- understand the equivalence between energy and mass as represented by $E = mc^2$ and recall and use this equation
- represent simple nuclear reactions by nuclear equations of the form $^{14}_{7}\text{N} + ^{4}_{2}\text{He} \rightarrow ^{17}_{8}\text{O} + ^{1}_{1}\text{H}$
- define and use the terms mass defect and binding energy
- sketch the variation of binding energy per nucleon with nucleon number
- explain what is meant by nuclear fusion and nuclear fission
- explain the relevance of binding energy per nucleon to nuclear reactions, including nuclear fusion and nuclear fission
- calculate the energy released in nuclear reactions using $E = c^2 \Delta m$
Nuclear physics
A-Level Physics · Topic 23
23.1
Mass-energy equivalence
Syllabus
Source: Cambridge International syllabus
Einstein's special relativity gives the famous link (mass-energy equivalence 质能等价):
Mass and energy are equivalent and can change into each other
$$E = m c^{2},$$
where $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$. A mass $m$ matches an energy 能量 $E$ — the two can change into each other. For a mass change $\Delta m$:
$$\Delta E = c^{2} \Delta m.$$
In nuclear physics the masses are tiny but $c^{2}$ is huge, so a small mass change means a large energy. A mass change of $1\ \text{u}$ ($1.661 \times 10^{-27}\ \text{kg}$) matches $\Delta E \approx 1.49 \times 10^{-10}\ \text{J}$. This gives a conversion you will use again and again:
$$1\ \text{u} \ \longleftrightarrow\ 931\ \text{MeV}.$$
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| mass-energy equivalence | 质能等价 | zhì néng děng jià |
| energy | 能量 | néng liàng |
23.1
Nuclear reactions
A nuclear reaction 核反应 is written like
$$^{14}_{7}\text{N} + {}^{4}_{2}\text{He} \to {}^{17}_{8}\text{O} + {}^{1}_{1}\text{H},$$
with nucleon number 核子数 conserved (top numbers: $14 + 4 = 17 + 1$) and charge conserved (bottom numbers: $7 + 2 = 8 + 1$) — this is conservation of charge 电荷守恒. Use these to fill in an unknown: identify the species, then balance the top and bottom numbers.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| nuclear reaction | 核反应 | hé fǎn yìng |
| nucleon number | 核子数 | hé zǐ shù |
| conservation of charge | 电荷守恒 | diàn hè shǒu héng |
23.1
Mass defect and binding energy
The mass of a nucleus 原子核 is less than the total mass of its separate protons 质子 and neutrons 中子. The difference is the mass defect 质量亏损 $\Delta m$:
$$\Delta m = (Z m_{\text{p}} + N m_{\text{n}}) - m_{\text{nucleus}}.$$
By $E = mc^{2}$, this "missing" mass was released as energy when the nucleus formed. To pull the nucleus fully apart you must put that energy back — the binding energy 结合能 $B$:
$$B = \Delta m \cdot c^{2}.$$
The assembled nucleus has less mass than its separate nucleons; the missing mass is released as binding energy
Worked example. A helium-4 nucleus has a mass defect of $\Delta m = 0.0304\ \text{u}$. Find its binding energy. ($1\ \text{u}$ corresponds to $931\ \text{MeV}$.)
$$B = \Delta m \cdot c^{2} = 0.0304 \times 931 \approx 28\ \text{MeV}.$$
A more tightly bound nucleus has a larger mass defect and larger binding energy. The binding energy per nucleon 比结合能 is $B/A$ (usually in MeV per nucleon) — a measure of how tightly each nucleon is held, useful for comparing nuclides.
Binding energy per nucleon vs nucleon number
A graph of $B/A$ against $A$ has a typical shape:
- for light nuclei ($A < 20$), $B/A$ rises quickly (with a spike at the very stable $^{4}_{2}\text{He}$).
- around $A \sim 56$ (iron), $B/A$ reaches its maximum of about $8.8\ \text{MeV}$. Iron-56 is the most stable nucleus.
- for heavy nuclei ($A > 100$), $B/A$ falls slowly, to about $7.5\ \text{MeV}$ for uranium.
So the curve is dome-shaped, rising to iron then falling.
Binding energy per nucleon peaks near iron ($A \approx 56$); lighter and heavier nuclei are less tightly bound
Explore
Mass defect energy lab
E = delta m c^2
Change mass defect and see binding energy rise with E = mc^2.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| nucleus | 原子核 | yuán zǐ hé |
| proton | 质子 | zhì zi |
| neutron | 中子 | zhōng zi |
| mass defect | 质量亏损 | zhì liàng kuī sǔn |
| binding energy | 结合能 | jié hé néng |
| binding energy per nucleon | 比结合能 | bǐ jié hé néng |
23.1
Nuclear fusion and fission
A nuclear power station releases energy by nuclear fission.
Energy is released when nuclei move towards the iron peak — by joining light nuclei or splitting heavy ones.
Fusion joins light nuclei; fission splits a heavy nucleus — both release energy by moving towards the iron peak
Nuclear fusion
Nuclear fusion 核聚变 joins two light nuclei into one heavier nucleus:
$$^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \to {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n} + \text{energy}.$$
The product has greater binding energy per nucleon than the reactants, so energy is released. Fusion powers stars. It needs very high temperatures (millions of kelvin) so the nuclei have enough kinetic energy 动能 to beat their electrostatic 静电 repulsion and get close enough for the strong nuclear force 强核力 to take over.
Nuclear fission
Nuclear fission 核裂变 splits a heavy nucleus into two lighter ones:
$$^{235}_{92}\text{U} + {}^{1}_{0}\text{n} \to {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\, {}^{1}_{0}\text{n} + \text{energy}.$$
The products have higher binding energy per nucleon than $^{235}$U, so energy is released. The extra neutrons can cause more fissions — a chain reaction 链式反应 in a large enough mass of fuel (the critical mass 临界质量). This is the basis of nuclear power and weapons.
In an uncontrolled chain reaction each fission of uranium-235 releases neutrons that cause more fissions
Calculating the energy released
- find the total mass of the reactants.
- find the total mass of the products.
- mass change $\Delta m = m_{\text{reactants}} - m_{\text{products}}$ (positive when energy is released).
- energy released $\Delta E = c^{2} \Delta m$.
In kg this gives joules; in atomic mass units use $\Delta E\ (\text{MeV}) = \Delta m\ (\text{u}) \times 931$.
Worked example. In a nuclear reaction the total mass decreases by $0.020\ \text{u}$. Find the energy released.
$$\Delta E = \Delta m\ (\text{u}) \times 931 = 0.020 \times 931 \approx 19\ \text{MeV}.$$
Explore
Nuclear fission chain reaction
A neutron splits a heavy nucleus, releasing energy and more neutrons — which split more nuclei.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| nuclear fusion | 核聚变 | hé jù biàn |
| kinetic energy | 动能 | dòng néng |
| electrostatic | 静电 | jìng diàn |
| strong nuclear force | 强核力 | qiáng hé lì |
| nuclear fission | 核裂变 | hé liè biàn |
| chain reaction | 链式反应 | liàn shì fǎn yìng |
| critical mass | 临界质量 | lín jiè zhì liàng |
23.2
Radioactive decay
Syllabus
- understand that fluctuations in count rate provide evidence for the random nature of radioactive decay
- understand that radioactive decay is both spontaneous and random
- define activity and decay constant, and recall and use $A = \lambda N$
- define half-life
- use $\lambda = 0.693 / t_{\frac{1}{2}}$
- understand the exponential nature of radioactive decay, and sketch and use the relationship $x = x_0 e^{-\lambda t}$, where $x$ could represent activity, number of undecayed nuclei or received count rate
Source: Cambridge International syllabus
Random and spontaneous
Radioactive decay is:
- spontaneous 自发 — it happens with no outside trigger, and the rate is not changed by temperature, pressure or chemical state; and
- random 随机 — you cannot predict when a given nucleus will decay, only the probability that it decays in a time.
Evidence for randomness: the count rate fluctuates. A Geiger counter 盖革计数器 next to a source clicks at uneven intervals — never a steady stream — although the long-run mean rate is well-defined.
Each beta particle from the source leaves a thin track in a cloud chamber -- direct evidence of separate, random decays
Activity and decay constant
For $N$ undecayed nuclei of a radionuclide 放射性核素, the rate of decay is
$$A = \lambda N.$$
- $A$ is the activity 活度 — decays per unit time. Unit: becquerel 贝克勒尔 (Bq) $= \text{s}^{-1}$.
- $\lambda$ is the decay constant 衰变常数 — the probability per unit time that a nucleus decays. Unit: $\text{s}^{-1}$.
$\lambda$ is fixed for a nuclide; a larger sample (larger $N$) has proportionally larger activity.
Exponential decay
Since $\lambda$ is the fractional decay rate, $\dfrac{dN}{dt} = -\lambda N$, whose solution is an exponential decay 指数衰减:
$$N = N_{0} e^{-\lambda t}.$$
Because $A = \lambda N$, the activity (and any count rate 计数率 proportional to it) decays the same way:
$$A = A_{0} e^{-\lambda t}.$$
Why exponential? For each nucleus, $\lambda$ is a fixed probability per unit time, independent of the others and of the nucleus's age. So the same fraction decays in each time interval, which gives exponential decay.
Half-life
The half-life 半衰期 $t_{1/2}$ is the time for the number of undecayed nuclei (or the activity, or the count rate) to fall to half. From $N = N_{0} e^{-\lambda t}$ with $N = N_{0}/2$:
$$\ln 2 = \lambda t_{1/2}, \qquad \lambda = \frac{\ln 2}{t_{1/2}} \approx \frac{0.693}{t_{1/2}}.$$
Worked example. A radioactive isotope has a half-life of $6.0$ hours. Find its decay constant.
Convert the half-life to seconds: $6.0\ \text{h} = 21\,600\ \text{s}$. Then
$$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{21\,600} \approx 3.2 \times 10^{-5}\ \text{s}^{-1}.$$
A larger decay constant means a shorter half-life. After $n$ half-lives the surviving fraction is $(1/2)^{n}$; after 5 half-lives only about 3% remains.
The number of undecayed nuclei falls by half in each half-life
Finding $\lambda$ from data
Given $A_{0}$ and $A$ at time $t$:
$$\lambda = \frac{1}{t} \ln\frac{A_{0}}{A}, \qquad t_{1/2} = \frac{\ln 2}{\lambda}.$$
Taking logs of $A = A_{0} e^{-\lambda t}$ gives $\ln A = \ln A_{0} - \lambda t$, so a plot of $\ln A$ against $t$ is a straight line with gradient $-\lambda$. Use this with several data points.
A graph of $\ln A$ against time is a straight line of gradient $-\lambda$
Explore
Decay equations (α, β, γ)
Choose a decay type; the daughter nuclide is fixed so the nucleon number A and the proton number Z both balance.
Explore
Half-life — watch the nuclei decay
Each nucleus has a fixed chance of decaying, at random. Move time forward: about half the remaining nuclei decay every half-life — so the count halves, then halves again.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| spontaneous | 自发 | zì fā |
| random | 随机 | suí jī |
| Geiger counter | 盖革计数器 | gài gé jì shù qì |
| radionuclide | 放射性核素 | fàng shè xìng hé sù |
| activity | 活度 | huó dù |
| becquerel | 贝克勒尔 | bèi kè lēi ěr |
| decay constant | 衰变常数 | shuāi biàn cháng shù |
| exponential decay | 指数衰减 | zhǐ shù shuāi jiǎn |
| count rate | 计数率 | jì shù lǜ |
| half-life | 半衰期 | bàn shuāi qī |
23.2
Exam tips
- $E = mc^2$; the mass defect is the mass lost when nucleons bind, released as binding energy.
- Binding energy per nucleon peaks near iron — both fusion (light nuclei) and fission (heavy nuclei) move towards it and release energy.
- Activity $A = \lambda N$; decay is exponential $N = N_0 e^{-\lambda t}$; half-life $t_{1/2} = \ln 2 / \lambda$.
- Balance nuclear equations by conserving nucleon number and proton number.