Deriving the Exponential Decay Law and Half-life in A-Level Physics Unit 5 (June 2022 Mark Scheme Insights) | A-Level物理单元5：指数衰变定律与半衰期推导（2022年6月评分方案解析）

📚 Deriving the Exponential Decay Law and Half-life in A-Level Physics Unit 5 (June 2022 Mark Scheme Insights) | A-Level物理单元5：指数衰变定律与半衰期推导（2022年6月评分方案解析）

Radioactive decay is a cornerstone of A-Level Physics Unit 5, and the June 2022 mark scheme for this unit reveals how examiners expect students to confidently derive and apply the exponential decay law. Mastering these derivations is not simply about memorising the final formula – it requires a clear understanding of the random nature of decay, the decay constant λ, and the link between activity and the number of undecayed nuclei. In this article we unpack the key steps that often appear in mark schemes, from setting up the differential equation dN/dt = –λN to obtaining N = N₀e⁻^λ^t and the half‑life relation T½ = ln2/λ. We will also extend the derivation to binding energy and mass defect, which are frequently examined in the nuclear physics section of Unit 5.

放射性衰变是A-Level物理单元5的核心内容，2022年6月的评分方案明确展示了考官期望学生如何自信地推导并应用指数衰变定律。掌握这些推导不仅仅是记住最终公式——它需要清晰理解衰变的随机性、衰变常量λ以及活度与未衰变原子核数量之间的联系。本文将拆解评分方案中常见的核心步骤，从建立微分方程 dN/dt = –λN 到得出 N = N₀e⁻^λ^t 和半衰期关系式 T½ = ln2/λ。我们还会将推导延伸至结合能与质量亏损，这些知识点在单元5核物理部分也经常被考查。

1. The Random Nature of Decay and the Decay Constant | 衰变的随机性与衰变常量

Every radioactive nucleus has a fixed probability of decaying per unit time. This probability is captured by the decay constant λ, which has units of s⁻¹. For a large sample containing N undecayed nuclei, the expected number of decays per second – the activity A – is proportional to N. The mark scheme for June 2022 emphasises that candidates must state clearly: “the rate of decay is directly proportional to the number of undecayed nuclei.” This proportionality is the foundation of the entire derivation.

每个放射性原子核在单位时间内衰变的概率是固定的。这个概率用衰变常量λ来描述，其单位为 s⁻¹。对于一个包含 N 个未衰变原子核的大样本，每秒预期的衰变次数——即活度 A——与 N 成正比。2022年6月的评分方案强调，考生必须清晰表达：“衰变速率与未衰变原子核的数量成正比。”这一正比关系是整个推导的基石。

2. Relating Activity to the Rate of Change of N | 活度与 N 的变化率的关系

Activity A is defined as the number of decays per unit time. Because each decay removes one nucleus from the undecayed collection, A is equal to the rate at which N decreases, taken with a positive sign. Mathematically we write A = –dN/dt. The negative sign arises because N decreases with time, so dN/dt is negative, but activity is a positive quantity. Substituting the proportionality A = λN gives the fundamental differential equation: –dN/dt = λN, or dN/dt = –λN. The June 2022 mark scheme awards marks for correctly explaining the origin of the negative sign and for writing the differential equation in this form.

活度 A 定义为单位时间内的衰变次数。由于每一次衰变都会从未衰变群体中移除一个原子核，因此 A 等于 N 减少的速率，并取正号。数学上我们写作 A = –dN/dt。负号的出现是因为 N 随时间减小，所以 dN/dt 为负，但活度是一个正量。代入比例关系 A = λN 就得到基本的微分方程：–dN/dt = λN，即 dN/dt = –λN。2022年6月的评分方案对正确解释负号的来源并以该形式写出微分方程给予评分。

3. Solving the Differential Equation by Separation of Variables | 用分离变量法求解微分方程

To solve dN/dt = –λN, we separate the variables N and t: (1/N) dN = –λ dt. Integrating both sides yields ∫ (1/N) dN = –λ ∫ dt. The left‑hand integral gives ln N, and the right‑hand integral gives –λt plus a constant of integration. Thus ln N = –λt + C, where C is determined by initial conditions. This step is crucial; the mark scheme often expects the candidate to show the separation of variables and the evaluation of the constant using the initial number of nuclei N₀ at t = 0.

为了求解 dN/dt = –λN，我们分离变量 N 和 t：(1/N) dN = –λ dt。两边积分得到 ∫ (1/N) dN = –λ ∫ dt。左侧积分得到 ln N，右侧积分得到 –λt 加上积分常数。因此 ln N = –λt + C，其中 C 由初始条件确定。这一步至关重要；评分方案通常期望考生展示变量分离过程，并利用 t=0 时的初始原子核数目 N₀ 求出常数。

4. Obtaining the Exponential Decay Law N = N₀e⁻^λ^t | 得出指数衰变定律 N = N₀e⁻^λ^t

At t = 0, N = N₀. Substituting into ln N = –λt + C gives ln N₀ = C. Therefore ln N = –λt + ln N₀, which can be rearranged as ln(N/N₀) = –λt. Exponentiating both sides leads to the well‑known exponential decay law: N = N₀ e⁻^λ^t. In the June 2022 exam context, candidates were often required to show each algebraic step, including the manipulation of logarithms and the justification for removing the logarithm by exponentiating. The final answer must be expressed clearly with the decay constant in the exponent.

当 t=0 时，N=N₀。代入 ln N = –λt + C 得到 ln N₀ = C。因此 ln N = –λt + ln N₀，可重组为 ln(N/N₀) = –λt。两边取指数就得到著名的指数衰变定律：N = N₀ e⁻^λ^t。在2022年6月考试的情境中，考生通常需要展示每一个代数步骤，包括对数的运算以及通过取指数消去对数的理由。最终答案必须清晰表达，指数中包含衰变常量。

5. Deriving the Half‑life Formula T½ = ln2 / λ | 推导半衰期公式 T½ = ln2 / λ

The half‑life T½ is the time taken for the number of undecayed nuclei to fall to half its initial value. Setting N = N₀/2 at t = T½ in the decay law gives N₀/2 = N₀ e⁻^λ ^T½. Cancelling N₀ yields 1/2 = e⁻^λ ^T½. Taking natural logarithms on both sides: ln(1/2) = –λ T½. Since ln(1/2) = –ln2, we have –ln2 = –λ T½, so T½ = ln2 / λ. The mark scheme for June 2022 makes it clear that full marks are only awarded if the candidate explicitly uses the definition of half‑life and correctly handles the logarithmic algebra.

半衰期 T½ 是未衰变原子核数量减少到初始值一半所需的时间。在衰变定律中令 t=T½ 时 N = N₀/2，得到 N₀/2 = N₀ e⁻^λ ^T½。约去 N₀ 得到 1/2 = e⁻^λ ^T½。两边取自然对数：ln(1/2) = –λ T½。由于 ln(1/2) = –ln2，我们有 –ln2 = –λ T½，因此 T½ = ln2 / λ。2022年6月的评分方案明确指出，只有考生明确使用半衰期的定义并正确处理对数运算，才能获得满分。

6. Exponential Decay of Activity | 活度的指数衰变

Since activity A is directly proportional to N (A = λN), multiplying the decay law N = N₀ e⁻^λ^t by λ gives A = λN₀ e⁻^λ^t. The initial activity is A₀ = λN₀, so we obtain A = A₀ e⁻^λ^t. The activity of a radioactive source therefore follows the same exponential decay pattern as the number of parent nuclei. The mark scheme frequently tests this relationship by asking students to sketch or interpret activity–time graphs and to calculate the activity after a given number of half‑lives.

由于活度 A 与 N 成正比（A = λN），将衰变定律 N = N₀ e⁻^λ^t 乘以 λ 得到 A = λN₀ e⁻^λ^t。初始活度为 A₀ = λN₀，因此我们得到 A = A₀ e⁻^λ^t。放射性源的活度因此遵循与母核数量相同的指数衰变规律。评分方案经常通过要求学生绘制或解读活度–时间图、计算给定若干半衰期后的活度来检验这一关系。

7. Using the Half‑life to Find the Decay Constant – A Worked Example | 利用半衰期求衰变常量——一个计算实例

A common task in the Unit 5 paper is to determine λ from an experimentally measured half‑life. For instance, if a sample has a half‑life of 8.0 days, convert this to seconds: T½ = 8.0 × 24 × 3600 = 6.912 × 10⁵ s. Then λ = ln2 / T½ = 0.693 / (6.912 × 10⁵) ≈ 1.00 × 10⁻⁶ s⁻¹. Once λ is known, the number of undecayed nuclei at any time can be predicted using N = N₀ e⁻^λ^t. The mark scheme values careful unit conversions and the use of the correct value for ln2 (typically 0.693).

单元5试卷中一项常见任务是根据实验测得的半衰期确定λ。例如，若某样品的半衰期为8.0天，将其换算为秒：T½ = 8.0 × 24 × 3600 = 6.912 × 10⁵ s。那么 λ = ln2 / T½ = 0.693 / (6.912 × 10⁵) ≈ 1.00 × 10⁻⁶ s⁻¹。一旦知道λ，任意时刻未衰变原子核的数量就可以通过 N = N₀ e⁻^λ^t 预测。评分方案重视仔细的单位换算以及正确使用ln2的数值（通常取0.693）。

8. Mass Defect and Binding Energy – The E = Δm c² Derivation | 质量亏损与结合能——E = Δm c² 推导

Moving beyond decay, Unit 5 also requires students to derive the energy released in nuclear processes using the mass–energy equivalence. The mass defect Δm is the difference between the total mass of the individual nucleons and the mass of the assembled nucleus. The binding energy is the energy equivalent of this mass defect, given by E = Δm c², where c is the speed of light in vacuum (3.00 × 10⁸ m s⁻¹). In the June 2022 mark scheme, credit was given for clearly defining Δm and for converting atomic mass units (u) to energy units (MeV). One atomic mass unit, 1 u = 1.661 × 10⁻²⁷ kg, corresponds to 931.5 MeV of energy. This conversion factor is central to many calculations.

除衰变外，单元5还要求学生利用质能等价关系推导核过程中释放的能量。质量亏损 Δm 是单个核子的总质量与组成原子核的质量之差。结合能就是该质量亏损的能量当量，由 E = Δm c² 给出，其中 c 是真空中的光速（3.00 × 10⁸ m s⁻¹）。在2022年6月的评分方案中，明确写出 Δm 的定义以及将原子质量单位（u）换算为能量单位（MeV）都可以得分。1个原子质量单位 1 u = 1.661 × 10⁻²⁷ kg，相当于931.5 MeV的能量。这一换算因子是许多计算的核心。

9. Calculating Binding Energy per Nucleon | 计算每个核子的结合能

To compare the stability of different nuclei, we calculate the binding energy per nucleon. First, find the mass defect Δm for the nucleus. Multiply Δm in kg by c² to get the total binding energy in joules, then divide by the number of nucleons A. Alternatively, using MeV units: total binding energy = Δm (in u) × 931.5 MeV, then divide by A. The mark scheme often expects candidates to show these steps in a logical order and to give the final answer in MeV per nucleon. For example, for an iron‑56 nucleus with a mass defect of about 0.528 u, the total binding energy is 0.528 × 931.5 ≈ 492 MeV, and the binding energy per nucleon is 492 / 56 ≈ 8.8 MeV.

为了比较不同原子核的稳定性，我们计算每个核子的结合能。首先，求出该原子核的质量亏损 Δm。将以千克为单位的 Δm 乘以 c² 得到以焦耳为单位的总结合能，然后除以核子数 A。另一种方式使用 MeV 单位：总结合能 = Δm（以 u 为单位）× 931.5 MeV，再除以 A。评分方案通常期望考生按逻辑顺序展示这些步骤，并给出以 MeV/核子 为单位的最终答案。例如，一个铁‑56原子核的质量亏损约为0.528 u，总结合能为 0.528 × 931.5 ≈ 492 MeV，每个核子的结合能则为 492 / 56 ≈ 8.8 MeV。

10. Energy Released in Nuclear Reactions | 核反应中释放的能量

The energy released (Q‑value) in a nuclear reaction is calculated from the difference in total mass between the initial reactants and the final products: Q = (mass of reactants – mass of products) × c². If the total mass decreases, the reaction is exothermic and Q is positive. In the June 2022 paper, a typical task might involve balancing a nuclear equation, identifying the masses of all particles (including alpha particles, beta particles, or neutrinos), and then computing Δm in u before converting to MeV. The mark scheme insists on using consistent units and on stating whether energy is released or absorbed.

核反应中释放的能量（Q值）根据反应物与生成物的总质量之差计算：Q = (反应物质量 – 生成物质量) × c²。如果总质量减小，则反应为放热反应，Q为正。在2022年6月的试卷中，一项典型任务可能涉及配平核反应方程，确认所有粒子（包括α粒子、β粒子或中微子）的质量，然后计算以 u 为单位的 Δm，再换算为 MeV。评分方案要求单位一致，并说明能量是释放还是吸收。

11. Linking the Decay Constant to the Probability of Decay | 衰变常量与衰变概率的联系

A deeper insight required by the mark scheme is the probabilistic interpretation of λ. For a single nucleus, λ is the probability that it will decay per unit time. If λ = 0.01 s⁻¹, there is a 1% chance that the nucleus decays in the next second. This statistical view explains why the activity of a sample fluctuates around the expected value λN, and why the standard deviation on a count rate is √N. Understanding this probabilistic foundation helps students explain why the exponential decay law is only valid for large numbers of nuclei and why it is a statistical law rather than a deterministic one.

评分方案所要求的一个更深层次的见解是对 λ 的概率解释。对单个原子核而言，λ 是其在单位时间内衰变的概率。若 λ = 0.01 s⁻¹，则该原子核在下一秒内衰变的几率为1%。这种统计观点解释了为何样品的活度会在预期值 λN 附近波动，以及为何计数率的标准差为 √N。理解这一概率基础有助于学生解释为什么指数衰变定律仅对大量原子核有效，以及为什么它是一个统计规律而非确定性规律。

12. Examiner Tips from the June 2022 Mark Scheme | 2022年6月评分方案的考官提示

The June 2022 mark scheme highlights several common pitfalls. Candidates often forget to include the negative sign in dN/dt = –λN or mishandle the units of λ, writing s instead of s⁻¹. When deriving T½, some students directly quote T½ = ln2 / λ without showing the substitution N = N₀/2 and the logarithmic steps – this costs marks. For binding energy questions, a frequent error is using atomic masses rather than nuclear masses without subtracting the electron masses. The mark scheme also penalises answers where the conversion 1 u = 931.5 MeV is not clearly stated or used correctly. Always show your working step by step, and explicitly state any approximations or assumptions.

2022年6月的评分方案指出了几个常见失分点。考生经常忘记在 dN/dt = –λN 中加入负号，或者弄错 λ 的单位，写成 s 而非 s⁻¹。推导 T½ 时，一些学生直接引用 T½ = ln2 / λ，而未展示 N = N₀/2 的代入及对数步骤——这会导致丢分。对于结合能问题，一个常见错误是直接使用原子质量而非核质量，却没有减去电子质量。评分方案还会对未明确写出或正确使用 1 u = 931.5 MeV 换算关系的作答进行扣分。务必逐步展示你的运算过程，并明确说明任何近似或假设。

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