Nuclear_02

Nuclear Decay (核衰变)

Radioactivity (放射性)

Historical Discoveries (历史发现)

• 1895: Wilhelm Rontgen 发现 X-rays (X射线，1901年诺贝尔奖)
• 1896: Henri Becquerel, Pierre Curie 和 Marie Curie 发现铀的放射性 (1903年诺贝尔奖)
• 1897: Ernest Rutherford 识别 α-rays 和 β-rays (α射线与β射线，1908年诺贝尔奖)
• 1898-99: Marie Curie 发现钋(Polonium)和镭(Radium) (1911年诺贝尔奖)
• 1900: γ-ray (γ射线)的观测
• 1909: Ernest Rutherford 确立α射线是氦核，β射线是电子，γ射线是光子

Fundamental Process (基本过程)

放射性衰变(Radioactive decay)是不稳定原子核自发发射辐射(能量或粒子)以达到更稳定状态的过程。主要衰变类型包括：

• Alpha decay (α衰变)
• Beta decay (β衰变)
• Gamma decay (γ衰变)

示例衰变方程：

$${}^{220}_{86}\text{Rn} \rightarrow {}^{216}_{84}\text{Po} + \alpha$$

Stochastic Nature (随机性本质)

放射性衰变具有固有概率性：

• 单个原子的确切衰变时间不可预测
• 单位时间衰变概率由衰变常数(decay constant) $\lambda$ 定义且恒定
• 未衰变原子核没有"记忆"效应，其衰变概率始终保持不变

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Decay Rate (衰变速率)

Mathematical Formulation (数学表述)

• Decay Constant ($\lambda$, 衰变常数): 单位时间衰变概率

  $$-\frac{dN}{N} \frac{1}{dt} = \lambda \quad \Rightarrow \quad \frac{dN}{N} = -\lambda dt \quad \Rightarrow \quad N = N_0 e^{-\lambda t}$$

• Half-life ($T_{1/2}$, 半衰期): 半数原子核衰变所需时间

  $$N = \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \quad \Rightarrow \quad e^{-\lambda T_{1/2}} = \frac{1}{2}$$

  $$T_{1/2} = \frac{\ln 2}{\lambda}$$

• Mean Lifetime ($\tau$, 平均寿命):

  $$\tau = \frac{1}{\lambda}$$

  推导过程：

  $$\tau = -\frac{1}{N_0} \int_{N_0}^{0} t dN = \frac{\int_{0}^{\infty} t \lambda N dt}{N_0} = \lambda \int_{0}^{\infty} t e^{-\lambda t} dt = \frac{1}{\lambda}$$

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Radioactive Activity (放射性活度)

Activity ($A$, 活度): 单位时间衰变次数，可实验测量

$$A = -\frac{dN}{dt} = \lambda N$$

Relationship to Half-life (与半衰期的关系)

$$\lambda = \frac{A}{N} \quad \Rightarrow \quad T_{1/2} = \frac{\ln 2}{\lambda}$$

Example Calculation (示例计算)

对于 $1 \, \text{mg}$ ${}^{238}\text{U}$ 活度 $A = 740 \, \text{min}^{-1}$:

• 阿伏伽德罗常数 $N_A = 6.023 \times 10^{23} \, \text{mol}^{-1}$
• ${}^{238}\text{U}$ 摩尔质量 = $238 \, \text{g/mol}$
• 初始原子数：

  $$N_0 = \frac{0.001}{238} \times 6.023 \times 10^{23} = 2.53 \times 10^{18}$$

• 半衰期：

  $$T_{1/2} = \frac{\ln 2 \cdot N_0}{A} = \frac{0.693 \times 2.53 \times 10^{18}}{740} \, \text{min} = 4.5 \times 10^9 \, \text{years}$$

• 平均寿命：

  $$\tau = \frac{1}{\lambda} = \frac{T_{1/2}}{\ln 2} \approx 6.49 \times 10^9 \, \text{years}$$

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Radiocarbon Dating (放射性碳定年法)

Principles (原理)

• 大气 ${}^{14}\text{C}/{}^{12}\text{C}$ 比例：$1.3 \times 10^{-12}$ (生物存活期间恒定)
• 生物死亡后，${}^{14}\text{C}$ 以 $T_{1/2} = 5730$ 年衰变；比例呈可预测性下降

Example Calculation (示例计算)

$100 \, \text{g}$ 碳样品活度 $A = 900 \, \text{min}^{-1}$:

• 初始 ${}^{14}\text{C}$ 原子数：

  $$N_0 = \frac{100}{12} \times 6.023 \times 10^{23} \times 1.3 \times 10^{-12} = 6.52 \times 10^{12}$$

• 现存 ${}^{14}\text{C}$ 原子数：

  $$N = \frac{A T_{1/2}}{\ln 2} = \frac{900 \times 5730 \times 365 \times 24 \times 60}{\ln 2} = 3.91 \times 10^{12}$$

• 样品年代：

  $$t = -\frac{T_{1/2}}{\ln 2} \ln \left( \frac{N}{N_0} \right) = -\frac{5730}{\ln 2} \ln \left( \frac{3.91 \times 10^{12}}{6.52 \times 10^{12}} \right) = 4227 \, \text{years}$$

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Alpha Decay (α衰变)

Process (过程)

$${}^A_Z X \rightarrow {}^{A-4}_{Z-2} Y + \alpha \quad (\alpha = {}^4_2\text{He})$$

Energy Conservation (能量守恒)

释放能量 $Q$ (产物动能)：

$$Q = (M_X - M_Y - M_\alpha) c^2$$

Momentum Conservation (动量守恒)

$$\vec{p}_Y + \vec{p}_\alpha = 0 \quad \Rightarrow \quad T_\alpha = Q \frac{M_Y}{M_Y + M_\alpha}$$

Example: ${}^{238}_{92}\text{U} \rightarrow {}^{234}_{90}\text{Th} + \alpha$

• 质量：$M_U = 238.050788 \, \text{u}$, $M_{Th} = 234.043601 \, \text{u}$, $M_\alpha = 4.002603 \, \text{u}$
• $Q = (238.050788 - 234.043601 - 4.002603) \times 931.494 \, \text{MeV/u} = 4.27 \, \text{MeV}$
• α粒子动能：

  $$T_\alpha = 4.27 \times \frac{234}{234 + 4} \approx 4.20 \, \text{MeV}$$

Binding Energy Relation (结合能关系)

$$Q = B_Y + B_\alpha - B_X$$

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Beta Decay (β衰变)

Types (类型)

1. β⁻ Decay (β⁻衰变): $n \rightarrow p + e^- + \bar{\nu}_e$

   $${}^A_Z X \rightarrow {}^A_{Z+1} Y + e^- + \bar{\nu}_e$$

   能量释放：

   $$Q = (M_X - M_Y) c^2$$

2. β⁺ Decay (β⁺衰变): $p \rightarrow n + e^+ + \nu_e$

   $${}^A_Z X \rightarrow {}^A_{Z-1} Y + e^+ + \nu_e$$

   能量释放：

   $$Q = (M_X - M_Y - 2m_e) c^2$$

3. Electron Capture (EC, 电子俘获): $p + e^- \rightarrow n + \nu_e$

   $${}^A_Z X + e^- \rightarrow {}^A_{Z-1} Y + \nu_e$$

   能量释放：

   $$Q = (M_X - M_Y) c^2 - E_b \quad (E_b = \text{被俘获电子结合能})$$

Example: ${ }^{14}\text{C} \rightarrow {}^{14}\text{N} + e^- + \bar{\nu}_e$

• 质量：$M_{\text{C}} = 14.003242 \, \text{u}$, $M_{\text{N}} = 14.003074 \, \text{u}$
• $Q = (14.003242 - 14.003074) \times 931.494 \, \text{MeV/u} = 156.6 \, \text{keV}$