Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus E3.1 to E3.6考纲 E3.1 至 E3.6PHYSICS HL
A source emits alpha, beta-minus and gamma. (a) identity and charge of each; (b) order by increasing penetration and how ionising power compares.某源发射 α、β⁻ 与 γ。(a) 各自的本质与电荷；(b) 按穿透力递增排序及电离能力如何对比。
Alpha is a helium nucleus ${}^{4}_{2}\mathrm{He}$, two protons and two neutrons, charge $+2e$. (A1)
Beta-minus is a fast-moving electron ${}^{0}_{-1}e$, charge $-e$; gamma is a high-energy photon (electromagnetic radiation), no mass and no charge. (A1)
In order of increasing penetrating power: alpha (stopped by paper or a few cm of air), then beta (a few mm of aluminium), then gamma (only attenuated by thick lead or concrete). (A1)
Ionising power runs in exactly the reverse order, $\alpha > \beta > \gamma$: the most ionising radiation is the least penetrating. (A1)
α 是氦核 ${}^{4}_{2}\mathrm{He}$，含两个质子与两个中子，电荷 $+2e$。(A1)
β⁻ 是高速运动的电子 ${}^{0}_{-1}e$，电荷 $-e$；γ 是高能光子（电磁辐射），无质量、无电荷。(A1)
按穿透力递增：α（被纸或几厘米空气挡住），然后 β（几毫米铝），再到 γ（只被厚铅或混凝土减弱）。(A1)
电离能力恰为相反顺序，$\alpha > \beta > \gamma$：电离最强的辐射穿透最弱。(A1)
(a) alpha decay of ${}^{226}_{88}\mathrm{Ra}$; (b) beta-minus decay of ${}^{90}_{38}\mathrm{Sr}$ with the lepton; (c) the two conserved quantities and why an antineutrino is needed. (Rn $Z = 86$; Y $Z = 39$.)(a) ${}^{226}_{88}\mathrm{Ra}$ 的 α 衰变；(b) ${}^{90}_{38}\mathrm{Sr}$ 的 β⁻ 衰变含轻子；(c) 两个守恒量及为何需要反中微子。（Rn $Z = 86$；Y $Z = 39$。）
Alpha emission removes $4$ from the nucleon number and $2$ from the charge: $A = 226 - 4 = 222$, $Z = 88 - 2 = 86$. (M1)
$$ {}^{226}_{88}\mathrm{Ra} \to {}^{222}_{86}\mathrm{Rn} + {}^{4}_{2}\alpha. $$Element $86$ is radon, so the daughter is ${}^{222}_{86}\mathrm{Rn}$. (A1)
Beta-minus keeps the nucleon number fixed ($A = 90$) and raises the charge by one ($Z = 38 + 1 = 39$); the emitted electron carries $-1$ charge, and an antineutrino accompanies it. (M1)
$$ {}^{90}_{38}\mathrm{Sr} \to {}^{90}_{39}\mathrm{Y} + {}^{0}_{-1}\beta + \bar{\nu}_e. $$Element $39$ is yttrium, so the daughter is ${}^{90}_{39}\mathrm{Y}$. (A1)
Conserved across the arrow: the nucleon number $A$ (top numbers) and the charge $Z$ (bottom numbers). (A1)
The antineutrino is required because the emitted electron is a lepton, so an antilepton must appear to conserve lepton number, and it carries the variable balance of energy and momentum that gives beta particles a continuous energy spectrum. (R1)
α 发射使核子数减 $4$、电荷减 $2$：$A = 226 - 4 = 222$，$Z = 88 - 2 = 86$。(M1)
$$ {}^{226}_{88}\mathrm{Ra} \to {}^{222}_{86}\mathrm{Rn} + {}^{4}_{2}\alpha. $$第 $86$ 号元素为氡，故子核为 ${}^{222}_{86}\mathrm{Rn}$。(A1)
β⁻ 使核子数不变（$A = 90$）、电荷加 $1$（$Z = 38 + 1 = 39$）；所发射电子带 $-1$ 电荷，并伴随一个反中微子。(M1)
$$ {}^{90}_{38}\mathrm{Sr} \to {}^{90}_{39}\mathrm{Y} + {}^{0}_{-1}\beta + \bar{\nu}_e. $$第 $39$ 号元素为钇，故子核为 ${}^{90}_{39}\mathrm{Y}$。(A1)
箭头两侧守恒的是核子数 $A$（上标）与电荷 $Z$（下标）。(A1)
需要反中微子，是因为所发射电子是轻子，必须有一反轻子出现以守恒轻子数；它还带走可变的那份能量与动量，使 β 粒子呈现连续能谱。(R1)
$T_{1/2} = 6.0\ \mathrm{h}$, sample starts with $N_0$ nuclei. (a) meaning of half-life; (b) fraction remaining after $18\ \mathrm{h}$; (c) time to fall to $\tfrac{1}{16}N_0$ and why no log is needed.$T_{1/2} = 6.0\ \mathrm{h}$，样品起始含 $N_0$ 个核。(a) 半衰期含义；(b) $18\ \mathrm{h}$ 后剩余分数；(c) 降到 $\tfrac{1}{16}N_0$ 的时间及为何无需对数。
The half-life is the time taken for half the radioactive nuclei in a sample to decay (equivalently, the time for the activity or count rate to halve). (B1)
Number of half-lives: $n = \dfrac{t}{T_{1/2}} = \dfrac{18}{6.0} = 3$. (M1)
$$ \frac{N}{N_0} = \left(\tfrac{1}{2}\right)^{3} = \tfrac{1}{8}. $$(A1)
$\tfrac{1}{16} = \left(\tfrac{1}{2}\right)^{4}$, so $4$ half-lives have passed. (M1)
$$ t = 4 \times T_{1/2} = 4 \times 6.0 = 24\ \mathrm{h}. $$(A1)
No logarithm is needed because the target fraction is an exact power of $\tfrac{1}{2}$, so the time is a whole number of half-lives. (R1)
半衰期是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）。(B1)
半衰期个数：$n = \dfrac{t}{T_{1/2}} = \dfrac{18}{6.0} = 3$。(M1)
$$ \frac{N}{N_0} = \left(\tfrac{1}{2}\right)^{3} = \tfrac{1}{8}. $$(A1)
$\tfrac{1}{16} = \left(\tfrac{1}{2}\right)^{4}$，故已过 $4$ 个半衰期。(M1)
$$ t = 4 \times T_{1/2} = 4 \times 6.0 = 24\ \mathrm{h}. $$(A1)
无需对数，因为目标分数恰是 $\tfrac{1}{2}$ 的整数次幂，时间为整数倍半衰期。(R1)
$N = 8.0\times 10^{16}$, $\lambda = 3.0\times 10^{-9}\ \mathrm{s^{-1}}$. (a) activity, SI unit and meaning; (b) corrected count rate from $520\ \mathrm{s^{-1}}$ with $40\ \mathrm{s^{-1}}$ background; (c) two natural background sources and why decay is random and spontaneous.$N = 8.0\times 10^{16}$，$\lambda = 3.0\times 10^{-9}\ \mathrm{s^{-1}}$。(a) 活度、SI 单位及含义；(b) 由 $520\ \mathrm{s^{-1}}$、本底 $40\ \mathrm{s^{-1}}$ 求修正后计数率；(c) 两个天然本底来源及衰变为何随机且自发。
Use the data-booklet relation $A = \lambda N$: $A = (3.0\times 10^{-9})(8.0\times 10^{16})$. (M1)
$$ A = 2.4\times 10^{8}\ \mathrm{Bq}. $$(A1)
The SI unit of activity is the becquerel: $1\ \mathrm{Bq} = 1$ decay (disintegration) per second $= 1\ \mathrm{s^{-1}}$. (B1)
Subtract the background from the measured rate: $520 - 40 = 480\ \mathrm{s^{-1}}$. (A1)
Two natural sources of background radiation: radon gas, rocks and soil, cosmic rays, or radioactive isotopes in food and the body (any two). (B1)
Decay is random because it is impossible to predict which nucleus will decay or when (only the probability per unit time is fixed), and spontaneous because the decay rate is unaffected by temperature, pressure or chemical state. (B1)
用数据手册关系 $A = \lambda N$：$A = (3.0\times 10^{-9})(8.0\times 10^{16})$。(M1)
$$ A = 2.4\times 10^{8}\ \mathrm{Bq}. $$(A1)
活度的 SI 单位是贝克勒尔：$1\ \mathrm{Bq} = $ 每秒 1 次衰变 $= 1\ \mathrm{s^{-1}}$。(B1)
从测得计数率减去本底：$520 - 40 = 480\ \mathrm{s^{-1}}$。(A1)
本底辐射的两个天然来源：氡气、岩石与土壤、宇宙射线，或食物与人体中的放射性同位素（任两个）。(B1)
衰变是随机的，因为无法预测哪个核会衰变、何时衰变（只有每单位时间的概率固定）；是自发的，因为衰变速率不受温度、压强或化学状态影响。(B1)
$T_{1/2} = 15\ \mathrm{h}$, $N_0 = 5.0\times 10^{18}$, $\ln 2 = 0.693$. (a) why $N = N_0 e^{-\lambda t}$; (b) $\lambda$ in $\mathrm{s^{-1}}$; (c) fraction undecayed after $24\ \mathrm{h}$; (d) initial activity in Bq.$T_{1/2} = 15\ \mathrm{h}$，$N_0 = 5.0\times 10^{18}$，$\ln 2 = 0.693$。(a) 为何 $N = N_0 e^{-\lambda t}$；(b) $\lambda$（$\mathrm{s^{-1}}$）；(c) $24\ \mathrm{h}$ 后未衰变分数；(d) 初始活度（Bq）。
Each nucleus has a constant probability $\lambda$ of decaying per unit time, so the number decaying per second is proportional to the number remaining: $\dfrac{dN}{dt} = -\lambda N$. (M1)
A quantity whose rate of change is proportional to itself decays exponentially, giving $N = N_0 e^{-\lambda t}$. (A1)
Use $\lambda = \dfrac{\ln 2}{T_{1/2}}$ with the half-life in seconds: $T_{1/2} = 15 \times 3600 = 5.4\times 10^{4}\ \mathrm{s}$. (M1)
$$ \lambda = \frac{0.693}{5.4\times 10^{4}} \approx 1.28\times 10^{-5}\ \mathrm{s^{-1}}. $$(A1)
Work in hours, where $\lambda_{\mathrm{h}} = \dfrac{0.693}{15} = 0.0462\ \mathrm{h^{-1}}$. Then $\dfrac{N}{N_0} = e^{-\lambda_{\mathrm{h}} t} = e^{-(0.0462)(24)} = e^{-1.109}$. (M1)
$$ \frac{N}{N_0} = e^{-1.109} \approx 0.33. $$(A1)
Use $A_0 = \lambda N_0$ with $\lambda$ in $\mathrm{s^{-1}}$: $A_0 = (1.28\times 10^{-5})(5.0\times 10^{18})$. (M1)
$$ A_0 \approx 6.4\times 10^{13}\ \mathrm{Bq}. $$(A1)
每个核每单位时间有恒定衰变概率 $\lambda$，故每秒衰变数正比于剩余数：$\dfrac{dN}{dt} = -\lambda N$。(M1)
变化率正比于自身的量呈指数衰减，得 $N = N_0 e^{-\lambda t}$。(A1)
用 $\lambda = \dfrac{\ln 2}{T_{1/2}}$，半衰期取秒：$T_{1/2} = 15 \times 3600 = 5.4\times 10^{4}\ \mathrm{s}$。(M1)
$$ \lambda = \frac{0.693}{5.4\times 10^{4}} \approx 1.28\times 10^{-5}\ \mathrm{s^{-1}}. $$(A1)
以小时计算，$\lambda_{\mathrm{h}} = \dfrac{0.693}{15} = 0.0462\ \mathrm{h^{-1}}$。则 $\dfrac{N}{N_0} = e^{-\lambda_{\mathrm{h}} t} = e^{-(0.0462)(24)} = e^{-1.109}$。(M1)
$$ \frac{N}{N_0} = e^{-1.109} \approx 0.33. $$(A1)
用 $A_0 = \lambda N_0$，$\lambda$ 取 $\mathrm{s^{-1}}$：$A_0 = (1.28\times 10^{-5})(5.0\times 10^{18})$。(M1)
$$ A_0 \approx 6.4\times 10^{13}\ \mathrm{Bq}. $$(A1)
$C_{\text{meas}} = 340,180,100,60\ \mathrm{s^{-1}}$ at $t = 0,10,20,30\ \mathrm{min}$; background $20\ \mathrm{s^{-1}}$. (a) why subtract, and the corrected rates; (b) half-life from the halvings; (c) percentage uncertainty at $t = 30\ \mathrm{min}$ given $\pm 10\ \mathrm{s^{-1}}$; (d) relation between count-rate half-life and nuclei half-life.$t = 0,10,20,30\ \mathrm{min}$ 时 $C_{\text{meas}} = 340,180,100,60\ \mathrm{s^{-1}}$；本底 $20\ \mathrm{s^{-1}}$。(a) 为何扣除及修正后速率；(b) 由减半求半衰期；(c) 给定 $\pm 10\ \mathrm{s^{-1}}$ 时 $t = 30\ \mathrm{min}$ 处的百分比不确定度；(d) 计数率半衰期与核数半衰期的关系。
The background is present whether or not the source is there, so it must be removed to leave the true count rate due to the source alone; otherwise the curve would level off at the background instead of tending to zero. (R1)
Subtract $20\ \mathrm{s^{-1}}$ from each reading: (M1)
$$ 340-20,\ 180-20,\ 100-20,\ 60-20 = 320,\ 160,\ 80,\ 40\ \mathrm{s^{-1}}. $$(A1)
The corrected rate halves at each step: $320 \to 160 \to 80 \to 40$, which is $3$ halvings over the $30\ \mathrm{min}$ recorded. (M1)
So $3\,T_{1/2} = 30\ \mathrm{min}$ (A1), giving $T_{1/2} = 10\ \mathrm{min}$. (A1)
The corrected count rate there is $40\ \mathrm{s^{-1}}$, with absolute uncertainty $\pm 10\ \mathrm{s^{-1}}$ (background uncertainty negligible): (M1)
$$ \frac{10}{40}\times 100\% = 25\%. $$(A1)
The half-life of the count rate equals the half-life of the number of undecayed nuclei. (A1)
Because the count rate is proportional to the activity and $A = \lambda N$ with $\lambda$ constant, the count rate is proportional to $N$ at all times, so the two share the same half-life. (R1)
无论源是否在场，本底都存在，故须扣除以留下仅由源贡献的真实计数率；否则曲线会在本底处趋平，而非趋于零。(R1)
从每个读数减去 $20\ \mathrm{s^{-1}}$：(M1)
$$ 340-20,\ 180-20,\ 100-20,\ 60-20 = 320,\ 160,\ 80,\ 40\ \mathrm{s^{-1}}. $$(A1)
修正后速率每步减半：$320 \to 160 \to 80 \to 40$，即在记录的 $30\ \mathrm{min}$ 内减半 $3$ 次。(M1)
故 $3\,T_{1/2} = 30\ \mathrm{min}$（A1），得 $T_{1/2} = 10\ \mathrm{min}$。(A1)
该处修正后计数率为 $40\ \mathrm{s^{-1}}$，绝对不确定度 $\pm 10\ \mathrm{s^{-1}}$（本底不确定度可忽略）：(M1)
$$ \frac{10}{40}\times 100\% = 25\%. $$(A1)
计数率的半衰期等于未衰变核数的半衰期。(A1)
因为计数率正比于活度，而 $A = \lambda N$ 且 $\lambda$ 恒定，故计数率在任一时刻都正比于 $N$，二者半衰期相同。(R1)
Corrected $C = 1000,779,607,472,368\ \mathrm{s^{-1}}$ at $t = 0,10,20,30,40\ \mathrm{s}$. (a) show $\ln C$ vs $t$ is linear, gradient and intercept; (b) tabulate $\ln C$, find the gradient and $\lambda$; (c) half-life; (d) intercept and $C_0$; (e) one advantage of the log plot.$t = 0,10,20,30,40\ \mathrm{s}$ 时修正后 $C = 1000,779,607,472,368\ \mathrm{s^{-1}}$。(a) 证明 $\ln C$ 对 $t$ 为直线、斜率与截距；(b) 列 $\ln C$ 表、求斜率与 $\lambda$；(c) 半衰期；(d) 截距与 $C_0$；(e) 对数图的一个优点。
Take natural logs of $C = C_0 e^{-\lambda t}$: (M1)
$$ \ln C = \ln C_0 - \lambda t. $$This has the form $y = c + mx$ with $y = \ln C$ and $x = t$, so a plot of $\ln C$ against $t$ is a straight line. (A1)
The gradient is $-\lambda$ and the vertical intercept is $\ln C_0$. (A1)
Computing $\ln C$ for each reading: (M1)
| $t\ /\ \mathrm{s}$ | $0$ | $10$ | $20$ | $30$ | $40$ |
|---|---|---|---|---|---|
| $\ln C$ | $6.908$ | $6.658$ | $6.409$ | $6.157$ | $5.908$ |
The $\ln C$ values fall by about $0.25$ every $10\ \mathrm{s}$. Using the endpoints: (A1)
$$ \text{gradient} = \frac{5.908 - 6.908}{40 - 0} = \frac{-1.000}{40} = -0.025\ \mathrm{s^{-1}}. $$Since the gradient is $-\lambda$, the decay constant is $\lambda = 0.025\ \mathrm{s^{-1}}$. (A1)
Use $T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{0.025}$: (M1)
$$ T_{1/2} = 27.7 \approx 28\ \mathrm{s}. $$(A1)
The vertical intercept is $\ln C_0 = 6.908$. (A1)
Hence $C_0 = e^{6.908} \approx 1.0\times 10^{3}\ \mathrm{s^{-1}}$, which matches the $t = 0$ reading of $1000\ \mathrm{s^{-1}}$. (A1)
Plotting $\ln C$ against $t$ gives a straight line whose gradient uses every data point through a best-fit, rather than relying on a single halving read off a curve. (B1)
A straight-line fit averages out random scatter and makes any departure from exponential decay (curvature) immediately visible, so $\lambda$ is obtained more reliably. (R1)
对 $C = C_0 e^{-\lambda t}$ 取自然对数：(M1)
$$ \ln C = \ln C_0 - \lambda t. $$此式形如 $y = c + mx$，其中 $y = \ln C$、$x = t$，故 $\ln C$ 对 $t$ 作图为直线。(A1)
斜率为 $-\lambda$，纵截距为 $\ln C_0$。(A1)
对每个读数计算 $\ln C$：(M1)
| $t\ /\ \mathrm{s}$ | $0$ | $10$ | $20$ | $30$ | $40$ |
|---|---|---|---|---|---|
| $\ln C$ | $6.908$ | $6.658$ | $6.409$ | $6.157$ | $5.908$ |
$\ln C$ 值每 $10\ \mathrm{s}$ 约下降 $0.25$。用两端点：(A1)
$$ \text{斜率} = \frac{5.908 - 6.908}{40 - 0} = \frac{-1.000}{40} = -0.025\ \mathrm{s^{-1}}. $$因斜率为 $-\lambda$，衰变常数 $\lambda = 0.025\ \mathrm{s^{-1}}$。(A1)
用 $T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{0.025}$：(M1)
$$ T_{1/2} = 27.7 \approx 28\ \mathrm{s}. $$(A1)
纵截距为 $\ln C_0 = 6.908$。(A1)
故 $C_0 = e^{6.908} \approx 1.0\times 10^{3}\ \mathrm{s^{-1}}$，与 $t = 0$ 读数 $1000\ \mathrm{s^{-1}}$ 一致。(A1)
绘制 $\ln C$ 对 $t$ 得到直线，其斜率经最佳拟合用上每个数据点，而非依赖从曲线上读出的单次减半。(B1)
直线拟合平均掉随机散布，并使任何偏离指数衰变（弯曲）立即可见，故 $\lambda$ 求得更可靠。(R1)
${}^{14}\mathrm{C}$, $T_{1/2} = 5\,730\ \mathrm{yr}$, $\ln 2 = 0.693$; tool activity is $30\%$ of living wood. (a) the beta-minus equation to nitrogen; (b) $\lambda$ in $\mathrm{yr^{-1}}$; (c) age by taking logs; (d) why the age is between one and two half-lives; (e) why dating fails for very old samples.${}^{14}\mathrm{C}$，$T_{1/2} = 5\,730\ \mathrm{yr}$，$\ln 2 = 0.693$；工具活度为活木的 $30\%$。(a) 到氮的 β⁻ 方程；(b) $\lambda$（$\mathrm{yr^{-1}}$）；(c) 取对数求年代；(d) 为何年代介于一与两个半衰期之间；(e) 为何对很老样品失效。
Beta-minus keeps $A = 14$ and raises $Z$ from $6$ to $7$ (nitrogen), with an electron and an antineutrino emitted: (M1)
$$ {}^{14}_{6}\mathrm{C} \to {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}\beta + \bar{\nu}_e. $$(A1)
$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{5730}$: (M1)
$$ \lambda \approx 1.21\times 10^{-4}\ \mathrm{yr^{-1}}. $$(A1)
The activity is proportional to the number of ${}^{14}\mathrm{C}$ nuclei, so $\dfrac{A}{A_0} = e^{-\lambda t} = 0.30$. Take natural logs: (M1)
$$ -\lambda t = \ln(0.30) \;\Rightarrow\; t = \frac{1}{\lambda}\ln\!\frac{1}{0.30} = \frac{\ln(0.30)}{-\lambda}. $$(M1 for rearranging to $t$)
$$ t = \frac{1.204}{1.21\times 10^{-4}}. $$(A1)
$$ t \approx 9\,950\ \mathrm{yr} \;(\approx 9.9\times 10^{3}\ \mathrm{yr}). $$(A1)
One half-life leaves $50\%$ ($5\,730\ \mathrm{yr}$) and two half-lives leave $25\%$ ($11\,460\ \mathrm{yr}$). Since $30\%$ lies between $50\%$ and $25\%$, the age must lie between one and two half-lives. (R1)
The computed $9\,950\ \mathrm{yr}$ sits between $5\,730$ and $11\,460\ \mathrm{yr}$, confirming the answer. (A1)
After many half-lives the remaining ${}^{14}\mathrm{C}$ activity becomes very small, comparable with the background. (B1)
The corrected count rate is then dominated by uncertainty, so the measured fraction (and hence the age) cannot be determined reliably. (R1)
β⁻ 使 $A = 14$ 不变、$Z$ 从 $6$ 增到 $7$（氮），并发射一个电子与一个反中微子：(M1)
$$ {}^{14}_{6}\mathrm{C} \to {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}\beta + \bar{\nu}_e. $$(A1)
$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{5730}$：(M1)
$$ \lambda \approx 1.21\times 10^{-4}\ \mathrm{yr^{-1}}. $$(A1)
活度正比于 ${}^{14}\mathrm{C}$ 核数，故 $\dfrac{A}{A_0} = e^{-\lambda t} = 0.30$。取自然对数：(M1)
$$ -\lambda t = \ln(0.30) \;\Rightarrow\; t = \frac{1}{\lambda}\ln\!\frac{1}{0.30} = \frac{\ln(0.30)}{-\lambda}. $$(整理为 $t$ 得 M1)
$$ t = \frac{1.204}{1.21\times 10^{-4}}. $$(A1)
$$ t \approx 9\,950\ \mathrm{yr} \;(\approx 9.9\times 10^{3}\ \mathrm{yr}). $$(A1)
一个半衰期剩 $50\%$（$5\,730\ \mathrm{yr}$），两个半衰期剩 $25\%$（$11\,460\ \mathrm{yr}$）。因 $30\%$ 介于 $50\%$ 与 $25\%$ 之间，年代须介于一与两个半衰期之间。(R1)
所求 $9\,950\ \mathrm{yr}$ 落在 $5\,730$ 与 $11\,460\ \mathrm{yr}$ 之间，验证了答案。(A1)
经过许多半衰期后，剩余 ${}^{14}\mathrm{C}$ 活度变得很小，与本底相当。(B1)
此时修正后计数率被不确定度主导，故测得分数（进而年代）无法可靠确定。(R1)
Medical isotope, $T_{1/2} = 6.0\ \mathrm{h}$, $A_0 = 4.0\times 10^{10}\ \mathrm{Bq}$ at arrival. (a) meaning of the activity; (b) activity after $24\ \mathrm{h}$; (c) time to fall to $5.0\times 10^{9}\ \mathrm{Bq}$; (d) why use promptly, given a $1.0\times 10^{10}\ \mathrm{Bq}$ threshold.医用同位素，$T_{1/2} = 6.0\ \mathrm{h}$，到货时 $A_0 = 4.0\times 10^{10}\ \mathrm{Bq}$。(a) 活度含义；(b) $24\ \mathrm{h}$ 后活度；(c) 降到 $5.0\times 10^{9}\ \mathrm{Bq}$ 的时间；(d) 给定 $1.0\times 10^{10}\ \mathrm{Bq}$ 阈值，为何尽快使用。
An activity of $4.0\times 10^{10}\ \mathrm{Bq}$ means $4.0\times 10^{10}$ nuclei decay per second in the sample ($1\ \mathrm{Bq} = 1$ decay per second). (B1)
Number of half-lives: $n = \dfrac{24}{6.0} = 4$. (M1)
The activity falls by $\left(\tfrac{1}{2}\right)^{4} = \tfrac{1}{16}$: (M1)
$$ A = \frac{4.0\times 10^{10}}{16} = 2.5\times 10^{9}\ \mathrm{Bq}. $$(A1)
The ratio is $\dfrac{5.0\times 10^{9}}{4.0\times 10^{10}} = \dfrac{1}{8} = \left(\tfrac{1}{2}\right)^{3}$, so $3$ half-lives have passed. (M1·M1)
$$ t = 3 \times 6.0 = 18\ \mathrm{h}. $$(A1)
The threshold ratio is $\dfrac{1.0\times 10^{10}}{4.0\times 10^{10}} = \dfrac{1}{4} = \left(\tfrac{1}{2}\right)^{2}$, which is reached after $2$ half-lives, i.e. $t = 12\ \mathrm{h}$. (M1·A1)
Because $A = \lambda N$ with $\lambda$ fixed, the activity falls as the number of undecayed nuclei falls; after $12\ \mathrm{h}$ too few nuclei remain to meet the required activity, so the isotope must be used within about $12\ \mathrm{h}$ of arrival. (R1)
$4.0\times 10^{10}\ \mathrm{Bq}$ 的活度表示样品中每秒有 $4.0\times 10^{10}$ 个核衰变（$1\ \mathrm{Bq} = $ 每秒 1 次衰变）。(B1)
半衰期个数：$n = \dfrac{24}{6.0} = 4$。(M1)
活度降为 $\left(\tfrac{1}{2}\right)^{4} = \tfrac{1}{16}$：(M1)
$$ A = \frac{4.0\times 10^{10}}{16} = 2.5\times 10^{9}\ \mathrm{Bq}. $$(A1)
比值为 $\dfrac{5.0\times 10^{9}}{4.0\times 10^{10}} = \dfrac{1}{8} = \left(\tfrac{1}{2}\right)^{3}$，故已过 $3$ 个半衰期。(M1·M1)
$$ t = 3 \times 6.0 = 18\ \mathrm{h}. $$(A1)
阈值比为 $\dfrac{1.0\times 10^{10}}{4.0\times 10^{10}} = \dfrac{1}{4} = \left(\tfrac{1}{2}\right)^{2}$，在 $2$ 个半衰期后到达，即 $t = 12\ \mathrm{h}$。(M1·A1)
因 $A = \lambda N$ 且 $\lambda$ 固定，活度随未衰变核数减少而下降；$12\ \mathrm{h}$ 后剩余核太少，无法满足所需活度，故同位素须在到货后约 $12\ \mathrm{h}$ 内使用。(R1)
${}^{210}_{84}\mathrm{Po}$, alpha emitter, $T_{1/2} = 138\ \mathrm{days}$, decays to Pb $(Z = 82)$; $N_0 = 2.0\times 10^{20}$, $\ln 2 = 0.693$. (a) decay equation; (b) absorber that stops the alpha with reason; (c) $\lambda$ in $\mathrm{s^{-1}}$ and initial activity; (d) activity after $276\ \mathrm{days}$.${}^{210}_{84}\mathrm{Po}$，α 发射体，$T_{1/2} = 138\ \mathrm{天}$，衰变为铅 $(Z = 82)$；$N_0 = 2.0\times 10^{20}$，$\ln 2 = 0.693$。(a) 衰变方程；(b) 挡住 α 的吸收体及理由；(c) $\lambda$（$\mathrm{s^{-1}}$）与初始活度；(d) $276\ \mathrm{天}$ 后活度。
Alpha emission removes $4$ from $A$ and $2$ from $Z$: $A = 210 - 4 = 206$, $Z = 84 - 2 = 82$ (lead). (M1)
$$ {}^{210}_{84}\mathrm{Po} \to {}^{206}_{82}\mathrm{Pb} + {}^{4}_{2}\alpha. $$The lead isotope produced is ${}^{206}_{82}\mathrm{Pb}$. (A1)
A single sheet of paper (or a few centimetres of air) is sufficient to stop the alpha radiation. (A1)
Alpha particles are heavy and doubly charged, so they ionise strongly and lose their energy over a very short range, making them the least penetrating of the three radiations. (R1)
Convert the half-life to seconds: $T_{1/2} = 138 \times 86400 = 1.19\times 10^{7}\ \mathrm{s}$, then $\lambda = \dfrac{\ln 2}{T_{1/2}}$: (M1)
$$ \lambda = \frac{0.693}{1.19\times 10^{7}} \approx 5.8\times 10^{-8}\ \mathrm{s^{-1}}. $$Use $A_0 = \lambda N_0 = (5.8\times 10^{-8})(2.0\times 10^{20})$: (M1)
$$ A_0 \approx 1.2\times 10^{13}\ \mathrm{Bq}. $$(A1)
$276\ \mathrm{days} = 2\,T_{1/2}$, so the activity falls to $\left(\tfrac{1}{2}\right)^{2} = \tfrac{1}{4}$ of $A_0$: $A = \dfrac{1.2\times 10^{13}}{4} \approx 2.9\times 10^{12}\ \mathrm{Bq}$. (A1)
α 发射使 $A$ 减 $4$、$Z$ 减 $2$：$A = 210 - 4 = 206$，$Z = 84 - 2 = 82$（铅）。(M1)
$$ {}^{210}_{84}\mathrm{Po} \to {}^{206}_{82}\mathrm{Pb} + {}^{4}_{2}\alpha. $$生成的铅同位素为 ${}^{206}_{82}\mathrm{Pb}$。(A1)
一张纸（或几厘米空气）即足以挡住 α 辐射。(A1)
α 粒子重且带两个正电荷，电离强，能量在极短程内损失殆尽，故是三种辐射中穿透力最弱的。(R1)
把半衰期换成秒：$T_{1/2} = 138 \times 86400 = 1.19\times 10^{7}\ \mathrm{s}$，再 $\lambda = \dfrac{\ln 2}{T_{1/2}}$：(M1)
$$ \lambda = \frac{0.693}{1.19\times 10^{7}} \approx 5.8\times 10^{-8}\ \mathrm{s^{-1}}. $$用 $A_0 = \lambda N_0 = (5.8\times 10^{-8})(2.0\times 10^{20})$：(M1)
$$ A_0 \approx 1.2\times 10^{13}\ \mathrm{Bq}. $$(A1)
$276\ \mathrm{天} = 2\,T_{1/2}$，故活度降为 $A_0$ 的 $\left(\tfrac{1}{2}\right)^{2} = \tfrac{1}{4}$：$A = \dfrac{1.2\times 10^{13}}{4} \approx 2.9\times 10^{12}\ \mathrm{Bq}$。(A1)