common to all nuclei, and $A$ the nucleon number.

Notes

The closest-approach method uses a head-on collision: as an α particle approaches a much heavier, effectively stationary nucleus, its kinetic energy converts entirely to electric potential energy at the instant its speed is momentarily zero. This gives only an upper bound on nuclear radius, since electrostatic repulsion alone (with no nuclear contact) is enough to reverse the particle's motion.
Electron diffraction gives a more direct and more precise measurement: electrons are not repelled electrostatically and can probe the nucleus at much smaller de Broglie wavelengths using higher energies, so the diffraction pattern's first-minimum angle is a wave phenomenon dependent on the nucleus's actual physical size, treated as a circular obstructing object.
Both methods are consistent with $R = R_{0} A^{1/3}$ : nuclear radius scales with the cube root of nucleon number, matching a roughly constant nucleon density across all nuclei — a nucleus with twice the nucleons is not twice the radius, only $2^{1/3} \approx 1.26$ times the radius, since it fills a proportionally larger volume rather than being spread thinner.
Typical nuclear radii calculated this way are a few femtometres ( $1 0^{- 15} m$ ) — around $1 0^{4}$ – $1 0^{5}$ times smaller than a typical atomic radius, which is exactly why Rutherford scattering's rare large-angle events were needed to reveal the nucleus in the first place: it occupies only a tiny fraction of the atom's volume.

Mass, binding energy and energy release 3.8.1.6

Definitions

Mass defect, $Δ m$ : the difference between the mass of a nucleus and the sum of the masses of its separate, unbound constituent nucleons.
Binding energy: the energy needed to completely separate a nucleus into its individual free nucleons — equivalently, the energy released when those separate nucleons come together to form the nucleus.
Atomic mass unit, $1 u \approx 1.661 \times 1 0^{- 27} kg$ , equivalent to a rest energy of $1 u c^{2} \approx 931.5 MeV$ .

Key results

Mass–energy equivalence: $E = Δ m c^{2}$ applies to any nuclear energy change, not only binding energy.
Binding energy per nucleon rises steeply for light nuclei, peaks in the iron–nickel region ( $A \approx 56$ ), then declines slowly for heavier nuclei.
Example fission: $_{0}^{1} n +_{92}^{235} U \to_{56}^{141} Ba +_{36}^{92} Kr + 3_{0}^{1} n + energy$ .
Example fusion: $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He +_{0}^{1} n + energy$ .

Notes

Binding energy per nucleon (total binding energy divided by nucleon number $A$ ) is the quantity that actually measures relative nuclear stability, since it corrects for the trivial fact that a larger nucleus simply has more total binding energy by having more nucleons.
A nucleus near the peak of the binding-energy-per-nucleon curve (iron, nickel) is the most tightly bound per nucleon, and so the most stable configuration available to nuclear matter.
Both fission of a heavy nucleus and fusion of light nuclei move the resulting nuclei toward that peak — exactly why both processes release energy, despite pulling nuclei apart in opposite directions along the curve.
Energy is released whenever the total binding energy of the products exceeds that of the reactants, which corresponds directly to the products having lower total rest mass — the 'missing' mass has been converted into the kinetic energy of the products and any emitted radiation, via $E = Δ m c^{2}$ .
In the example fission reaction, a slow neutron is absorbed by uranium-235, which splits into two lighter daughter nuclei plus several free neutrons — those released neutrons are what allow a chain reaction, since each can go on to trigger further fission in other uranium nuclei.
In the example fusion reaction, deuterium and tritium (hydrogen isotopes) fuse into helium-4 plus a free neutron — fusion requires extremely high temperatures to give the positively charged nuclei enough kinetic energy to approach closely despite their mutual electrostatic repulsion, before the short-range strong nuclear force can bind them together.
Because $c^{2}$ is such a large number, even a mass difference that is a tiny fraction of a single atomic mass unit corresponds to an enormous energy release compared with any chemical reaction, which involves energy changes many orders of magnitude smaller for a comparable number of particles.

A controlled chain reaction 3.8.1.7–3.8.1.8

Definitions

Critical mass: the minimum mass (for a given geometry and material) at which a self-sustaining chain reaction becomes possible.
Moderator: a material that slows fast neutrons released by fission to the lower (thermal) speeds at which they are far more likely to cause further fission.

Key results

Critical (steady-state) chain reaction condition: on average, exactly one neutron from each fission event goes on to cause a further fission event. Fewer than one on average: sub-critical, the reaction dies out. More than one on average: super-critical, the reaction rate grows.

Notes

In a thermal fission reactor, fuel rods (typically containing uranium-235) undergo fission, releasing fast neutrons; these must be slowed by a moderator (commonly water or graphite) before they are likely to cause further fission, since slow ('thermal') neutrons are captured far more effectively by uranium-235 nuclei than fast ones.
A moderator works by elastic collisions between neutrons and light nuclei of similar mass to the neutron itself — a neutron can transfer a much larger fraction of its kinetic energy in a single such collision than it could to a much heavier nucleus, so repeated collisions with light nuclei slow it efficiently, exactly why light materials like water or graphite are chosen over heavy ones.
Control rods (commonly boron or cadmium, both strong neutron absorbers) are inserted or withdrawn to adjust the fission rate directly: inserting them further absorbs more neutrons, lowering the fission rate toward sub-critical; withdrawing them raises the fission rate toward critical.
Coolant (water or carbon dioxide, depending on reactor design) flows through the core to carry the heat released by fission away to generate steam and drive turbines, while shielding (typically thick concrete) absorbs escaping radiation to protect everything and everyone outside the core.
Reaching and holding the critical condition — on average exactly one neutron from each fission causing one further fission — is what gives a steady, controllable power output, deliberately different from the uncontrolled, rapidly runaway super-critical chain reaction used in a fission weapon.
Critical mass is not a fixed number for a given isotope: it depends on the fuel's geometry (more compact shapes lose fewer neutrons through their surface) and on how many neutrons are otherwise lost to absorption or escape rather than causing further fission.
Shutting a reactor down safely still requires ongoing management: control rods are inserted fully to stop the chain reaction, but the fuel continues generating significant heat from the decay of fission products for some time afterward, so removing this decay heat must continue after shutdown to prevent damage.
Spent fuel and other radioactive waste must be handled remotely, shielded and safely contained; some fission products remain radioactive for very long periods, so long-lived waste requires secure long-term isolation from the environment — a significant ongoing part of a reactor's overall cost and safety case.
The general principles for protecting people from ionising radiation exposure apply throughout nuclear power generation and medicine alike: minimise the time spent near a source, increase distance from it (exploiting the inverse-square fall-off in intensity), use shielding appropriate to the radiation type, and prevent contamination (radioactive material getting onto or into the body).
Weighing the benefits of nuclear power and medical uses of radiation against their risks (radiation exposure, waste management) is a recurring theme this specification examines directly — a well-reasoned balance of both sides is expected, not a one-sided answer.

Worked examples

Worked example 3.8.1 · 5 marks

In a Rutherford scattering experiment, an alpha particle of kinetic energy $4.8 MeV$ is fired directly at a gold nucleus ( $Z = 79$ ) and decelerates to rest at its distance of closest approach, before being repelled back the way it came.

Calculate this distance of closest approach. (:

e = 1.60 \times 1 0^{- 19} C

ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}

)

Show worked solution

At closest approach, all the alpha particle's kinetic energy has converted to electric potential energy:

E_{k} = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{r _{m i n}}

(the alpha particle carries charge $2 e$ ).

Converting $E_{k}$ to joules:

4.8 \times 1 0^{6} \times 1.60 \times 1 0^{- 19} = 7.68 \times 1 0^{- 13} J

Rearranging:

r_{m i n} = \frac{2 Z e ^{2}}{4 π ε _{0} E _{k}} = \frac{2 \times 79 \times ( 1.60 \times 1 0 ^{- 19} ) ^{2}}{4 π \times 8.85 \times 1 0 ^{- 12} \times 7.68 \times 1 0 ^{- 13}} \approx 4.75 \times 1 0^{- 14} m

Mark scheme · 5 marks

Recognises all kinetic energy converts to electric potential energy at closest approach 1 mark
Converts $E_{k}$ from MeV to joules, $7.68 \times 1 0^{- 13} J$ 1 mark
Selects and rearranges $E_{k} = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{r _{m i n}}$ for $r_{m i n}$ 1 mark
Substitutes all values correctly, including $Z = 79$ and charge $2 e$ for the alpha particle 1 mark
Calculates $r_{m i n} \approx 4.75 \times 1 0^{- 14} m$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.8.1 · 6 marks

$r$ / m	Corrected count rate $C$ / s⁻¹	$1/ r^{2}$ / m⁻²
0.10	184.0	100.0
0.20	46.0	25.0
0.30	20.4	11.1
0.40	11.5	6.25

In a Required Practical 12 investigation, a student measures the count rate from a sealed γ source at several source–detector distances, first subtracting a measured background count rate of $0.30 s^{- 1}$ from every reading to give the corrected count rate $C$ shown in the table.

(a) Show that the data is consistent with the inverse-square law.

(b) Use the data to predict the corrected count rate at $r = 0.50 m$ .

Show worked solution

(a) The inverse-square law predicts $C \propto 1/ r^{2}$ , so $C$ divided by $1/ r^{2}$ should be constant:

184.0/100.0 = 1.84

46.0/25.0 = 1.84

20.4/11.1 \approx 1.84

11.5/6.25 \approx 1.84

— constant across all four readings (to 3 s.f.), confirming the inverse-square law and giving a gradient of $1.84 s^{- 1} m^{2}$ for a graph of $C$ against $1/ r^{2}$ .

(b) At $r = 0.50 m$ ,

1/ r^{2} = 1/0.25 = 4.0 m^{- 2}

so:

C = 1.84 \times 4.0 \approx 7.4 s^{- 1}

Mark scheme · 6 marks

States the inverse-square law predicts $C \propto 1/ r^{2}$ , so $C / (1/ r^{2})$ should be constant 1 mark
Calculates $C / (1/ r^{2})$ for at least two rows and shows they agree, $\approx 1.84 s^{- 1} m^{2}$ 1 mark
States this constant value is the gradient of a $C$ against $1/ r^{2}$ graph 1 mark
Calculates $1/ r^{2} = 4.0 m^{- 2}$ at $r = 0.50 m$ 1 mark
Uses the gradient to predict $C$ at this distance 1 mark
Calculates $C \approx 7.4 s^{- 1}$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.8.1 · 5 marks

A radioactive isotope has a half-life of 8.0 days.

A sample initially contains $N_{0}$ nuclei of this isotope.

(a) Calculate the decay constant $λ$ , in $s^{- 1}$ .

(b) Calculate the fraction of the original nuclei remaining after 20 days.

Show worked solution

(a):

λ = \frac{ln 2}{T _{1/2}}

Converting the half-life to seconds:

8.0 \times 24 \times 3600 = 6.912 \times 1 0^{5} s

So:

λ = \frac{0.693}{6.912 \times 1 0 ^{5}} \approx 1.00 \times 1 0^{- 6} s^{- 1}

(b):

\frac{N}{N _{0}} = e^{- λ t}

Using $t = 20$ days directly with $T_{1/2} = 8.0$ days:

\frac{N}{N _{0}} = (\frac{1}{2})^{t / T_{1/2}} = (\frac{1}{2})^{20/8.0} = (\frac{1}{2})^{2.5} \approx 0.177

so about 17.7% of the original nuclei remain.

Mark scheme · 5 marks

Converts the half-life to seconds, $6.912 \times 1 0^{5} s$ 1 mark
Calculates $λ = ln 2/ T_{1/2} \approx 1.00 \times 1 0^{- 6} s^{- 1}$ 1 mark
Selects $N / N_{0} = (1/2)^{t / T_{1/2}}$ (or the equivalent exponential form with $λ$ and $t$ in consistent units) 1 mark
Calculates the exponent $t / T_{1/2} = 2.5$ 1 mark
Calculates the remaining fraction $\approx 0.177$ (17.7%) 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.8.1 · 4 marks

The nuclide $_{13}^{27} Al$ has an experimentally measured nuclear radius of $3.6 fm$ .

Use this to calculate a value for $R_{0}$ in the empirical relation $R = R_{0} A^{1/3}$ , and use your value of $R_{0}$ to predict the nuclear radius of $_{50}^{125} Sn$ .

Show worked solution

Rearranging $R = R_{0} A^{1/3}$ :

R_{0} = \frac{R}{A ^{1/3}} = \frac{3.6 \times 1 0 ^{- 15}}{2 7 ^{1/3}} = \frac{3.6 \times 1 0 ^{- 15}}{3.0} = 1.2 \times 1 0^{- 15} m

For $_{50}^{125} Sn$ :

R = R_{0} A^{1/3} = 1.2 \times 1 0^{- 15} \times 12 5^{1/3} = 1.2 \times 1 0^{- 15} \times 5.0 = 6.0 \times 1 0^{- 15} m

(6.0 fm).

Mark scheme · 4 marks

Rearranges $R = R_{0} A^{1/3}$ to make $R_{0}$ the subject 1 mark
Calculates $2 7^{1/3} = 3.0$ and hence $R_{0} = 1.2 \times 1 0^{- 15} m$ 1 mark
Calculates $12 5^{1/3} = 5.0$ 1 mark
Calculates the predicted radius of $_{50}^{125} Sn$ as $6.0 \times 1 0^{- 15} m$ (6.0 fm) 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.8.1 · 6 marks

In one fission reaction, a $_{92}^{235} U$ nucleus (mass $234.9935 u$ ) absorbs a neutron (mass $1.0087 u$ ) and splits into $_{56}^{141} Ba$ (mass $140.8833 u$ ), $_{36}^{92} Kr$ (mass $91.8973 u$ ) and 3 neutrons.

Calculate the mass defect of this reaction and the energy released, in MeV. (:

1 u = 1.66 \times 1 0^{- 27} kg

c = 3.00 \times 1 0^{8} m s^{- 1}

1 eV = 1.60 \times 1 0^{- 19} J

)

Show worked solution

Total mass before:

234.9935 + 1.0087 = 236.0022 u

Total mass after:

140.8833 + 91.8973 + 3 \times 1.0087 = 140.8833 + 91.8973 + 3.0261 = 235.8067 u

Mass defect:

Δ m = 236.0022 - 235.8067 = 0.1955 u = 0.1955 \times 1.66 \times 1 0^{- 27} \approx 3.245 \times 1 0^{- 28} kg

Energy released:

E = Δ m c^{2} = 3.245 \times 1 0^{- 28} \times (3.00 \times 1 0^{8})^{2} \approx 2.92 \times 1 0^{- 11} J

which is:

\frac{2.92 \times 1 0 ^{- 11}}{1.60 \times 1 0 ^{- 19}} \approx 1.83 \times 1 0^{8} eV \approx 183 MeV

Mark scheme · 6 marks

Calculates the total mass before the reaction, $236.0022 u$ 1 mark
Calculates the total mass after the reaction, including all 3 neutrons, $235.8067 u$ 1 mark
Calculates the mass defect $Δ m = 0.1955 u$ 1 mark
Converts $Δ m$ to kilograms, $\approx 3.245 \times 1 0^{- 28} kg$ 1 mark
Uses $E = Δ m c^{2}$ to calculate $E \approx 2.92 \times 1 0^{- 11} J$ 1 mark
Converts to MeV, $E \approx 183 MeV$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.8.1 · 4 marks

In a thermal nuclear reactor, explain the role of

(a) the moderator and

(b) the control rods, and explain what happens to the fission rate if the control rods are withdrawn slightly from a critical reactor.

Show worked solution

(a) The moderator (commonly water or graphite) slows the fast neutrons released by fission to much lower ('thermal') speeds, via repeated elastic collisions with light nuclei of similar mass to a neutron, since slow neutrons are far more likely to cause further fission in uranium-235 than fast ones.

(b) The control rods (commonly boron or cadmium, both strong neutron absorbers) are inserted or withdrawn to directly control the fission rate by absorbing a controlled fraction of the neutrons produced.

Withdrawing the rods slightly from a critical reactor reduces neutron absorption, so more neutrons are available to cause further fission; the fission rate rises, and the reactor becomes supercritical, until the rods are reinserted to restore a steady, critical rate.

Mark scheme · 4 marks

States the moderator slows fast neutrons to thermal speeds via elastic collisions with light nuclei 1 mark
States slow neutrons are far more likely to cause further fission than fast ones 1 mark
States control rods absorb neutrons to control the fission rate 1 mark
Explains that withdrawing the rods reduces absorption, raising the fission rate and making the reactor supercritical 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.8.1 · 4 marks

A student proposes that beta-plus ( $β^{+}$ ) decay could occur for a proton in complete isolation, in exactly the same way it occurs for a proton bound inside a radioactive nucleus.

Explain why this is not possible, and explain what makes $β^{+}$ decay possible when the same proton is bound inside a suitable nucleus.

Show worked solution

A free neutron has more rest mass-energy than a free proton, so converting an isolated free proton into a neutron by $β^{+}$ decay ( $p \to n + e^{+} + ν_{e}$ ) would require creating rest mass-energy from nothing, violating conservation of energy — it is therefore forbidden for an isolated proton.

Inside a suitable nucleus, however, the binding energy of the daughter nucleus can be sufficiently greater than that of the parent nucleus to supply the extra energy needed, making the overall decay energetically possible even though it is not for a free proton alone.

Mark scheme · 4 marks

States a free neutron has greater rest mass-energy than a free proton 1 mark
Explains an isolated proton undergoing $β^{+}$ decay would violate conservation of energy 1 mark
States this is why $β^{+}$ decay of a free proton is forbidden 1 mark
Explains that inside a nucleus, a sufficiently large binding-energy difference between parent and daughter can supply the needed extra energy, making the decay possible 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.8.1 · 4 marks

A patient is injected with a radioactive tracer of half-life 6.0 hours for a medical imaging scan.

Immediately after injection the activity is $4.0 \times 1 0^{7} Bq$ .

The scan itself takes place 12 hours after injection.

Calculate the activity at the time of the scan, and suggest why a short half-life is desirable for this application.

Show worked solution

Since 12 hours is exactly $2$ half-lives ( $12/6.0 = 2$ ), the activity halves twice:

A = 4.0 \times 1 0^{7} \times (\frac{1}{2})^{2} = 1.0 \times 1 0^{7} Bq

A short half-life is desirable because the tracer decays away to a low, safer activity soon after the scan is complete, minimising the total radiation dose the patient receives from material that remains in the body once it is no longer medically useful.

Mark scheme · 4 marks

Recognises 12 hours corresponds to exactly 2 half-lives 1 mark
Calculates the activity as $A_{0} \times (1/2)^{2}$ 1 mark
Calculates $A = 1.0 \times 1 0^{7} Bq$ 1 mark
Explains a short half-life minimises the patient's radiation dose from tracer remaining after the scan 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.