IB Physics HL · 鼎睿学苑

Unit E.3: Radioactive decay单元 E.3：放射性衰变

The decay unit of Theme E "Nuclear and quantum physics". Radioactivity as a random and spontaneous process, the three radiations (alpha, beta, gamma) and how they differ in charge, penetration and ionising power, balancing nuclear decay equations with the neutrino and antineutrino, half-life and decay curves, and activity measured in becquerels against a background. The HL extension formalises all of this into the exponential decay law $N = N_0 e^{-\lambda t}$ and the decay-constant relation $\lambda = \ln 2 / T_{1/2}$, letting you solve decay problems with logarithms.主题 E"核物理与量子物理"中的衰变单元。放射性是随机且自发的过程；三种辐射（α、β、γ）在电荷、穿透力与电离能力上的区别；用中微子与反中微子配平核衰变方程；半衰期与衰变曲线；以及以贝克勒尔为单位、相对本底辐射测量的活度。HL 扩展将上述内容形式化为指数衰变定律 $N = N_0 e^{-\lambda t}$ 与衰变常数关系 $\lambda = \ln 2 / T_{1/2}$，让你用对数求解衰变问题。

IB Physics · Theme E.3 · First Assessment 2025 Papers 1 · 2 6 Topics · SL + HL mix6 个核心专题 · SL + HL 混合
Read me first先读这里

How to use this guide本指南使用说明

E.3 splits cleanly into two halves. The SL half is conceptual and bookkeeping: know what alpha, beta and gamma are, balance the conservation of nucleon number and charge in a decay equation, read a half-life off a decay curve, and state what activity means. The HL half is algebraic: the exponential decay law turns "two half-lives have passed" into a logarithm you can solve for any time. Master the conservation rules first, then layer the exponential law on top.E.3 可清晰地分为两半。SL 部分是概念与记账：知道 α、β、γ 是什么，在衰变方程中配平核子数与电荷守恒，从衰变曲线读出半衰期，并能说明活度的含义。HL 部分是代数：指数衰变定律把"已经过两个半衰期"变成可对任意时间求解的对数方程。先掌握守恒规则，再叠加指数定律。

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If you are cramming如果你在临阵磨枪

Memorise the three radiations: alpha $\left({}^{4}_{2}\alpha\right)$ heavy and stopped by paper; beta $\left({}^{0}_{-1}\beta\right)$ light and stopped by aluminium; gamma (massless photon) only attenuated by lead. In every decay equation, top numbers (nucleon number) and bottom numbers (charge) must each balance. After $n$ half-lives, a sample is $\left(\tfrac{1}{2}\right)^{n}$ of its start. Activity $A = \lambda N$ in becquerels (1 decay per second).

背熟三种辐射：α 粒子 $\left({}^{4}_{2}\alpha\right)$ 重、被纸挡住；β 粒子 $\left({}^{0}_{-1}\beta\right)$ 轻、被铝挡住；γ（无质量光子）只被铅减弱。每个衰变方程中，上标（核子数）与下标（电荷）必须各自平衡。经过 $n$ 个半衰期后，样品剩 $\left(\tfrac{1}{2}\right)^{n}$。活度 $A = \lambda N$，单位贝克勒尔（每秒 1 次衰变）。

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If you are going for a 7如果你目标是 7 分

Be fluent moving between $N = N_0 e^{-\lambda t}$, $A = A_0 e^{-\lambda t}$ and the link λ = ln 2 / T₁/₂. Take logs cleanly to solve for $t$ or $\lambda$. Explain why decay is exponential (constant probability per nucleus per unit time) and why an antineutrino appears in beta-minus decay (lepton-number conservation, energy spectrum of the emitted electron). State that randomness means you cannot predict which nucleus decays when, only the average rate.

能在 $N = N_0 e^{-\lambda t}$、$A = A_0 e^{-\lambda t}$ 与关系式 λ = ln 2 / T₁/₂ 之间自如转换。干净地取对数以求 $t$ 或 $\lambda$。解释衰变为何是指数式（每个核每单位时间衰变概率恒定），以及 β⁻ 衰变中为何出现反中微子（轻子数守恒、所发射电子的能谱）。说明随机性意味着无法预测哪个核在何时衰变，只能预测平均速率。

HL flagHL 标记说明 Sections E3.5 (the exponential decay law) and E3.6 (the decay constant and its link to half-life, solving decay problems with logarithms) are HL extension content. SL students may safely skip the HL-flagged blocks; E3.1 to E3.4 are common SL + HL material.E3.5（指数衰变定律）与 E3.6（衰变常数及其与半衰期的关系、用对数求解衰变问题）为 HL 扩展内容。SL 学生可跳过带 HL 标记的段落；E3.1 至 E3.4 为 SL + HL 共有内容。
Topic E3.1 · Randomness, Spontaneity & Radiation TypesTopic E3.1 · 随机性、自发性与辐射类型

Randomness, Spontaneity, and the Three Radiations随机性、自发性与三种辐射 E.3 SL+HL

Two defining properties. Radioactive decay is:
  • Random — you cannot predict which nucleus decays, or when. Only the probability per unit time is fixed.
  • Spontaneous — unaffected by temperature, pressure, or chemical state. An unstable nucleus decays on its own.
The three radiations.
  • Alpha $\alpha = {}^{4}_{2}\mathrm{He}$ nucleus, charge $+2e$, heavy. Most ionising, least penetrating (stopped by paper / few cm of air).
  • Beta-minus $\beta^{-} = {}^{0}_{-1}e$ (fast electron), charge $-e$. Medium ionising, stopped by a few mm of aluminium.
  • Beta-plus $\beta^{+} = {}^{0}_{+1}e$ (positron), charge $+e$. Similar penetration to $\beta^{-}$; annihilates with an electron.
  • Gamma $\gamma$, a high-energy photon, no charge, no mass. Least ionising, most penetrating (only attenuated by thick lead / concrete).
Ranking. Ionising power: $\alpha > \beta > \gamma$. Penetrating power: $\gamma > \beta > \alpha$ (the reverse order).
两条本质性质。放射性（radioactivity）衰变是：
  • 随机的（random）——无法预测哪个核衰变、何时衰变。只有每单位时间的概率是确定的。
  • 自发的（spontaneous）——不受温度、压强或化学状态影响。不稳定核自行衰变。
三种辐射。
  • α 衰变（alpha） $\alpha = {}^{4}_{2}\mathrm{He}$ 核，电荷 $+2e$，重。电离能力最强、穿透力最弱（被纸或几厘米空气挡住）。
  • β⁻ 衰变（beta-minus） $\beta^{-} = {}^{0}_{-1}e$（高速电子），电荷 $-e$。电离能力中等，被几毫米铝挡住。
  • β⁺ 衰变（beta-plus） $\beta^{+} = {}^{0}_{+1}e$（正电子），电荷 $+e$。穿透力与 β⁻ 相近；与电子湮灭。
  • γ 衰变（gamma） $\gamma$，高能光子，无电荷、无质量。电离能力最弱、穿透力最强（只被厚铅或混凝土减弱）。
排序。电离能力：$\alpha > \beta > \gamma$。穿透力：$\gamma > \beta > \alpha$（顺序相反）。
Radiation辐射 Identity本质 Charge电荷 Stopped by被…挡住 Ionising电离
$\alpha$ ${}^{4}_{2}\mathrm{He}$ nucleus${}^{4}_{2}\mathrm{He}$ 核 $+2e$ paper, few cm air纸、几厘米空气 strongest最强
$\beta^{-}$ fast electron高速电子 $-e$ few mm aluminium几毫米铝 medium中等
$\gamma$ high-energy photon高能光子 $0$ thick lead / concrete厚铅 / 混凝土 weakest最弱
Worked Example E3.1 (penetration experiment)E3.1 例题（穿透实验）

A radioactive source gives a count rate of $480\ \mathrm{s^{-1}}$. Placing a sheet of paper in front of it drops the rate to $300\ \mathrm{s^{-1}}$; adding $5\ \mathrm{mm}$ of aluminium drops it further to $40\ \mathrm{s^{-1}}$ (the background). Identify the radiation(s) emitted.某放射源计数率为 $480\ \mathrm{s^{-1}}$。在其前方放一张纸后降至 $300\ \mathrm{s^{-1}}$；再加 $5\ \mathrm{mm}$ 铝后进一步降至 $40\ \mathrm{s^{-1}}$（即本底）。判断所发射的辐射。

Identify. Compare each absorber against what stops each radiation. Background is $40\ \mathrm{s^{-1}}$, so subtract it mentally when judging "fully stopped".

识别。把每种吸收体与各辐射的阻挡能力对照。本底为 $40\ \mathrm{s^{-1}}$，判断"完全挡住"时心算扣除它。

Set up — paper. Paper drops the rate by $480 - 300 = 180\ \mathrm{s^{-1}}$. Only alpha is stopped by paper, so alpha is present (it carries $180\ \mathrm{s^{-1}}$ of the count).

列式——纸。纸使计数率降低 $480 - 300 = 180\ \mathrm{s^{-1}}$。只有 α 被纸挡住，故含有 α（它占 $180\ \mathrm{s^{-1}}$）。

Execute — aluminium. Aluminium drops the remaining $300\ \mathrm{s^{-1}}$ down to the background $40\ \mathrm{s^{-1}}$, removing $260\ \mathrm{s^{-1}}$. A few mm of aluminium stops beta, so beta is present and accounts for the rest. Nothing survives to suggest gamma.

执行——铝。铝把剩余的 $300\ \mathrm{s^{-1}}$ 降到本底 $40\ \mathrm{s^{-1}}$，去除 $260\ \mathrm{s^{-1}}$。几毫米铝能挡住 β，故含 β，占其余部分。没有残余信号提示 γ。

Evaluate. The source emits both $\alpha$ and $\beta$, but no detectable $\gamma$ (if gamma were present, some count would survive the aluminium above background).

评估。该源同时发射 $\alpha$ 与 $\beta$，但无可检测的 $\gamma$（若有 γ，铝后计数会高于本底）。

Going deeper: why decay is unaffected by chemistry深入：为何衰变不受化学影响

Chemical reactions rearrange outer electrons; energies involved are a few eV. Nuclear binding energies are millions of eV (MeV). Heating a sample, dissolving it, or putting it under pressure changes the electron cloud but leaves the nucleus untouched, so the decay rate is a fixed property of the nuclide. This is why a half-life is a reliable clock (radiometric dating) and why you cannot "switch off" a radioactive source by cooling it.

化学反应重排外层电子，涉及能量仅几 eV；核结合能为数百万 eV（MeV）。加热、溶解或加压改变电子云，却不触及核，故衰变速率是核素的固定属性。这正是半衰期可作可靠时钟（放射性测年）的原因，也是为何无法靠降温"关闭"放射源。

Which list orders the radiations by increasing penetrating power?哪个排列按穿透力递增排序？
E3.1 · Q1
$\gamma,\ \beta,\ \alpha$
$\beta,\ \alpha,\ \gamma$
$\alpha,\ \beta,\ \gamma$
$\alpha,\ \gamma,\ \beta$
Penetration increases $\alpha \to \beta \to \gamma$: alpha is stopped by paper, beta by a few mm of aluminium, gamma only by thick lead. Ionising power runs the opposite way.穿透力按 $\alpha \to \beta \to \gamma$ 递增：α 被纸挡，β 被几毫米铝挡，γ 只被厚铅挡。电离能力顺序相反。
The most ionising radiation ($\alpha$) is the least penetrating. So increasing penetration goes $\alpha,\ \beta,\ \gamma$.电离最强的 $\alpha$ 穿透最弱。故穿透递增顺序为 $\alpha,\ \beta,\ \gamma$。
Doubling the temperature of a radioactive sample changes its decay rate by:把放射性样品的温度加倍，其衰变速率改变：
E3.1 · Q2
It doubles.加倍。
It halves.减半。
It increases slightly.略微增大。
No change — decay is spontaneous.不变——衰变是自发的。
Decay is spontaneous: it is a nuclear process independent of temperature, pressure, or chemical environment. The decay constant $\lambda$ is fixed for a given nuclide.衰变是自发的：属核过程，与温度、压强或化学环境无关。给定核素的衰变常数 $\lambda$ 固定。
Temperature affects electron-cloud (chemical) energies of a few eV, not the MeV-scale nucleus. The decay rate is unchanged.温度影响几 eV 的电子云（化学）能量，而非 MeV 级的核。衰变速率不变。
Topic E3.2 · Balancing Nuclear Decay EquationsTopic E3.2 · 配平核衰变方程

Balancing Nuclear Equations and the (Anti)neutrino配平核方程与（反）中微子 E.3 SL+HL

Two conservation rules. In any nuclear decay ${}^{A}_{Z}X \to \dots$, both quantities are conserved across the arrow:
  • Nucleon number $A$ (top): sum before $=$ sum after.
  • Proton number / charge $Z$ (bottom): sum before $=$ sum after.
Standard decays. $$ \text{Alpha: } {}^{A}_{Z}X \to {}^{A-4}_{Z-2}Y + {}^{4}_{2}\alpha $$ $$ \text{Beta-minus: } {}^{A}_{Z}X \to {}^{A}_{Z+1}Y + {}^{0}_{-1}\beta + \bar{\nu}_e $$ $$ \text{Beta-plus: } {}^{A}_{Z}X \to {}^{A}_{Z-1}Y + {}^{0}_{+1}\beta + \nu_e $$ The (anti)neutrino. $\beta^{-}$ decay emits an electron antineutrino $\bar{\nu}_e$; $\beta^{+}$ decay emits an electron neutrino $\nu_e$. They have (almost) no mass and no charge, so they do not change the $A$/$Z$ bookkeeping, but they are required to conserve energy and lepton number. Gamma emission $\left({}^{A}_{Z}X^{*} \to {}^{A}_{Z}X + \gamma\right)$ changes neither $A$ nor $Z$.
两条守恒规则。任何核衰变 ${}^{A}_{Z}X \to \dots$ 中，箭头两侧两个量都守恒：
  • 核子数 $A$（上标）：前 $=$ 后。
  • 质子数 / 电荷 $Z$（下标）：前 $=$ 后。
标准衰变。 $$ \text{α 衰变：} {}^{A}_{Z}X \to {}^{A-4}_{Z-2}Y + {}^{4}_{2}\alpha $$ $$ \text{β⁻ 衰变：} {}^{A}_{Z}X \to {}^{A}_{Z+1}Y + {}^{0}_{-1}\beta + \bar{\nu}_e $$ $$ \text{β⁺ 衰变：} {}^{A}_{Z}X \to {}^{A}_{Z-1}Y + {}^{0}_{+1}\beta + \nu_e $$ （反）中微子（neutrino）。β⁻ 衰变发射电子反中微子 $\bar{\nu}_e$；β⁺ 衰变发射电子中微子 $\nu_e$。它们（几乎）无质量、无电荷，不改变 $A$/$Z$ 记账，但为守恒能量与轻子数所必需。γ 发射 $\left({}^{A}_{Z}X^{*} \to {}^{A}_{Z}X + \gamma\right)$ 既不改变 $A$ 也不改变 $Z$。
Worked Example E3.2 (find the daughter)E3.2 例题（求子核）

Carbon-14 undergoes beta-minus decay. Write the full decay equation, identifying the daughter nuclide. (Carbon is $Z = 6$; nitrogen is $Z = 7$.)碳-14 发生 β⁻ 衰变。写出完整衰变方程并指出子核素。（碳 $Z = 6$；氮 $Z = 7$。）

Identify. Parent ${}^{14}_{6}\mathrm{C}$. Beta-minus emits ${}^{0}_{-1}\beta$ plus an antineutrino $\bar{\nu}_e$.

识别。母核 ${}^{14}_{6}\mathrm{C}$。β⁻ 发射 ${}^{0}_{-1}\beta$ 与一个反中微子 $\bar{\nu}_e$。

Set up — balance the top (nucleon number). $14 = A + 0$, so the daughter has $A = 14$.

列式——配平上标（核子数）。$14 = A + 0$，故子核 $A = 14$。

Execute — balance the bottom (charge). $6 = Z + (-1)$, so $Z = 7$. Element $Z = 7$ is nitrogen, N.

执行——配平下标（电荷）。$6 = Z + (-1)$，故 $Z = 7$。$Z = 7$ 为氮 N。

$$ {}^{14}_{6}\mathrm{C} \to {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}\beta + \bar{\nu}_e. $$

Evaluate. Top: $14 = 14 + 0$. Bottom: $6 = 7 + (-1)$. Both balance. A neutron in the nucleus has effectively become a proton plus an emitted electron, raising $Z$ by one while $A$ stays fixed — the signature of beta-minus.

评估。上标：$14 = 14 + 0$。下标：$6 = 7 + (-1)$。均平衡。核内一个中子实质上变成质子加发射电子，$Z$ 增 1 而 $A$ 不变——这是 β⁻ 的特征。

Going deeper: why the antineutrino had to exist深入：反中微子为何必须存在

If beta-minus were a simple two-body decay $n \to p + e^{-}$, the emitted electron would always carry a single fixed energy (fixed by momentum and energy conservation between two bodies). Experiment instead shows a continuous spectrum of electron energies up to a maximum. Pauli proposed a third, nearly undetectable particle — the antineutrino — to carry the missing, variable share of energy and momentum. It also balances lepton number: the electron is a lepton ($+1$), so an antilepton ($-1$, the antineutrino) must accompany it.

若 β⁻ 是简单的二体衰变 $n \to p + e^{-}$，所发射电子将总带固定能量（由二体的动量与能量守恒确定）。但实验显示电子能量为连续谱，直至某最大值。泡利提出第三个几乎不可探测的粒子——反中微子——以带走缺失且可变的那部分能量与动量。它也使轻子数守恒：电子是轻子（$+1$），故须有一反轻子（$-1$，即反中微子）伴随。

A nuclide ${}^{238}_{92}\mathrm{U}$ emits an alpha particle. The daughter nuclide is:核素 ${}^{238}_{92}\mathrm{U}$ 发射一个 α 粒子。子核素为：
E3.2 · Q1
${}^{238}_{90}\mathrm{Th}$
${}^{234}_{90}\mathrm{Th}$
${}^{234}_{92}\mathrm{U}$
${}^{242}_{94}\mathrm{Pu}$
Alpha removes $A$ by 4 and $Z$ by 2: $238 - 4 = 234$, $92 - 2 = 90$. Element 90 is thorium: ${}^{234}_{90}\mathrm{Th}$.α 使 $A$ 减 4、$Z$ 减 2：$238 - 4 = 234$，$92 - 2 = 90$。第 90 号元素为钍：${}^{234}_{90}\mathrm{Th}$。
An alpha is ${}^{4}_{2}\mathrm{He}$, so subtract 4 from the top and 2 from the bottom. Check both numbers balance.α 是 ${}^{4}_{2}\mathrm{He}$，故上标减 4、下标减 2。检查两个数都平衡。
In the decay ${}^{22}_{11}\mathrm{Na} \to {}^{22}_{10}\mathrm{Ne} + X$, the particle $X$ (with its accompanying lepton) is:在衰变 ${}^{22}_{11}\mathrm{Na} \to {}^{22}_{10}\mathrm{Ne} + X$ 中，粒子 $X$（连同伴随轻子）为：
E3.2 · Q2
$\beta^{-}$ and an antineutrino $\bar{\nu}_e$$\beta^{-}$ 与反中微子 $\bar{\nu}_e$
an alpha particle一个 α 粒子
$\beta^{+}$ and a neutrino $\nu_e$$\beta^{+}$ 与中微子 $\nu_e$
a gamma photon only仅一个 γ 光子
$A$ is unchanged ($22 \to 22$) but $Z$ drops by one ($11 \to 10$), the signature of beta-plus. So $X = {}^{0}_{+1}\beta$ accompanied by an electron neutrino $\nu_e$.$A$ 不变（$22 \to 22$）而 $Z$ 减 1（$11 \to 10$），是 β⁺ 的特征。故 $X = {}^{0}_{+1}\beta$，伴随电子中微子 $\nu_e$。
$Z$ decreases by 1 with $A$ fixed — that is beta-plus, which emits a positron and a neutrino (not an antineutrino).$A$ 不变而 $Z$ 减 1——这是 β⁺，发射正电子与中微子（不是反中微子）。
Topic E3.3 · Half-life and Decay CurvesTopic E3.3 · 半衰期与衰变曲线

Half-life and Reading Decay Curves半衰期与读取衰变曲线 E.3 SL+HL

Half-life $T_{1/2}$. The time for half the radioactive nuclei in a sample to decay — equivalently, the time for the activity (or count rate) to halve. It is constant for a given nuclide. The fraction rule. After $n$ whole half-lives: $$ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{n}, \qquad n = \frac{t}{T_{1/2}}. $$ So after 1, 2, 3 half-lives a sample is at $\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}$ of its start. Reading a graph. From a decay (count-rate vs time) curve, find $T_{1/2}$ by reading the time at which the curve falls to half its initial value. Best practice: read several halvings ($N_0 \to N_0/2 \to N_0/4$) and average the intervals — they should all be equal.
半衰期 $T_{1/2}$（half-life）。样品中一半放射性核衰变所需的时间——等价地，活度（或计数率）减半所需时间。对给定核素为常数。 分数法则。经过 $n$ 个整数半衰期后： $$ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{n}, \qquad n = \frac{t}{T_{1/2}}. $$ 故经过 1、2、3 个半衰期后，样品分别为初始的 $\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}$。 读图。在衰变（计数率-时间）曲线上，读出曲线降到初值一半所对应的时间即 $T_{1/2}$。最佳做法：读多次减半（$N_0 \to N_0/2 \to N_0/4$）并对各时间区间取平均——它们应相等。
Worked Example E3.3 (fraction remaining)E3.3 例题（剩余分数）

Iodine-131 has a half-life of $8.0\ \mathrm{days}$. A hospital sample starts with an activity of $640\ \mathrm{MBq}$. Find (a) the activity after $24\ \mathrm{days}$, and (b) the time for the activity to fall to $40\ \mathrm{MBq}$.碘-131 半衰期为 $8.0\ \mathrm{天}$。某医院样品初始活度为 $640\ \mathrm{MBq}$。求 (a) $24\ \mathrm{天}$ 后的活度，(b) 活度降到 $40\ \mathrm{MBq}$ 所需时间。

Identify. Whole-number half-lives, so use the fraction rule $\left(\tfrac{1}{2}\right)^{n}$ rather than the exponential law.

识别。都是整数倍半衰期，故用分数法则 $\left(\tfrac{1}{2}\right)^{n}$，无需指数定律。

Set up (a). $n = 24 / 8.0 = 3$ half-lives.

列式 (a)。$n = 24 / 8.0 = 3$ 个半衰期。

$$ A = A_0 \left(\tfrac{1}{2}\right)^{3} = 640 \times \tfrac{1}{8} = 80\ \mathrm{MBq}. $$

Execute (b). Halve repeatedly: $640 \to 320 \to 160 \to 80 \to 40$, i.e. 4 halvings. So $t = 4 \times 8.0 = 32\ \mathrm{days}$.

执行 (b)。反复减半：$640 \to 320 \to 160 \to 80 \to 40$，即 4 次减半。故 $t = 4 \times 8.0 = 32\ \mathrm{天}$。

Evaluate. Both answers land on whole half-lives, so no logarithms are needed. If the target activity had been (say) $50\ \mathrm{MBq}$, it would fall between halvings and the HL exponential law would be required.

评估。两问都落在整数半衰期上，无需对数。若目标活度为（例如）$50\ \mathrm{MBq}$，则落在两次减半之间，须用 HL 指数定律。

Going deeper: the constant-ratio test for exponential decay深入：指数衰变的恒比检验

A decay is exponential if equal time intervals always multiply the quantity by the same factor. To verify from data without plotting logs: pick equally spaced times and check that the ratio of successive readings is constant. If readings at $t = 0, 10, 20, 30\ \mathrm{s}$ are $800, 560, 392, 274$, the ratios are $560/800 = 0.70$, $392/560 = 0.70$, $274/392 = 0.70$ — constant, so the decay is exponential. The half-life is then $T_{1/2} = 10 \times \ln 2 / \ln(1/0.70) \approx 19.4\ \mathrm{s}$, anticipating the HL relation in E3.6.

若相等时间间隔总把量乘以相同因子，则衰变为指数式。无需画对数即可验证：取等间隔时间，检验相邻读数之比是否恒定。若 $t = 0, 10, 20, 30\ \mathrm{s}$ 的读数为 $800, 560, 392, 274$，则比值 $560/800 = 0.70$、$392/560 = 0.70$、$274/392 = 0.70$——恒定，故为指数衰变。半衰期 $T_{1/2} = 10 \times \ln 2 / \ln(1/0.70) \approx 19.4\ \mathrm{s}$，预示 E3.6 的 HL 关系。

A sample has a half-life of $5.0\ \mathrm{years}$. The fraction of the original nuclei remaining after $20\ \mathrm{years}$ is:某样品半衰期为 $5.0\ \mathrm{年}$。$20\ \mathrm{年}$ 后剩余原始核的分数为：
E3.3 · Q1
$\tfrac{1}{4}$
$\tfrac{1}{8}$
$\tfrac{1}{20}$
$\tfrac{1}{16}$
$n = 20 / 5.0 = 4$ half-lives. Fraction $= \left(\tfrac{1}{2}\right)^{4} = \tfrac{1}{16}$.$n = 20 / 5.0 = 4$ 个半衰期。分数 $= \left(\tfrac{1}{2}\right)^{4} = \tfrac{1}{16}$。
Count the half-lives: $n = t / T_{1/2} = 4$. Then the remaining fraction is $(1/2)^{n}$.数半衰期个数：$n = t / T_{1/2} = 4$。剩余分数为 $(1/2)^{n}$。
A count rate falls from $480\ \mathrm{s^{-1}}$ to $60\ \mathrm{s^{-1}}$ in $9.0\ \mathrm{hours}$ (background negligible). The half-life is:计数率在 $9.0\ \mathrm{小时}$ 内从 $480\ \mathrm{s^{-1}}$ 降到 $60\ \mathrm{s^{-1}}$（本底可忽略）。半衰期为：
E3.3 · Q2
$3.0\ \mathrm{hours}$
$1.5\ \mathrm{hours}$
$4.5\ \mathrm{hours}$
$9.0\ \mathrm{hours}$
$480 \to 240 \to 120 \to 60$ is 3 halvings. So $3 \, T_{1/2} = 9.0\ \mathrm{h} \Rightarrow T_{1/2} = 3.0\ \mathrm{h}$.$480 \to 240 \to 120 \to 60$ 为 3 次减半。故 $3 \, T_{1/2} = 9.0\ \mathrm{h} \Rightarrow T_{1/2} = 3.0\ \mathrm{h}$。
Count how many times the rate halves from 480 to 60. That count times $T_{1/2}$ equals the elapsed time.数从 480 到 60 减半的次数。该次数乘以 $T_{1/2}$ 等于经过时间。
Topic E3.4 · Activity, the Becquerel and BackgroundTopic E3.4 · 活度、贝克勒尔与本底辐射

Activity, the Becquerel, and Background Radiation活度、贝克勒尔与本底辐射 E.3 SL+HL

Activity. The number of nuclei decaying per unit time. It is proportional to how many undecayed nuclei remain: $$ A = \lambda N $$ where A = λN (data booklet), $N$ is the number of undecayed nuclei and $\lambda$ is the decay constant (the probability per nucleus per unit time of decaying). The becquerel. The SI unit of activity. $1\ \mathrm{Bq} = 1$ decay per second $= 1\ \mathrm{s^{-1}}$. (The older curie, $1\ \mathrm{Ci} = 3.7 \times 10^{10}\ \mathrm{Bq}$, is not required.) Background radiation. Radiation always present from natural and artificial sources: radon gas, rocks and soil, cosmic rays, food and the body, plus medical and nuclear-industry contributions. A measured count rate must have the background subtracted to give the source's true count rate (the corrected count rate).
活度（activity）。单位时间内衰变的核数。它正比于剩余未衰变核数： $$ A = \lambda N $$ 其中 A = λN（数据手册），$N$ 为未衰变核数，$\lambda$ 为衰变常数（每个核每单位时间衰变的概率）。 贝克勒尔（becquerel）。活度的 SI 单位。$1\ \mathrm{Bq} = $ 每秒 1 次衰变 $= 1\ \mathrm{s^{-1}}$。（旧单位居里 $1\ \mathrm{Ci} = 3.7 \times 10^{10}\ \mathrm{Bq}$ 不作要求。） 本底辐射（background radiation）。来自天然与人造源、始终存在的辐射：氡气、岩石与土壤、宇宙射线、食物与人体，以及医疗和核工业的贡献。测得的计数率须减去本底才能得到源的真实计数率（修正后计数率）。
Worked Example E3.4 (activity from number of nuclei)E3.4 例题（由核数求活度）

A sample contains $N = 6.0 \times 10^{18}$ undecayed nuclei of a nuclide whose decay constant is $\lambda = 2.5 \times 10^{-9}\ \mathrm{s^{-1}}$. Find the activity in becquerels. A nearby detector reads $1\,250\ \mathrm{s^{-1}}$ while the background alone is $30\ \mathrm{s^{-1}}$; state the corrected count rate.某样品含 $N = 6.0 \times 10^{18}$ 个未衰变核，其衰变常数 $\lambda = 2.5 \times 10^{-9}\ \mathrm{s^{-1}}$。求活度（贝克勒尔）。附近探测器读数为 $1\,250\ \mathrm{s^{-1}}$，单独本底为 $30\ \mathrm{s^{-1}}$；给出修正后计数率。

Identify. Use the data-booklet relation A = λN.

识别。用数据手册关系 A = λN。

Set up. $A = \lambda N = (2.5 \times 10^{-9})(6.0 \times 10^{18})\ \mathrm{s^{-1}}$.

列式。$A = \lambda N = (2.5 \times 10^{-9})(6.0 \times 10^{18})\ \mathrm{s^{-1}}$。

$$ A = 1.5 \times 10^{10}\ \mathrm{Bq}. $$

Execute — corrected count rate. Subtract the background: $1\,250 - 30 = 1\,220\ \mathrm{s^{-1}}$.

执行——修正后计数率。减去本底：$1\,250 - 30 = 1\,220\ \mathrm{s^{-1}}$。

Evaluate. The activity $1.5 \times 10^{10}\ \mathrm{Bq}$ is the true decay rate inside the sample; the detector's count rate is smaller because it only intercepts a fraction of the emissions and must be background-corrected before any half-life is extracted from it.

评估。活度 $1.5 \times 10^{10}\ \mathrm{Bq}$ 是样品内部真实衰变速率；探测器计数率更小，因为它只截获一部分发射，且须先扣除本底，才能据以求半衰期。

Going deeper: why activity itself decays exponentially深入：活度本身为何指数衰减

Since $A = \lambda N$ and $\lambda$ is constant, the activity is directly proportional to $N$ at every instant. As $N$ falls exponentially (E3.5), so does $A$ — they share the same half-life. This is why we can measure half-life from a count-rate curve rather than counting nuclei directly: the count rate is proportional to the (corrected) activity. It also means a freshly delivered medical isotope is most active on arrival and must be used promptly.

由于 $A = \lambda N$ 且 $\lambda$ 恒定，活度在任一时刻都正比于 $N$。当 $N$ 指数下降（E3.5）时，$A$ 亦然——二者半衰期相同。这正是我们能从计数率曲线（而非直接数核）测半衰期的原因：计数率正比于（修正后）活度。这也意味着新送达的医用同位素到货时活度最高，须尽快使用。

An activity of $1\ \mathrm{Bq}$ corresponds to:$1\ \mathrm{Bq}$ 的活度对应：
E3.4 · Q1
1 nucleus in the sample样品中 1 个核
1 joule released per second每秒释放 1 焦耳
1 decay per second每秒 1 次衰变
1 half-life per second每秒 1 个半衰期
The becquerel is the SI unit of activity: $1\ \mathrm{Bq} = 1$ disintegration per second $= 1\ \mathrm{s^{-1}}$.贝克勒尔是活度的 SI 单位：$1\ \mathrm{Bq} = $ 每秒 1 次衰变 $= 1\ \mathrm{s^{-1}}$。
Activity counts decays, not energy or nuclei. $1\ \mathrm{Bq} = 1$ decay per second.活度计的是衰变次数，不是能量或核数。$1\ \mathrm{Bq} = $ 每秒 1 次衰变。
A detector reads $210\ \mathrm{s^{-1}}$ near a source and $30\ \mathrm{s^{-1}}$ with the source removed. The corrected count rate from the source is:探测器靠近源时读数 $210\ \mathrm{s^{-1}}$，移走源后读数 $30\ \mathrm{s^{-1}}$。来自源的修正后计数率为：
E3.4 · Q2
$240\ \mathrm{s^{-1}}$
$180\ \mathrm{s^{-1}}$
$210\ \mathrm{s^{-1}}$
$30\ \mathrm{s^{-1}}$
Subtract background: $210 - 30 = 180\ \mathrm{s^{-1}}$. The $30\ \mathrm{s^{-1}}$ is the ever-present background, present whether or not the source is there.减去本底：$210 - 30 = 180\ \mathrm{s^{-1}}$。$30\ \mathrm{s^{-1}}$ 是始终存在的本底，无论源是否在场。
The background must be subtracted, not added. Corrected rate $=$ measured $-$ background.本底应减去而非加上。修正后计数率 $=$ 测得 $-$ 本底。
Topic E3.5 · Exponential Decay LawTopic E3.5 · 指数衰变定律

The Exponential Decay Law指数衰变定律 HL only E.3 AHL

Why exponential. Each nucleus has a fixed probability $\lambda$ of decaying per unit time, so the number decaying per second is proportional to how many remain: $\dfrac{dN}{dt} = -\lambda N$. The solution is exponential. Decay law (data booklet). $$ N = N_0 e^{-\lambda t} $$ where N = N₀e^(−λt), $N_0$ is the initial number of undecayed nuclei and $\lambda$ is the decay constant. Activity follows the same law. Since $A = \lambda N$, $$ A = A_0 e^{-\lambda t}, \qquad A_0 = \lambda N_0. $$ Count rate, being proportional to activity, also decays as $C = C_0 e^{-\lambda t}$. Use this when the elapsed time is not a whole number of half-lives.
为何是指数。每个核每单位时间有固定衰变概率 $\lambda$，故每秒衰变数正比于剩余数：$\dfrac{dN}{dt} = -\lambda N$。其解为指数式。 衰变定律（数据手册）。 $$ N = N_0 e^{-\lambda t} $$ 其中 N = N₀e^(−λt)，$N_0$ 为初始未衰变核数，$\lambda$ 为衰变常数。 活度服从同一定律。由 $A = \lambda N$， $$ A = A_0 e^{-\lambda t}, \qquad A_0 = \lambda N_0. $$ 计数率正比于活度，同样按 $C = C_0 e^{-\lambda t}$ 衰减。当经过时间不是整数倍半衰期时使用本式。
Worked Example E3.5 (non-integer time)E3.5 例题（非整数倍时间） HL

A source has decay constant $\lambda = 0.040\ \mathrm{s^{-1}}$ and initial activity $A_0 = 8.0 \times 10^{5}\ \mathrm{Bq}$. Find the activity after $30\ \mathrm{s}$.某源衰变常数 $\lambda = 0.040\ \mathrm{s^{-1}}$，初始活度 $A_0 = 8.0 \times 10^{5}\ \mathrm{Bq}$。求 $30\ \mathrm{s}$ 后的活度。

Identify. Time is given directly, not as a number of half-lives, so use $A = A_0 e^{-\lambda t}$ from the data booklet.

识别。时间直接给出，而非半衰期个数，故用数据手册 $A = A_0 e^{-\lambda t}$。

Set up. Exponent: $-\lambda t = -(0.040)(30) = -1.2$.

列式。指数：$-\lambda t = -(0.040)(30) = -1.2$。

$$ A = (8.0 \times 10^{5}) \, e^{-1.2}. $$

Execute. $e^{-1.2} \approx 0.301$, so $A \approx (8.0 \times 10^{5})(0.301) \approx 2.4 \times 10^{5}\ \mathrm{Bq}$.

执行。$e^{-1.2} \approx 0.301$，故 $A \approx (8.0 \times 10^{5})(0.301) \approx 2.4 \times 10^{5}\ \mathrm{Bq}$。

Evaluate. The activity has fallen to about $30\%$ of its start. Sanity check: one half-life is $T_{1/2} = \ln 2 / \lambda \approx 17.3\ \mathrm{s}$, so $30\ \mathrm{s}$ is roughly $1.7$ half-lives — between $\tfrac{1}{2}$ and $\tfrac{1}{4}$ of $A_0$, consistent with $0.30$.

评估。活度降到初值约 $30\%$。验算：一个半衰期 $T_{1/2} = \ln 2 / \lambda \approx 17.3\ \mathrm{s}$，故 $30\ \mathrm{s}$ 约 $1.7$ 个半衰期——介于 $A_0$ 的 $\tfrac{1}{2}$ 与 $\tfrac{1}{4}$ 之间，与 $0.30$ 一致。

Going deeper: solving $dN/dt = -\lambda N$深入：求解 $dN/dt = -\lambda N$

The decay law is the equation $\dfrac{dN}{dt} = -\lambda N$. Separating variables:

衰变定律即方程 $\dfrac{dN}{dt} = -\lambda N$。分离变量：

$$ \int_{N_0}^{N} \frac{dN}{N} = -\lambda \int_{0}^{t} dt \;\Rightarrow\; \ln\!\frac{N}{N_0} = -\lambda t. $$

Exponentiating both sides gives $N = N_0 e^{-\lambda t}$. The IB syllabus quotes this result rather than requiring the integration, but recognising that "rate proportional to amount present" forces an exponential is exactly the reasoning that connects E3.5 to the random, constant-probability picture of E3.1.

两边取指数得 $N = N_0 e^{-\lambda t}$。IB 大纲直接引用此结果，不要求积分，但要理解"速率正比于现存量"必然导致指数式——这正是将 E3.5 与 E3.1 中随机、恒定概率图景相连的推理。

HL The decay law $N = N_0 e^{-\lambda t}$ arises because the rate of decay is:HL 衰变定律 $N = N_0 e^{-\lambda t}$ 成立是因为衰变速率：
E3.5 · Q1
constant in time随时间恒定
proportional to the number of nuclei present正比于现存核数
proportional to the temperature正比于温度
proportional to the square of the time正比于时间的平方
$dN/dt = -\lambda N$: the decay rate is proportional to the number remaining. "Rate $\propto$ amount present" is precisely the condition that produces exponential decay.$dN/dt = -\lambda N$：衰变速率正比于剩余数。"速率 $\propto$ 现存量"正是产生指数衰变的条件。
Exponential decay comes from $dN/dt \propto N$ (with a minus sign). A constant rate would give a straight line, not an exponential.指数衰变源于 $dN/dt \propto N$（带负号）。恒定速率给出直线而非指数。
HL A nuclide has $\lambda = 0.10\ \mathrm{s^{-1}}$. The fraction of nuclei remaining after $10\ \mathrm{s}$ is closest to:HL 某核素 $\lambda = 0.10\ \mathrm{s^{-1}}$。$10\ \mathrm{s}$ 后剩余核数分数最接近：
E3.5 · Q2
$0.90$
$0.10$
$0.37$
$0.50$
$N/N_0 = e^{-\lambda t} = e^{-(0.10)(10)} = e^{-1} \approx 0.37$. (At $t = 1/\lambda$, the "mean lifetime", the fraction is always $e^{-1}$.)$N/N_0 = e^{-\lambda t} = e^{-(0.10)(10)} = e^{-1} \approx 0.37$。（在 $t = 1/\lambda$，即"平均寿命"处，分数恒为 $e^{-1}$。）
Compute $\lambda t = 1$, then $N/N_0 = e^{-1} \approx 0.37$, not $0.10$ or $0.50$.先算 $\lambda t = 1$，再 $N/N_0 = e^{-1} \approx 0.37$，不是 $0.10$ 或 $0.50$。
Topic E3.6 · Decay Constant and Half-lifeTopic E3.6 · 衰变常数与半衰期

Decay Constant, Half-life, and Logarithms衰变常数、半衰期与对数 HL only E.3 AHL

The link. Setting $N = \tfrac{1}{2} N_0$ in $N = N_0 e^{-\lambda t}$ at $t = T_{1/2}$ gives the data-booklet relation: $$ \lambda = \frac{\ln 2}{T_{1/2}} $$ where λ = ln2 / T₁/₂. A large $\lambda$ (high decay probability) means a short half-life, and vice versa. Solving for time. To find $t$, take natural logs of the decay law: $$ t = \frac{1}{\lambda}\ln\!\frac{N_0}{N} = \frac{T_{1/2}}{\ln 2}\ln\!\frac{N_0}{N}. $$ Units. $\lambda$ has units of inverse time ($\mathrm{s^{-1}}$, $\mathrm{yr^{-1}}$, …); $\lambda$ and $T_{1/2}$ must use matching time units. $\ln 2 \approx 0.693$.
关系式。在 $N = N_0 e^{-\lambda t}$ 中令 $t = T_{1/2}$、$N = \tfrac{1}{2} N_0$，得数据手册关系： $$ \lambda = \frac{\ln 2}{T_{1/2}} $$ 其中 λ = ln2 / T₁/₂。$\lambda$ 大（衰变概率高）意味着半衰期短，反之亦然。 解时间。求 $t$ 时对衰变定律取自然对数： $$ t = \frac{1}{\lambda}\ln\!\frac{N_0}{N} = \frac{T_{1/2}}{\ln 2}\ln\!\frac{N_0}{N}. $$ 单位。$\lambda$ 的单位为时间倒数（$\mathrm{s^{-1}}$、$\mathrm{yr^{-1}}$…）；$\lambda$ 与 $T_{1/2}$ 须用一致的时间单位。$\ln 2 \approx 0.693$。
Worked Example E3.6 (carbon dating)E3.6 例题（碳测年） HL

Carbon-14 has a half-life of $5\,730\ \mathrm{years}$. A wooden artefact shows a ${}^{14}\mathrm{C}$ activity that is $40\%$ of that in living wood. Estimate its age. (Take $\ln 2 = 0.693$.)碳-14 半衰期为 $5\,730\ \mathrm{年}$。某木制文物的 ${}^{14}\mathrm{C}$ 活度为活木的 $40\%$。估计其年代。（取 $\ln 2 = 0.693$。）

Identify. $A / A_0 = 0.40$ is not a whole number of halvings, so use the exponential law with logs.

识别。$A / A_0 = 0.40$ 不是整数倍减半，故用指数定律配合对数。

Set up — decay constant. $\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{5730} = 1.21 \times 10^{-4}\ \mathrm{yr^{-1}}$.

列式——衰变常数。$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{5730} = 1.21 \times 10^{-4}\ \mathrm{yr^{-1}}$。

Execute — take logs. From $A = A_0 e^{-\lambda t}$, $\;t = \dfrac{1}{\lambda}\ln\dfrac{A_0}{A} = \dfrac{1}{\lambda}\ln\dfrac{1}{0.40}$.

执行——取对数。由 $A = A_0 e^{-\lambda t}$，$\;t = \dfrac{1}{\lambda}\ln\dfrac{A_0}{A} = \dfrac{1}{\lambda}\ln\dfrac{1}{0.40}$。

$$ t = \frac{\ln(2.5)}{1.21 \times 10^{-4}} = \frac{0.916}{1.21 \times 10^{-4}} \approx 7\,600\ \mathrm{years}. $$

Evaluate. $40\%$ lies between one half-life ($50\%$, $5\,730\ \mathrm{yr}$) and two ($25\%$, $11\,460\ \mathrm{yr}$), so $\sim 7\,600\ \mathrm{yr}$ is in the right range. Always express the answer to a sensible number of significant figures and keep the time unit consistent with $\lambda$.

评估。$40\%$ 介于一个半衰期（$50\%$，$5\,730\ \mathrm{yr}$）与两个（$25\%$，$11\,460\ \mathrm{yr}$）之间，故约 $7\,600\ \mathrm{yr}$ 合理。答案应取合理有效数字，且时间单位与 $\lambda$ 保持一致。

Going deeper: deriving $\lambda = \ln 2 / T_{1/2}$深入：推导 $\lambda = \ln 2 / T_{1/2}$

By definition, after one half-life the number is halved: $N = \tfrac{1}{2} N_0$ when $t = T_{1/2}$. Substitute into $N = N_0 e^{-\lambda t}$:

按定义，经一个半衰期后核数减半：$t = T_{1/2}$ 时 $N = \tfrac{1}{2} N_0$。代入 $N = N_0 e^{-\lambda t}$：

$$ \tfrac{1}{2} N_0 = N_0 e^{-\lambda T_{1/2}} \;\Rightarrow\; \tfrac{1}{2} = e^{-\lambda T_{1/2}}. $$

Taking natural logs: $-\ln 2 = -\lambda T_{1/2}$, hence $\lambda = \ln 2 / T_{1/2}$. The same step backwards lets you find $T_{1/2}$ from a measured $\lambda$: $T_{1/2} = \ln 2 / \lambda$.

取自然对数：$-\ln 2 = -\lambda T_{1/2}$，故 $\lambda = \ln 2 / T_{1/2}$。反向同理可由测得的 $\lambda$ 求 $T_{1/2}$：$T_{1/2} = \ln 2 / \lambda$。

HL A nuclide has half-life $T_{1/2} = 14\ \mathrm{days}$. Its decay constant is (take $\ln 2 = 0.693$):HL 某核素半衰期 $T_{1/2} = 14\ \mathrm{天}$。其衰变常数为（取 $\ln 2 = 0.693$）：
E3.6 · Q1
$0.050\ \mathrm{day^{-1}}$
$14\ \mathrm{day^{-1}}$
$0.693\ \mathrm{day^{-1}}$
$20\ \mathrm{day^{-1}}$
$\lambda = \ln 2 / T_{1/2} = 0.693 / 14 \approx 0.050\ \mathrm{day^{-1}}$. Note the unit is inverse days, matching the half-life unit.$\lambda = \ln 2 / T_{1/2} = 0.693 / 14 \approx 0.050\ \mathrm{day^{-1}}$。单位为天的倒数，与半衰期单位匹配。
Use $\lambda = \ln 2 / T_{1/2}$, dividing $0.693$ by the half-life. The result is small and has units of inverse time.用 $\lambda = \ln 2 / T_{1/2}$，把 $0.693$ 除以半衰期。结果很小，单位为时间倒数。
HL A source's activity falls to $20\%$ of its initial value. In terms of half-lives $T_{1/2}$, the elapsed time is closest to:HL 某源活度降到初值的 $20\%$。以半衰期 $T_{1/2}$ 计，经过时间最接近：
E3.6 · Q2
$1.0\,T_{1/2}$
$5.0\,T_{1/2}$
$0.20\,T_{1/2}$
$2.3\,T_{1/2}$
$t = \dfrac{1}{\lambda}\ln\dfrac{A_0}{A} = \dfrac{T_{1/2}}{\ln 2}\ln 5 = T_{1/2}\dfrac{1.609}{0.693} \approx 2.3\,T_{1/2}$. (Two half-lives give $25\%$, just above $20\%$, so a little over $2$ is right.)$t = \dfrac{1}{\lambda}\ln\dfrac{A_0}{A} = \dfrac{T_{1/2}}{\ln 2}\ln 5 = T_{1/2}\dfrac{1.609}{0.693} \approx 2.3\,T_{1/2}$。（两个半衰期为 $25\%$，略高于 $20\%$，故略多于 $2$ 个正确。）
Solve $0.20 = (1/2)^{t/T_{1/2}}$, i.e. $t/T_{1/2} = \ln 5 / \ln 2 \approx 2.3$. It must exceed 2 (which gives $25\%$).解 $0.20 = (1/2)^{t/T_{1/2}}$，即 $t/T_{1/2} = \ln 5 / \ln 2 \approx 2.3$。须大于 2（对应 $25\%$）。
Exam Tactics考试技巧

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Balancing equations (every paper)配平方程（每张试卷）
  • Balance the top row (nucleon number) and bottom row (charge) separately. Both must match across the arrow; an unbalanced equation loses the A1 mark.
  • 分别配平上排（核子数）与下排（电荷）。箭头两侧都须相等；不平衡的方程会丢 A1。
  • Remember the (anti)neutrino in beta decay. $\beta^{-}$ takes $\bar{\nu}_e$, $\beta^{+}$ takes $\nu_e$. It carries no $A$ or $Z$ but markschemes expect it named.
  • 记得 β 衰变中的（反）中微子。$\beta^{-}$ 配 $\bar{\nu}_e$，$\beta^{+}$ 配 $\nu_e$。它不带 $A$ 或 $Z$，但评分要求写出。
Whole vs partial half-lives整数 vs 非整数半衰期
  • If the time is a whole number of half-lives, just halve repeatedly — no logs needed. Faster and less error-prone.
  • 若时间是整数倍半衰期，反复减半即可——无需对数。更快也更不易出错。
  • If not, reach for the HL law $N = N_0 e^{-\lambda t}$ and take natural logs to solve for $t$ or $\lambda$. HL
  • 否则用 HL 定律 $N = N_0 e^{-\lambda t}$ 并取自然对数求 $t$ 或 $\lambda$。HL
Background and count rate (data-response)本底与计数率（数据题）
  • Always subtract the background before extracting a half-life from count-rate data. Forgetting it skews every reading and the curve never reaches zero.
  • 从计数率数据求半衰期前务必先减去本底。遗漏会使每个读数偏差，且曲线永不归零。
  • Activity and count rate are proportional, so they share the same half-life — you can read $T_{1/2}$ from either.
  • 活度与计数率成正比，故半衰期相同——可从任一读出 $T_{1/2}$。
Conceptual traps概念陷阱
  • "Random" and "spontaneous" are distinct. Random = unpredictable which/when; spontaneous = unaffected by external conditions. Questions ask for both words.
  • "随机"与"自发"含义不同。随机 = 无法预测哪个/何时；自发 = 不受外界条件影响。题目要求两个词都答到。
  • Ionising power and penetrating power run in opposite orders. $\alpha$ ionises most but penetrates least.
  • 电离能力与穿透力顺序相反。$\alpha$ 电离最强、穿透最弱。
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Two defining properties of radioactive decay?放射性衰变的两条本质性质？
Random and spontaneous.随机且自发。
Identity and charge of an alpha particle?α 粒子的本质与电荷？
${}^{4}_{2}\mathrm{He}$ nucleus, charge $+2e$.${}^{4}_{2}\mathrm{He}$ 核，电荷 $+2e$。
Penetrating power, increasing order?穿透力递增顺序？
$$\alpha < \beta < \gamma$$
Two quantities conserved in nuclear decay?核衰变中守恒的两个量？
Nucleon number $A$ and charge $Z$.核子数 $A$ 与电荷 $Z$。
Beta-minus decay equation?β⁻ 衰变方程？
$${}^{A}_{Z}X \to {}^{A}_{Z+1}Y + {}^{0}_{-1}\beta + \bar{\nu}_e$$
Which lepton accompanies $\beta^{+}$ decay?β⁺ 衰变伴随哪种轻子？
An electron neutrino $\nu_e$.电子中微子 $\nu_e$。
Definition of half-life?半衰期定义？
Time for half the nuclei (or activity) to decay.一半核（或活度）衰变所需时间。
Fraction left after $n$ half-lives?$n$ 个半衰期后剩余分数？
$$\frac{N}{N_0} = \left(\tfrac{1}{2}\right)^{n}$$
Activity formula and its unit?活度公式及单位？
$$A = \lambda N$$unit: becquerel, $1\ \mathrm{Bq} = 1\ \mathrm{s^{-1}}$.单位：贝克勒尔，$1\ \mathrm{Bq} = 1\ \mathrm{s^{-1}}$。
What is background radiation?什么是本底辐射？
Ever-present radiation (radon, rocks, cosmic, food); subtract it from readings.始终存在的辐射（氡、岩石、宇宙、食物）；读数须扣除它。
Exponential decay law? HL指数衰变定律？HL
$$N = N_0 e^{-\lambda t}$$
Decay constant ↔ half-life link? HL衰变常数与半衰期关系？HL
$$\lambda = \frac{\ln 2}{T_{1/2}}$$
Mixed Practice综合练习

Unit E.3 Practice Quiz单元 E.3 练习测验

A nucleus ${}^{210}_{84}\mathrm{Po}$ decays to ${}^{206}_{82}\mathrm{Pb}$. The emitted particle is:核 ${}^{210}_{84}\mathrm{Po}$ 衰变为 ${}^{206}_{82}\mathrm{Pb}$。所发射粒子为：
Q1
a $\beta^{-}$ particleβ⁻ 粒子
a $\beta^{+}$ particleβ⁺ 粒子
an $\alpha$ particleα 粒子
a $\gamma$ photonγ 光子
$A$ drops by 4 ($210 \to 206$) and $Z$ by 2 ($84 \to 82$) — exactly the change from emitting ${}^{4}_{2}\alpha$.$A$ 减 4（$210 \to 206$）、$Z$ 减 2（$84 \to 82$）——正是发射 ${}^{4}_{2}\alpha$ 的变化。
Find the change in $A$ and $Z$: $-4$ and $-2$. Only an alpha particle does that.求 $A$ 与 $Z$ 的变化：$-4$ 与 $-2$。只有 α 粒子如此。
A source of half-life $6.0\ \mathrm{hours}$ has an initial activity of $800\ \mathrm{Bq}$. Its activity after $18\ \mathrm{hours}$ is:半衰期 $6.0\ \mathrm{小时}$ 的源初始活度为 $800\ \mathrm{Bq}$。$18\ \mathrm{小时}$ 后活度为：
Q2
$200\ \mathrm{Bq}$
$100\ \mathrm{Bq}$
$400\ \mathrm{Bq}$
$50\ \mathrm{Bq}$
$n = 18 / 6.0 = 3$ half-lives. $A = 800 \times (1/2)^{3} = 800 / 8 = 100\ \mathrm{Bq}$.$n = 18 / 6.0 = 3$ 个半衰期。$A = 800 \times (1/2)^{3} = 100\ \mathrm{Bq}$。
Three half-lives means three halvings: $800 \to 400 \to 200 \to 100$.三个半衰期即减半三次：$800 \to 400 \to 200 \to 100$。
Why must an antineutrino be emitted in beta-minus decay?为何 β⁻ 衰变必须发射反中微子？
Q3
To conserve energy and lepton number (the electron energy spectrum is continuous)为守恒能量与轻子数（电子能谱连续）
To conserve nucleon number为守恒核子数
To carry away the nuclear charge为带走核电荷
To make the daughter nucleus heavier为使子核更重
The emitted electron's energy varies up to a maximum (a continuous spectrum), so a third particle must carry the balance of energy and momentum. The antineutrino also conserves lepton number alongside the electron.所发射电子能量在最大值以下变化（连续谱），故须有第三个粒子带走能量与动量的余量。反中微子也与电子一起守恒轻子数。
The antineutrino carries no charge and no nucleon number; its role is energy/momentum and lepton-number conservation.反中微子不带电荷与核子数；其作用是守恒能量/动量与轻子数。
HL A nuclide has decay constant $\lambda = 0.20\ \mathrm{s^{-1}}$. Its half-life is closest to:HL 某核素衰变常数 $\lambda = 0.20\ \mathrm{s^{-1}}$。其半衰期最接近：
Q4
$0.14\ \mathrm{s}$
$5.0\ \mathrm{s}$
$0.20\ \mathrm{s}$
$3.5\ \mathrm{s}$
$T_{1/2} = \ln 2 / \lambda = 0.693 / 0.20 \approx 3.5\ \mathrm{s}$.$T_{1/2} = \ln 2 / \lambda = 0.693 / 0.20 \approx 3.5\ \mathrm{s}$。
Rearrange $\lambda = \ln 2 / T_{1/2}$ to $T_{1/2} = \ln 2 / \lambda = 0.693 / 0.20$.把 $\lambda = \ln 2 / T_{1/2}$ 变形为 $T_{1/2} = \ln 2 / \lambda = 0.693 / 0.20$。
HL A sample of $N_0 = 4.0 \times 10^{20}$ nuclei has $\lambda = 1.5 \times 10^{-2}\ \mathrm{s^{-1}}$. Its initial activity is:HL 含 $N_0 = 4.0 \times 10^{20}$ 个核的样品，$\lambda = 1.5 \times 10^{-2}\ \mathrm{s^{-1}}$。其初始活度为：
Q5
$2.7 \times 10^{22}\ \mathrm{Bq}$
$1.5 \times 10^{-2}\ \mathrm{Bq}$
$6.0 \times 10^{18}\ \mathrm{Bq}$
$4.0 \times 10^{20}\ \mathrm{Bq}$
$A_0 = \lambda N_0 = (1.5 \times 10^{-2})(4.0 \times 10^{20}) = 6.0 \times 10^{18}\ \mathrm{Bq}$.$A_0 = \lambda N_0 = (1.5 \times 10^{-2})(4.0 \times 10^{20}) = 6.0 \times 10^{18}\ \mathrm{Bq}$。
Use $A = \lambda N$ with the initial $N_0$. Multiply the decay constant by the number of nuclei.用 $A = \lambda N$ 代入初始 $N_0$。把衰变常数乘以核数。
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Readiness Checklist备考清单

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IB Physics HL · Unit E.3 Radioactive decay · v1 · First Assessment 2025IB 物理 HL · 单元 E.3 放射性衰变 · v1 · 首次考核 2025