Unit E3 · Nuclear and Quantum PhysicsUnit E3 · 核物理与量子物理
Radioactive Decay放射性衰变
IB-Style Practice QuestionsIB 风格练习题
MEDIUMHARDPaper 1Paper 1BPaper 2HL ONLY
Syllabus E3.1 to E3.6考纲 E3.1 至 E3.6PHYSICS HL
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PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 30 marks短结构题 · 可用计算器 · 30 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. In every nuclear equation, balance nucleon number and charge across the arrow. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。每个核方程都要在箭头两侧配平核子数与电荷。数值答案保留适当的有效数字。
A radioactive source emits a mixture of alpha, beta-minus and gamma radiation.某放射源同时发射 α、β⁻ 与 γ 辐射的混合。
(a)State the identity (what each particle or wave actually is) and the electric charge of each of the three radiations.写出三种辐射各自的本质（每种粒子或波到底是什么）及其电荷。[2]
(b)List the three radiations in order of increasing penetrating power, and state how the order of ionising power compares with this.将三种辐射按穿透力递增排列，并说明电离能力的排序与之如何对比。[2]
Q2MEDIUMPaper 1balancing nuclear equations + the (anti)neutrino配平核方程与（反）中微子[6 marks]
Use the conservation rules for nuclear decay. (Element symbols: radon Rn has $Z = 86$; yttrium Y has $Z = 39$.)运用核衰变的守恒规则。（元素符号：氡 Rn 的 $Z = 86$；钇 Y 的 $Z = 39$。）
(a)Radium-226 $\left({}^{226}_{88}\mathrm{Ra}\right)$ decays by alpha emission. Write the complete decay equation, identifying the daughter nuclide.镭-226 $\left({}^{226}_{88}\mathrm{Ra}\right)$ 发生 α 衰变。写出完整衰变方程并指出子核素。[2]
(b)Strontium-90 $\left({}^{90}_{38}\mathrm{Sr}\right)$ decays by beta-minus emission. Write the complete decay equation, including the accompanying lepton.锶-90 $\left({}^{90}_{38}\mathrm{Sr}\right)$ 发生 β⁻ 衰变。写出完整衰变方程，包含伴随的轻子。[2]
(c)State the two quantities conserved in a nuclear decay equation, and explain in one sentence why an antineutrino must be included in (b).写出核衰变方程中守恒的两个量，并用一句话说明 (b) 中为何必须包含反中微子。[2]
Q3MEDIUMPaper 1half-life: the fraction rule半衰期：分数法则[6 marks]
A radioactive nuclide has a half-life of $6.0\ \mathrm{hours}$. A freshly prepared sample contains $N_0$ undecayed nuclei.某放射性核素半衰期为 $6.0\ \mathrm{小时}$。一份新制样品含 $N_0$ 个未衰变核。
(a)State what is meant by the half-life of a radioactive nuclide.说明放射性核素半衰期的含义。[1]
(b)Calculate the fraction of the original nuclei remaining after $18\ \mathrm{hours}$.计算 $18\ \mathrm{小时}$ 后剩余原始核的分数。[2]
(c)Determine the time taken for the number of undecayed nuclei to fall to $\tfrac{1}{16}$ of $N_0$, and explain why no logarithm is needed here.求未衰变核数降到 $N_0$ 的 $\tfrac{1}{16}$ 所需时间，并说明此处为何无需对数。[3]
Q4HARDPaper 1activity, the becquerel, background活度、贝克勒尔、本底[6 marks]
A sealed sample contains $N = 8.0\times 10^{16}$ undecayed nuclei of a nuclide whose decay constant is $\lambda = 3.0\times 10^{-9}\ \mathrm{s^{-1}}$.一份密封样品含 $N = 8.0\times 10^{16}$ 个未衰变核，其衰变常数 $\lambda = 3.0\times 10^{-9}\ \mathrm{s^{-1}}$。
(a)Calculate the activity of the sample, and state the SI unit of activity together with what one unit represents.计算样品的活度，并写出活度的 SI 单位及一个单位代表的含义。[3]
(b)A detector placed near the sample reads $520\ \mathrm{s^{-1}}$. With the sample removed it reads $40\ \mathrm{s^{-1}}$. Calculate the corrected count rate from the sample.放在样品附近的探测器读数为 $520\ \mathrm{s^{-1}}$。移走样品后读数为 $40\ \mathrm{s^{-1}}$。计算来自样品的修正后计数率。[1]
(c)State two natural sources of the background radiation, and state one reason why radioactive decay is described as both random and spontaneous.写出本底辐射的两个天然来源，并说明放射性衰变被描述为既随机又自发的一个理由。[2]
Q5HARDPaper 1HL ONLYexponential law + decay constant指数定律与衰变常数[8 marks]
A radioactive nuclide has a half-life of $15\ \mathrm{hours}$. A sample of it contains $N_0 = 5.0\times 10^{18}$ undecayed nuclei at $t = 0$. Take $\ln 2 = 0.693$.某放射性核素半衰期为 $15\ \mathrm{小时}$。在 $t = 0$ 时其样品含 $N_0 = 5.0\times 10^{18}$ 个未衰变核。取 $\ln 2 = 0.693$。
(a)Explain why the number of undecayed nuclei follows the exponential law $N = N_0 e^{-\lambda t}$.解释未衰变核数为何服从指数定律 $N = N_0 e^{-\lambda t}$。[2]
(b)Calculate the decay constant $\lambda$ of the nuclide, in $\mathrm{s^{-1}}$.计算该核素的衰变常数 $\lambda$，单位 $\mathrm{s^{-1}}$。[2]
(c)Using the decay law with the time expressed in hours, determine the fraction of the nuclei still undecayed after $24\ \mathrm{hours}$.用衰变定律（时间以小时表示），求 $24\ \mathrm{小时}$ 后仍未衰变的核数分数。[2]
(d)Calculate the initial activity of the sample, in becquerels.计算样品的初始活度，单位贝克勒尔。[2]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 22 marks图像 · 数据 · 不确定度 · 22 分
Graph and Data Questions图像与数据题
These items reward careful background subtraction and correct handling of decay curves and logarithms. Subtract the background before extracting any half-life or decay constant from count-rate data.这些题考查细致的本底扣除以及对衰变曲线与对数的正确处理。从计数率数据求任何半衰期或衰变常数前，先减去本底。
Q6HARDPaper 1Bbackground subtraction + half-life from a decay curve本底扣除与由衰变曲线求半衰期[10 marks]
A student measures the count rate $C_{\text{meas}}$ from a radioactive source at several times $t$. A separate measurement with the source removed gives a steady background of $20\ \mathrm{s^{-1}}$.一名学生在若干时刻 $t$ 测量放射源的计数率 $C_{\text{meas}}$。移走源后单独测得稳定本底为 $20\ \mathrm{s^{-1}}$。
$t\ /\ \mathrm{min}$
$0$
$10$
$20$
$30$
$C_{\text{meas}}\ /\ \mathrm{s^{-1}}$
$340$
$180$
$100$
$60$
(a)Explain why the background must be subtracted from each reading, and state the corrected count rate at each of the four times.解释为何须从每个读数中减去本底，并写出四个时刻各自的修正后计数率。[3]
(b)Using the corrected count rates, determine the half-life of the source, showing how the repeated halvings give your value.用修正后计数率求源的半衰期，写明各次减半如何给出你的结果。[3]
(c)Each measured count rate has an absolute uncertainty of $\pm 10\ \mathrm{s^{-1}}$. For the reading at $t = 30\ \mathrm{min}$, calculate the percentage uncertainty in the corrected count rate. (The background uncertainty is negligible.)每个测得计数率的绝对不确定度为 $\pm 10\ \mathrm{s^{-1}}$。对 $t = 30\ \mathrm{min}$ 的读数，计算修正后计数率的百分比不确定度。（本底不确定度可忽略。）[2]
(d)State what the half-life of the count rate tells you about the half-life of the number of undecayed nuclei, justifying your answer in one sentence.说明计数率的半衰期反映未衰变核数的半衰期如何，并用一句话给出理由。[2]
Q7HARDPaper 1BHL ONLYlinearising the decay law: $\ln C$ vs $t$衰变定律线性化：$\ln C$ 对 $t$[12 marks]
For a different source the background-corrected count rate $C$ is recorded against time $t$:对另一个源，将本底修正后的计数率 $C$ 对时间 $t$ 记录如下：
$t\ /\ \mathrm{s}$
$0$
$10$
$20$
$30$
$40$
$C\ /\ \mathrm{s^{-1}}$
$1000$
$779$
$607$
$472$
$368$
(a)Starting from $C = C_0 e^{-\lambda t}$, show that a graph of $\ln C$ against $t$ should be a straight line, and state what its gradient and its vertical intercept represent.从 $C = C_0 e^{-\lambda t}$ 出发，证明 $\ln C$ 对 $t$ 的图应为直线，并说明其斜率与纵截距分别代表什么。[3]
(b)Complete a table of $\ln C$ for the five data points, then determine the gradient of the $\ln C$ against $t$ line and hence the decay constant $\lambda$.为五个数据点列出 $\ln C$ 表，然后求 $\ln C$ 对 $t$ 直线的斜率，由此求衰变常数 $\lambda$。[3]
(c)Use your value of $\lambda$ to calculate the half-life of the source. Take $\ln 2 = 0.693$.用你求得的 $\lambda$ 计算源的半衰期。取 $\ln 2 = 0.693$。[2]
(d)Read off the vertical intercept of the line and use it to confirm the initial count rate $C_0$.读出直线的纵截距，并据以确认初始计数率 $C_0$。[2]
(e)State one advantage of plotting $\ln C$ against $t$ rather than reading the half-life directly off the raw $C$ against $t$ curve.写出绘制 $\ln C$ 对 $t$ 相比直接从原始 $C$ 对 $t$ 曲线读半衰期的一个优点。[2]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 30 marks长结构题 · 可用计算器 · 30 分
Extended Structured Problems长结构问题
Set up each problem by stating the law you are using. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer. Keep the time units of $\lambda$ and $T_{1/2}$ consistent.每题先写明所用定律。长题中方法分占比最大；中间值多保留几位，仅在最终答案处取舍有效数字。保持 $\lambda$ 与 $T_{1/2}$ 的时间单位一致。
Q8HARDPaper 2HL ONLYcarbon dating with the exponential law用指数定律进行碳测年[12 marks]
Living material maintains a fixed proportion of carbon-14, which decays by beta-minus emission with a half-life of $5\,730\ \mathrm{years}$. A wooden tool recovered from an archaeological site shows a ${}^{14}\mathrm{C}$ activity equal to $30\%$ of that found in living wood of the same mass. Take $\ln 2 = 0.693$.活体物质维持碳-14 的固定比例，碳-14 通过 β⁻ 衰变，半衰期为 $5\,730\ \mathrm{年}$。某考古遗址出土的木制工具，其 ${}^{14}\mathrm{C}$ 活度为同质量活木的 $30\%$。取 $\ln 2 = 0.693$。
(a)Write the nuclear equation for the beta-minus decay of carbon-14 to nitrogen $\left(Z = 7\right)$, including the antineutrino.写出碳-14 经 β⁻ 衰变为氮 $\left(Z = 7\right)$ 的核方程，包含反中微子。[2]
(b)Calculate the decay constant of carbon-14, in $\mathrm{yr^{-1}}$.计算碳-14 的衰变常数，单位 $\mathrm{yr^{-1}}$。[2]
(c)By taking natural logarithms of the decay law, estimate the age of the wooden tool.对衰变定律取自然对数，估计木制工具的年代。[4]
(d)Without further calculation, explain why the estimated age must lie between one and two half-lives, and use this to check your answer in (c).不再计算，解释为何所估年代必介于一个与两个半衰期之间，并据以核对 (c) 的答案。[2]
(e)State one reason why carbon dating becomes unreliable for samples that are many tens of thousands of years old.说明碳测年对年代达数万年的样品为何变得不可靠的一个原因。[2]
Q9HARDPaper 2medical isotope: activity over several half-lives医用同位素：跨多个半衰期的活度[10 marks]
A hospital receives a sample of a medical radioisotope with a half-life of $6.0\ \mathrm{hours}$. On arrival ($t = 0$) the sample has an activity of $4.0\times 10^{10}\ \mathrm{Bq}$.某医院收到一份半衰期为 $6.0\ \mathrm{小时}$ 的医用放射性同位素样品。到货时（$t = 0$）样品活度为 $4.0\times 10^{10}\ \mathrm{Bq}$。
(a)State what an activity of $4.0\times 10^{10}\ \mathrm{Bq}$ means in terms of decays per second.用每秒衰变次数说明 $4.0\times 10^{10}\ \mathrm{Bq}$ 的活度含义。[1]
(b)Calculate the activity of the sample $24\ \mathrm{hours}$ after arrival, using the fraction rule.用分数法则计算到货后 $24\ \mathrm{小时}$ 的样品活度。[3]
(c)Determine the time after arrival at which the activity has fallen to $5.0\times 10^{9}\ \mathrm{Bq}$.求到货后活度降到 $5.0\times 10^{9}\ \mathrm{Bq}$ 的时间。[3]
(d)A procedure requires the activity to be at least $1.0\times 10^{10}\ \mathrm{Bq}$. Using the relation between activity and the number of undecayed nuclei, explain why the isotope must be used promptly after arrival.某项操作要求活度至少为 $1.0\times 10^{10}\ \mathrm{Bq}$。用活度与未衰变核数的关系，解释为何同位素到货后须尽快使用。[3]
Polonium-210 $\left({}^{210}_{84}\mathrm{Po}\right)$ is an alpha emitter with a half-life of $138\ \mathrm{days}$. It decays to an isotope of lead $\left(\mathrm{Pb},\ Z = 82\right)$. A source initially contains $N_0 = 2.0\times 10^{20}$ undecayed polonium-210 nuclei. Take $\ln 2 = 0.693$.钋-210 $\left({}^{210}_{84}\mathrm{Po}\right)$ 是 α 发射体，半衰期为 $138\ \mathrm{天}$。它衰变为铅 $\left(\mathrm{Pb},\ Z = 82\right)$ 的一种同位素。某源初始含 $N_0 = 2.0\times 10^{20}$ 个未衰变钋-210 核。取 $\ln 2 = 0.693$。
(a)Write the complete decay equation, identifying the lead isotope produced.写出完整衰变方程，指出生成的铅同位素。[2]
(b)State, with a reason, what thickness of absorber would be sufficient to stop the alpha radiation emitted.写出能挡住所发射 α 辐射的吸收体厚度，并说明理由。[2]
(c)Calculate the decay constant of polonium-210 in $\mathrm{s^{-1}}$, and hence the initial activity of the source in becquerels.计算钋-210 的衰变常数（单位 $\mathrm{s^{-1}}$），由此求源的初始活度（单位贝克勒尔）。[3]
(d)Determine the activity of the source after $276\ \mathrm{days}$.求 $276\ \mathrm{天}$ 后源的活度。[1]