📚 The Mathematics of Radioactive Decay | 放射性衰变的数学
Radioactive decay provides one of the cleanest applications of exponential functions and differential equations in A-level Mathematics. The same mathematics is used in physics, archaeology, medicine and nuclear engineering, so it is an excellent cross-topic revision area. This article rebuilds the model from its assumptions, solves the differential equation step by step, and then applies it to the standard exam tasks: half-life, activity, log-linear graphs and carbon dating.
放射性衰变是 A-level 数学中指数函数与微分方程最清晰的应用之一。同样的数学在物理、考古、医学和核工程中都有使用，因此它是一个很好的跨主题复习领域。本文将从头重建这一模型，逐步求解微分方程，并把它应用到标准考题中：半衰期、活度、对数线性图像以及碳定年。
1. Why Decay Is a Mathematical Model | 为什么衰变是一个数学模型
A single radioactive nucleus is unpredictable. We cannot say when it will decay, only that it might decay in the next second. Once a sample contains a very large number of nuclei, the randomness averages out and the behaviour becomes remarkably regular. That transition from individual randomness to collective regularity is exactly what turns radioactive decay into a mathematical topic.
单个放射性原子核是不可预测的。我们无法说出它会在何时衰变，只能说它有可能在下一秒衰变。但当一个样品包含数量极大的原子核时，随机性会被平均掉，整体行为就变得非常有规律。正是这种从个体随机到集体规律的转变，使放射性衰变成为一个数学课题。
The mathematics involved is drawn from several parts of the A-level specification: exponential functions, rates of change and differentiation, integration by separation of variables, logarithms and log-linear graphs, and the language of modelling assumptions. A question on decay can therefore test pure mathematics and mechanics-style reasoning at the same time.
其中涉及的数学来自 A-level 大纲的多个部分：指数函数、变化率与微分、分离变量积分、对数与对数线性图像，以及建模假设的表达方式。因此一道衰变题可以同时考查纯数学和力学式的推理能力。
Real applications are everywhere. Archaeologists date bone and charcoal with carbon-14; hospitals use technetium-99m and fluorine-18 as tracers; geologists date rocks with potassium-40 and rubidium-87; nuclear engineers must calculate how quickly waste becomes less radioactive. All of these rely on the same single differential equation.
现实应用随处可见。考古学家用碳-14 为骨骼和木炭定年；医院用锝-99m 和氟-18 作为示踪剂；地质学家用钾-40 和铷-87 为岩石定年；核工程师必须计算废料放射性降低的速度。所有这些都依赖同一个微分方程。
2. The Decay Model and Assumptions | 衰变模型与假设
Radioactive decay is modelled as a continuous process in which the rate of decay is proportional to the number of undecayed nuclei present. This single sentence is the whole physical content of the model, and everything else in this article follows from it by mathematics alone.
放射性衰变被建模为一个连续过程，其中衰变速率与当前未衰变的原子核数成正比。这一句话就是模型的全部物理内容，本文其余的一切都仅由数学从它推导出来。
ere are about 6.02 × 10²³ nuclei, so the statistical variation is far too small to notice.
当原子核数 N 非常大时，这个假设成立，随机涨落可以忽略不计。一摩尔的物质中约有 6.02 × 10²³ 个原子核，因此统计涨落小到无法察觉。
We also assume that every nucleus has the same fixed probability of decaying per unit time, independent of external conditions. Temperature, pressure, chemical bonding and the age of the nucleus have no measurable effect on the decay probability.
我们还假设每个原子核在单位时间内都有相同的固定衰变概率，且与外部条件无关。温度、压强、化学键合状态以及原子核已存在的时间，对衰变概率都没有可测量的影响。
Two further modelling points are worth stating explicitly in an exam answer. First, N is treated as a continuous and differentiable function of time even though it is really an integer, which is acceptable only because N is huge. Second, the sample is assumed to be closed: no new radioactive nuclei enter or leave, and no daughter nuclei feed back into the count.
在考试作答中，还有两个建模要点值得明确指出。第一，尽管 N 实际上是整数，我们仍把它当作时间的连续可微函数，这只因为 N 非常巨大才成立。第二，假设样品是封闭的：没有新的放射性原子核进入或离开，也没有子体核素回馈到计数中。
3. The Differential Equation dN/dt = −λN | 微分方程 dN/dt = −λN
/dt. The minus sign appears because N is decreasing, while activity is naturally quoted as a positive number.
设 N(t) 为 t 时刻未衰变原子核的数量。活度，即单位时间内的衰变次数，由 N 的减少速率给出，即 A = −dN/dt。负号出现是因为 N 在减少，而活度通常以正数表示。
The fundamental law states that this rate is proportional to N itself, which gives the first-order differential equation
基本定律指出该速率与 N 本身成正比，因此得一阶微分方程
dN/dt = −λN
Here λ is the decay constant, a positive quantity with units of s⁻¹ (or min⁻¹, h⁻¹, y⁻¹ depending on the time unit used). It represents the probability per unit time that a given nucleus decays.
其中 λ 是衰变常数，为正值，单位为 s⁻¹（或用 min⁻¹、h⁻¹、y⁻¹，取决于时间单位）。它表示某个给定原子核在单位时间内发生衰变的概率。
The equation is first-order because only the first derivative appears, linear because N appears only to the power one, and separable because it can be rearranged with all the N terms on one side and all the t terms on the other. Recognising these three labels is often worth a mark on its own.
该方程是一阶的，因为只出现一阶导数；是线性的，因为 N 只以一次幂出现；也是可分离变量的，因为可以重新整理使所有含 N 的项在一侧、所有含 t 的项在另一侧。识别出这三个标签本身往往就能得一分。
e is simply its decay constant multiplied by the number of undecayed nuclei still present.
把 A = −dN/dt 与衰变定律结合，就得到非常有用的关系式 A = λN：样品的活度等于其衰变常数乘以仍存在的未衰变原子核数。
4. Solving by Separation of Variables | 用分离变量法求解
Start by separating the variables, keeping in mind that N is positive for any physically meaningful sample so that dividing by N is legitimate.
首先分离变量，注意对于任何有物理意义的样品 N 都为正，因此除以 N 是合法的。
∫ (1/N) dN = ∫ −λ dt
t + c, where c is an arbitrary constant of integration. Since N is positive we do not need a modulus sign.
两边积分得 ln N = −λt + c，其中 c 为任意积分常数。由于 N 为正，无需加绝对值符号。
Exponentiating both sides turns the logarithm back into a power: N = e−λt + c = ec × e−λt. The constant ec is best renamed, because it has a direct physical meaning.
两边取指数，把对数还原为幂：N = e−λt + c = ec × e−λt。常数 ec 最好换一个名字，因为它有直接的物理意义。
Apply the initial condition. At t = 0, N = N₀, so ec = N₀. Substituting back gives the exponential decay law
代入初始条件。当 t = 0 时 N = N₀，故 ec = N₀。代回即得指数衰变定律
N = N₀ e−λt = N₀ exp(−λt)
Count rate and mass of undecayed material behave identically, which is why a single exponential curve describes so many different measured quantities.
因为 A = λN 且常数 λ 在所有时刻都相同，活度遵循完全相同的定律：A = A₀ exp(−λt)。计数率和未衰变物质的质量表现完全一致，这就是为什么同一条指数曲线能描述这么多不同的测量量。
pest at t = 0, and N tends to zero asymptotically without ever reaching it in finite time.
该解具有指数衰减的典型特征：在每个相等的时间间隔内 N 都按相同倍数下降；曲线在 t = 0 处最陡；N 渐近趋于零，但在有限时间内永远不会达到零。
5. Decay Constant and Half-Life | 衰变常数与半衰期
ts initial value. Substituting N = N₀/2 at t = T½ into the decay law gives a clean result.
半衰期 T½ 是未衰变原子核数（或活度、或计数率）降到其初始值一半所需的时间。在 t = T½ 时令 N = N₀/2 代入衰变定律，可得简洁的结果。
N₀/2 = N₀ e−λT½ ⇒ T½ = ln 2 / λ ≈ 0.693 / λ
that N₀ cancels, so the half-life is completely independent of the sample size: a kilogram and a microgram of the same isotope have identical half-lives.
整理得 λ = ln 2 / T½。注意 N₀ 被约掉了，因此半衰期完全与样品大小无关：同一种同位素的1千克和1微克具有相同的半衰期。
The enormous range shows why decay mathematics is useful across such different timescales.
半衰期是同位素的固定属性。碳-14 的 T½ ≈ 5730 年，铀-238 的 T½ ≈ 4.5 × 10⁹ 年，锝-99m 的 T½ ≈ 6 小时。这一巨大的范围说明为什么衰变数学能应用于如此不同的时间尺度。
pears in physics questions and is worth remembering as a check: the mean lifetime is always slightly longer than the half-life.
另一个相关量是平均寿命 τ = 1/λ ≈ 1.44 T½。它偶尔出现在物理题中，值得记住以便核对：平均寿命总是略大于半衰期。
6. Activity, Count Rate and Units | 活度、计数率与单位
he number of undecayed nuclei at that instant, so it also decays exponentially with the same constant λ.
活度以贝克勒尔为单位，1 Bq = 每秒 1 次衰变。由于 A = λN，活度与该瞬间未衰变的原子核数成正比，因此它也以相同的常数 λ 呈指数衰减。
| Quantity | Symbol | SI unit |
| Number of nuclei | N | dimensionless |
| Decay constant | λ | s⁻¹ |
| Activity | A | Bq = s⁻¹ |
| Half-life | T½ | s (or y) |
hat occur, because of geometry, absorption and detector efficiency. If the efficiency is 8% erved count rate of 40 counts per second corresponds to an activity of 40 ÷ 0.08 = 500 Bq.
一个关键区别是：活度与计数率并不是一回事。由于几何位置、吸收和探测器效率的原因，探测器只能记录到实际发生衰变的一部分。若效率为 8%，则观测到的计数率 40 计数每秒对应的活度为 40 ÷ 0.08 = 500 Bq。
The number of nuclei can be found from the mass: N = (m / M) × NA, where M is the molar mass in g mol⁻¹ and NA ogadro constant. This conversion often provides the missing link in a multi-step question.
原子核数可由质量求得：N = (m / M) × NA，其中 M 是以 g mol⁻¹ 为单位的摩尔质量，NA = 6.02 × 10²³ mol⁻¹ 为阿伏加德罗常数。这一换算常常是多步骤题目中缺失的一环。
7. Log-Linear Form and Graph Work | 对数线性形式与图像
lysed in log-linear form. Taking natural logarithms of both sides of N = N₀ exp(−λt) gives
指数形式便于理论分析，但实验数据几乎总是用对数线性形式处理。对 N = N₀ exp(−λt) 两边取自然对数得
ln N = ln N₀ − λt
ln N against t gives a straight line with gradient −λ and intercept ln N₀. This is the standard method for extracting the decay constant from experimental data, and it is far more reliable than reading values off a curve by eye.
它具有 y = mx + c 的形式，因此以 ln N 对 t 作图得到一条直线，斜率 −λ，截距 ln N₀。这是从实验数据中求衰变常数的标准方法，比用肉眼从曲线上读数可靠得多。
Alternatively, using base-10 logarithms gives a gradient of −λ / ln 10, or one can plot log₁₀N against t and multiply the gradient by ln 10 to recover λ. Always state which base you used, and never mix the two within one calculation.
也可以用常用对数，此时斜率为 −λ / ln 10；或者以 log₁₀N 对 t 作图，再把斜率乘以 ln 10 得到 λ。务必说明你使用的是哪种底数，并且绝不要在同一次计算中混用两种。
A third and very intuitive form is the halving formula, which uses the half-life directly:
第三种非常直观的形式是减半公式，直接使用半衰期：
N = N₀ (½)t/T½
This version is handy for quick mental estimates. After n half-lives the remaining fraction is (½)ⁿ: after 3 half-lives one eighth remains, after 10 half-lives roughly one thousandth. Choosing the right form of the equation saves time in an exam.
这个版本便于快速估算。经过 n 个半衰期后剩余比例为 (½)ⁿ：3 个半衰期后剩八分之一，10 个半衰期后约剩千分之一。在考试中选择正确形式的方程能节省时间。
8. Worked Examples | 例题精讲
of 800 Bq at t = 0 and 200 Bq after 12.0 hours. Find the half-life and the decay constant.
例 1。某样品在 t = 0 时活度为 800 Bq，12.0 小时后为 200 Bq。求半衰期与衰变常数。
= 6.0 hours. Then λ = ln 2 / T½ = 0.693 / 6.0 = 0.116 h⁻¹ (to three significant figures).
活度降为原来的四分之一，即减半两次，故半衰期为 12.0 ÷ 2 = 6.0 小时。于是 λ = ln 2 / T½ = 0.693 / 6.0 = 0.116 h⁻¹（三位有效数字）。
tant of 3.0 × 10⁻⁶ s⁻¹. Find the initial activity and the number remaining after one hour.
例 2。某放射源含 5.0 × 10²⁰ 个未衰变原子核，衰变常数为 3.0 × 10⁻⁶ s⁻¹。求初始活度以及一小时后剩余的原子核数。
) ≈ 4.95 × 10²⁰. The sample has barely changed, which is typical for a long-lived isotope.
由 A₀ = λN₀ = 3.0 × 10⁻⁶ × 5.0 × 10²⁰ = 1.5 × 10¹⁵ Bq。再取 t = 3600 s，λt = 3.0 × 10⁻⁶ × 3600 = 0.0108，故 N = 5.0 × 10²⁰ × exp(−0.0108) ≈ 4.95 × 10²⁰。样品几乎没有变化，这对长寿命同位素而言很典型。
3 / 0.0231 ≈ 30.0 minutes. Note how the gradient of a log-linear plot is used directly as λ without any further conversion.
例 3。以 ln A 对 t 作图得到直线，斜率为 −0.0231 min⁻¹。求半衰期。此处 λ = 0.0231 min⁻¹，故 T½ = 0.693 / 0.0231 ≈ 30.0 分钟。注意对数线性图像的斜率可以直接作为 λ 使用，无需再做换算。
9. Radiocarbon Dating and Other Applications | 放射性碳定年与其他应用
Living organisms continually exchange carbon with the atmosphere, so the ratio of carbon-14 to carbon-12 in their tissues stays constant. When the organism dies, the exchange stops and the carbon-14 begins to decay with a half-life of about 5730 years.
活着的生物不断与大气交换碳，因此其组织中碳-14 与碳-12 的比例保持恒定。当生物死亡后，这种交换停止，碳-14 便开始以约 5730 年的半衰期衰变。
he same size has activity R₀, then R = R₀ exp(−λt) and the age is obtained by rearranging:
若某样品今天的活度为 R，而同尺寸的活体样品活度为 R₀，则 R = R₀ exp(−λt)，整理即得年龄：
t = (1/λ) ln(R₀/R) = (T½ / ln 2) ln(R₀/R) ≈ 8270 × ln(R₀/R) years
11 500 years, which is exactly two half-lives as expected. If the ratio is 0.60, then t ≈
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