题型概述：放射性衰变与核能Question type: radioactive decay & nuclear energy
Top Gold放射性衰变和光电效应一样，是 Section 1 近代物理里套路固定的送分题：2021–2023 连续三年都有（Q1(t) 6 分、Q1(w) 5 分、Q1(j) 4 分），2025 年又出了两道。物理内容只有一个指数公式、一个"λ 与半衰期"的换算，再加上摩尔换算或 E = Δmc²。失分主要来自四处：λ 和 t 的时间单位不一致（天、年、秒混用）；把"衰变掉的"和"剩下的"弄反；α 衰变的动能忘了按动量分配；把"母核少掉的质量"当成"样品少掉的质量"。Like the photoelectric effect, radioactive decay is a fixed-pattern modern-physics item in Section 1: it appeared three years running in 2021–2023 (Q1(t) 6 marks, Q1(w) 5 marks, Q1(j) 4 marks) and twice more in 2025. The physics is one exponential formula, the link between λ and half-life, plus a mole conversion or E = Δmc². Marks are lost in four places: mismatched time units for λ and t (days, years, seconds); mixing up "decayed" with "remaining"; forgetting to share the α-decay energy by momentum; and treating "mass lost by the parent nuclide" as "mass lost by the sample".
答案可用科学计数法输入，例如 3.76e-5 或 3.76×10^-5。Answers can be typed in scientific notation, e.g. 3.76e-5 or 3.76×10^-5.
1衰变定律与半衰期The decay law and half-life
每个原子核在单位时间内衰变的概率是同一个常数 λ（衰变常量），与它"已经存在了多久"无关。于是 dN/dt = −λN，解出来就是指数衰减。经过一个半衰期 N 恰好减半，代入 e−λt½ = ½ 就得到 λ 与半衰期的关系。Every nucleus has the same constant probability per unit time of decaying, the decay constant λ, however long it has already existed. So dN/dt = −λN, whose solution is an exponential decay. After one half-life N has exactly halved; putting e−λt½ = ½ gives the link between λ and the half-life.
N = N0e−λt = N0(½)t/t½ λ = ln 2 / t½ ≈ 0.693 / t½ t = ln(N0/N) / λ两种写法等价。比例恰好是 ½ 的整数次幂时用 (½)n 心算最快（1/8 ⇒ 3 个半衰期）；否则取对数。λ 和 t 的时间单位必须一致。若 λt ≪ 1，则衰变掉的比例 1 − e−λt ≈ λt。The two forms are equivalent. When the ratio is a whole power of ½, (½)n is quickest (1/8 ⇒ 3 half-lives); otherwise take logs. λ and t must use the same time unit. If λt ≪ 1, the fraction decayed 1 − e−λt ≈ λt.
2活度、原子数与质量Activity, number of nuclei and mass
活度 A 是每秒衰变的次数（单位 Bq = s⁻¹）。因为每个核衰变的概率相同，活度与剩下的核数成正比，所以活度、核数、母核质量、计数率都按同一个指数衰减，题目给哪个比值都可以直接代入。核数与质量之间用摩尔质量和阿伏伽德罗常数换算。Activity A is the number of decays per second (unit Bq = s⁻¹). Because every nucleus has the same decay probability, activity is proportional to the number of nuclei left, so activity, number of nuclei, parent mass and count rate all fall with the same exponential; whichever ratio a question gives can be used directly. Convert between number and mass with the molar mass and Avogadro's constant.
A = λN N = (m/M)NA A/A0 = N/N0 = e−λt碳-14 测年就是这一条：比较样品与活体的"比活度"（每 kg 的活度），t = ln(A0/A)/λ。Carbon-14 dating is exactly this: compare the specific activity (activity per kg) of the sample with living material, t = ln(A0/A)/λ.
3衰变方程与质能关系Decay equations and mass–energy
核反应中质量数 A（上标）和电荷数 Z（下标）分别守恒。α 衰变：A − 4，Z − 2；β⁻ 衰变（n → p + e⁻ + ν̄）：A 不变，Z + 1；γ 是光子：两者都不变。反应后总质量比反应前略小，差值 Δm 以动能形式释放。母核静止时，两个产物动量等大反向，动能 p²/2m 与质量成反比。In a nuclear reaction the mass number A (top) and the charge number Z (bottom) are each conserved. α decay: A − 4, Z − 2; β⁻ decay (n → p + e⁻ + ν̄): A unchanged, Z + 1; γ is a photon: neither changes. The products are slightly lighter than the parent; the difference Δm is released as kinetic energy. For a parent at rest the two products have equal and opposite momenta, so their kinetic energies p²/2m are inversely proportional to their masses.
AZX → A−4Z−2Y + 42α Q = (mX − mY − mα)c² Kα = Q · mY/(mY + mα)1 u ≈ 931.5 MeV/c²，1 MeV = 1.602×10⁻¹³ J。质量差只有总质量的 10⁻⁵ 量级，相减前先统一数量级、保留全部有效数字。1 u ≈ 931.5 MeV/c², 1 MeV = 1.602×10⁻¹³ J. The mass difference is only ~10⁻⁵ of the total, so bring the masses to the same power of ten and keep every significant figure before subtracting.
| 子题型Sub-type | 识别特征How to spot it | 核心步骤Core step | 真题Past paper |
|---|---|---|---|
| 衰变了多少How many decay | "how many … will decay", t½ | λ = ln2/t½，ΔN = N₀(1 − e−λt) ≈ N₀λtλ = ln2/t½, ΔN = N₀(1 − e−λt) ≈ N₀λt | 2023 Q1(j) |
| α 粒子动能α kinetic energy | 给出各核质量（kg 或 u）Nuclear masses given (kg or u) | Q = Δmc²，再按动量守恒分配Q = Δmc², then share by momentum | 2022 Q1(w) |
| 样品质量变化Change in sample mass | "mass … reduced", mg | 衰变比例 × 摩尔数 × 离开样品的那部分的摩尔质量Fraction decayed × moles × molar mass of what actually leaves | 2021 Q1(t) |
| 半衰期 + 方程配平Half-life + balancing | "fell to one eighth", a, b, c … | (½)n；A、Z 守恒(½)n; conserve A and Z | 2016 Q1(m) |
| 碳-14 测年Carbon-14 dating | "specific activity", "living wood" | t = ln(A₀/A)/λt = ln(A₀/A)/λ | 2014 Q1(f) |