Calculating Energy Released in Radioactive Decay | 放射性衰变释放能量的计算

📚 Calculating Energy Released in Radioactive Decay | 放射性衰变释放能量的计算

Radioactive decay is a spontaneous nuclear process in which an unstable nucleus transforms into a more stable configuration, often releasing energy in the form of kinetic energy of particles and electromagnetic radiation. This energy release is a direct consequence of the mass-energy equivalence principle, and its calculation is central to nuclear physics.

放射性衰变是一种自发的核过程，在此过程中不稳定的原子核转变为更稳定的形态，通常以粒子的动能和电磁辐射的形式释放能量。这种能量释放是质能等价原理的直接结果，其计算是核物理学的核心内容。

1. The Principle of Mass-Energy Equivalence | 质能等价原理

In 1905, Albert Einstein proposed that mass and energy are interchangeable. The famous equation \(E = mc^2\) is not a conversion formula between two separate quantities, but rather a statement that mass itself is a form of energy. In nuclear reactions, the total mass of the products is slightly less than the total mass of the reactants; this mass difference, known as the mass defect, is converted into energy.

1905年，阿尔伯特·爱因斯坦提出质量与能量可以相互转换。著名的方程 \(E = mc^2\) 并非两个独立量之间的换算公式，而是表明质量本身就是能量的一种形式。在核反应中，产物的总质量略小于反应物的总质量；这一质量差称为质量亏损，它被转化为能量。

ΔE = Δm × c²

where \(\Delta m\) is the mass defect in kilograms and \(c\) is the speed of light in vacuum, \(2.998 × 10⁸ m/s\).

其中 Δm 是以千克为单位的质量亏损，c 是真空中的光速，大小为 \(2.998 × 10⁸\) m/s。

2. Atomic Mass Unit and Energy Equivalents | 原子质量单位与能量当量

In nuclear physics, masses are conveniently expressed in atomic mass units (u), where \(1 u = 1.66054 × 10^{-27} kg\). The corresponding energy equivalent is calculated as:

在核物理学中，质量通常以原子质量单位（u）表示，其中 \(1 u = 1.66054 × 10^{-27}\) kg。其对应的能量当量计算如下：

1 u × c² = 1.66054 × 10⁻²⁷ × (2.998 × 10⁸)² = 1.492 × 10⁻¹⁰ J

This is often expressed in electron volts: \(1 u = 931.5 MeV\). Note that in many CIE A-Level problems, the value of 930 MeV is used as an approximation for the energy equivalent of one atomic mass unit.

这一能量通常以电子伏特表示：\(1 u = 931.5\) MeV。值得注意的是，在许多 CIE A-Level 题目中，常使用约 930 MeV 作为一原子质量单位的能量当量近似值。

Mass in kg: \(m = N × 1.66054 × 10^{-27}\) kg, where \(N\) is the mass in u.
Mass in kg： \(m = N × 1.66054 × 10^{-27}\) kg，其中 N 是以 u 为单位的质量数值。

Energy in MeV: \(E = N × 931.5\) MeV, where \(N\) is the mass defect in u.
能量以 MeV 表示： \(E = N × 931.5\) MeV，其中 N 是以 u 为单位的质量亏损数值。

3. The General Procedure for Calculating Energy Released | 计算释放能量的一般步骤

For any radioactive decay, the energy released can be calculated using a systematic three-step procedure. This approach ensures accuracy and is directly applicable to examination questions.

对于任何放射性衰变，释放的能量都可以通过系统化的三步步骤来计算。这种方法能够确保准确性，并可直接应用于考试题目。

Step 1: Write the balanced nuclear equation for the decay, ensuring that both mass number (A) and atomic number (Z) are conserved.
步骤 1： 写出平衡的核反应方程，确保质量数（A）和原子序数（Z）均守恒。

Step 2: Calculate the mass defect \(\Delta m\) by subtracting the total initial mass from the total final mass.
步骤 2： 用反应前的总质量减去反应后的总质量，计算质量亏损 Δm。

Step 3: Convert the mass defect into energy using \(\Delta E = \Delta m c²\). Ensure that the mass is expressed in appropriate units (kg or u) before performing the calculation.
步骤 3： 使用 \(\Delta E = \Delta m c²\) 将质量亏损转换为能量。计算前需确保质量采用适当单位（kg 或 u）。

It is critical to distinguish between the mass of the neutral atom and the mass of the nucleus. In A-Level calculations, the masses of electrons cancel out when both sides of the equation are treated consistently using neutral atomic masses. Therefore, it is standard practice to use the masses of neutral atoms for both parent and daughter nuclides.

区分中性原子质量和原子核质量至关重要。在 A-Level 计算中，若方程两边均一致地使用中性原子质量，电子的质量会相互抵消。因此，标准做法是母核和子核均采用中性原子的质量。

4. Energy Released in Alpha Decay | α 衰变释放的能量

In alpha decay, a nucleus emits an alpha particle (a helium-4 nucleus, \(^4_2He\)). The general equation is:

在 α 衰变中，原子核发射一个 α 粒子（氦-4 原子核，\(^4_2He\)）。一般方程为：

\(^A_Z X \rightarrow ^{A-4}_{Z-2} Y + ^4_2 He + Q\)

where \(Q\) is the energy released. This energy appears as the kinetic energy of the alpha particle and the daughter nucleus, with the lighter alpha particle carrying away the majority of the kinetic energy.

其中 Q 是释放的能量。该能量以 α 粒子和子核的动能形式出现，由于动量守恒，更轻的 α 粒子携带了大部分动能。

Worked Example 1: Consider the alpha decay of polonium-210: \(^{210}_{84}Po \rightarrow ^{206}_{82}Pb + ^4_2He\). Given masses: \(m(Po-210) = 209.98286 u\), \(m(Pb-206) = 205.97445 u\), \(m(He-4) = 4.00260 u\). Calculate the energy released in MeV.

示例 1： 考虑钋-210 的 α 衰变：\(^{210}_{84}Po \rightarrow ^{206}_{82}Pb + ^4_2He\)。已知质量：\(m(Po-210) = 209.98286 u\)，\(m(Pb-206) = 205.97445 u\)，\(m(He-4) = 4.00260 u\)。计算以 MeV 为单位的释放能量。

Solution: The mass defect is:

解答： 质量亏损为：

\(\Delta m = 209.98286 – (205.97445 + 4.00260) = 0.00581 u\)

Using \(1 u = 931.5 MeV/c²\):

利用 \(1 u = 931.5 MeV/c²\)：

\(E = 0.00581 × 931.5 = 5.41 MeV\)

5. Energy Released in Beta-minus Decay | β⁻ 衰变释放的能量

In beta-minus decay, a neutron transforms into a proton, an electron and an antineutrino. The daughter nucleus has the same mass number but the atomic number increases by one. The general equation is:

在 β⁻ 衰变中，一个中子转变为质子、一个电子和一个反中微子。子核的质量数不变，但原子序数增加 1。一般方程为：

\(^A_Z X \rightarrow ^A_{Z+1} Y + ^0_{-1}e + \bar{\nu}_e + Q\)

The energy released is shared among the electron, the antineutrino and the daughter nucleus. Because the antineutrino is nearly massless, the electron energy spectrum is continuous.

释放的能量分布在电子、反中微子和子核之间。由于反中微子几乎无质量，电子的能谱是连续的。

Worked Example 2: For the decay \(^{32}_{15}P \rightarrow ^{32}_{16}S + ^0_{-1}e + \bar{\nu}_e\), given masses: \(m(P-32) = 31.97391 u\) and \(m(S-32) = 31.97207 u\). Calculate the maximum kinetic energy of the emitted electron.

示例 2： 对于衰变 \(^{32}_{15}P \rightarrow ^{32}_{16}S + ^0_{-1}e + \bar{\nu}_e\)，已知质量：\(m(P-32) = 31.97391 u\)，\(m(S-32) = 31.97207 u\)。计算发射电子的最大动能。

Solution: The mass defect is:

解答： 质量亏损为：

\(\Delta m = 31.97391 – 31.97207 = 0.00184 u\)

Converting to energy:

转换为能量：

\(E = 0.00184 × 931.5 = 1.71 MeV\)

This is the maximum kinetic energy of the electron. The actual energy is variable because the antineutrino can carry away any share of the total energy.

这是电子的最大动能。实际能量是变化的，因为反中微子可以带走总能量中的任意份额。

6. Energy Released in Beta-plus Decay | β⁺ 衰变释放的能量

In beta-plus decay, a proton transforms into a neutron, a positron and a neutrino. The daughter nucleus has the same mass number but the atomic number decreases by one. The general equation is:

在 β⁺ 衰变中，一个质子转变为中子、一个正电子和一个中微子。子核的质量数不变，但原子序数减少 1。一般方程为：

\(^A_Z X \rightarrow ^A_{Z-1} Y + ^0_{+1}e + \nu_e + Q\)

An important detail in beta-plus decay is that the mass of the positron must be accounted for. When using neutral atomic masses, the mass of the electron in the parent atom must be subtracted and the positron mass must be added to the daughter product, leading to a ‘two-electron’ correction factor of \(2m_e\).

β⁺ 衰变中的一个重要细节是必须考虑正电子的质量。当使用中性原子质量时，需要减去母原子中的电子质量，并在子产物中加上正电子质量，因此引入 \(2m_e\) 的修正因子。

Worked Example 3: For the decay \(^{18}_{9}F \rightarrow ^{18}_{8}O + ^0_{+1}e + \nu_e\), given masses: \(m(F-18) = 18.00094 u\), \(m(O-18) = 17.99916 u\), and \(m_e = 0.00055 u\). Calculate the energy released.

示例 3： 对于衰变 \(^{18}_{9}F \rightarrow ^{18}_{8}O + ^0_{+1}e + \nu_e\)，已知质量：\(m(F-18) = 18.00094 u\)，\(m(O-18) = 17.99916 u\)，\(m_e = 0.00055 u\)。计算释放的能量。

Solution: The corrected mass defect is:

解答： 修正后的质量亏损为：

\(\Delta m = 18.00094 – [17.99916 + 2 × 0.00055] = 0.00068 u\)

Thus:

因此：

\(E = 0.00068 × 931.5 = 0.633 MeV\)

7. Energy Released in Gamma Decay | γ 衰变释放的能量

Gamma decay involves the emission of high-energy photons from an excited nucleus. There is no change in mass number or atomic number; the nucleus simply transitions to a lower energy state. The energy of the gamma photon equals the energy difference between the two nuclear energy levels:

γ 衰变涉及激发态原子核发射高能光子。质量数和原子序数均不改变；原子核只是跃迁到更低的能态。γ 光子的能量等于两个核能级之间的能量差：

\(E_\gamma = hf = \frac{hc}{\lambda}\)

In a typical gamma decay, the mass change is so small that it is usually not measurable. Instead, the photon energy is determined from the nuclear energy level diagram or from the photon’s frequency and wavelength. For example, if a nucleus emits a gamma photon of wavelength \(1.0 × 10^{-12}\) m, the photon energy is:

在典型的 γ 衰变中，质量变化极小，通常无法测量。此时，光子能量由核能级图或光子的频率和波长确定。例如，若原子核发射波长为 \(1.0 × 10^{-12}\) m 的 γ 光子，光子能量为：

\(E = \frac{6.63 × 10^{-34} × 3.00 × 10^8}{1.0 × 10^{-12}} = 1.99 × 10^{-13} J = 1.24 MeV\)

8. Using Binding Energy per Nucleon | 利用比结合能计算

The binding energy of a nucleus is the energy required to separate it into its individual nucleons. The binding energy per nucleon is a measure of nuclear stability. The energy released in a nuclear decay can also be calculated as the difference between the total binding energy of the products and that of the reactants:

原子核的结合能是将核分离为独立核子所需的能量。比结合能是核稳定性的度量。核衰变释放的能量也可以计算为产物的总结合能与反应物的总结合能之差：

\(Q = (BE_{products}) – (BE_{reactants})\)

For a decay to be energetically possible, the total binding energy of the products must be greater than that of the parent nucleus. This means that the products are more tightly bound and therefore more stable.

要使衰变在能量上可行，产物的总结合能必须大于母核的总结合能。这意味着产物结合得更紧密，因此更稳定。

Worked Example 4: The binding energy per nucleon of U-238 is 7.57 MeV/nucleon, and that of Th-234 is 7.62 MeV/nucleon. For the alpha decay \(^{238}U \rightarrow ^{234}Th + ^4He\), the alpha particle has a binding energy per nucleon of 7.08 MeV/nucleon. Calculate the energy released.

示例 4： U-238 的比结合能为 7.57 MeV/核子，Th-234 为 7.62 MeV/核子。对于 α 衰变 \(^{238}U \rightarrow ^{234}Th + ^4He\)，α 粒子的比结合能为 7.08 MeV/核子。计算释放的能量。

Solution: The total binding energy of the parent is:

解答： 母核的总结合能为：

\(BE_U = 238 × 7.57 = 1801.66 MeV\)

The total binding energy of the products is:

产物的总结合能为：

\(BE_{products} = (234 × 7.62) + (4 × 7.08) = 1783.08 + 28.32 = 1811.40 MeV\)

Therefore, the energy released is:

因此，释放的能量为：

\(Q = 1811.40 – 1801.66 = 9.74 MeV\)

9. Kinetic Energy Distribution Among Decay Products | 衰变产物间的动能分配

When a stationary nucleus decays, the total momentum before decay is zero. By the principle of conservation of momentum, the products must move in opposite directions with equal and opposite momenta:

当静止原子核衰变时，衰变前的总动量为零。根据动量守恒原理，产物必须沿相反方向运动，且动量大小相等、方向相反：

\(m_1 v_1 = m_2 v_2\)

Applying this to alpha decay, if the daughter nucleus has mass \(M\) and the alpha particle has mass \(m\), the kinetic energies are related by:

将此应用于 α 衰变，若子核质量为 M，α 粒子质量为 m，则动能关系为：

\(\frac{KE_\alpha}{KE_{daughter}} = \frac{M}{m} = \frac{A-4}{4}\)

Since \(M \gg m\), the alpha particle carries nearly all the kinetic energy. The fraction of the total energy carried by the alpha particle is:

由于 M 远大于 m，α 粒子携带了几乎所有的动能。α 粒子携带的总能量份额为：

\(f_\alpha = \frac{M}{M+m} = \frac{A-4}{A}\)

where \(A\) is the mass number of the parent nucleus.

其中 A 是母核的质量数。

10. Units and Conversion Factors | 单位与换算因子

Examination questions often require conversion between joules and electron volts. The following conversion relationships are essential:

考试题目经常需要在焦耳和电子伏特之间进行换算。以下换算关系至关重要：

Quantity \| 量	Value \| 数值
\(1 eV\)	\(1.60 × 10^{-19} J\)
\(1 MeV\)	\(1.60 × 10^{-13} J\)
\(1 u\)	\(1.66 × 10^{-27} kg\)
\(1 u\) energy equivalent \| 能量当量	\(931.5 MeV = 1.49 × 10^{-10} J\)

In CIE examinations, take care to check whether masses are given in kilograms or atomic mass units. When energy is required in joules, use \(E = \Delta m c²\) with mass in kilograms. When energy is required in MeV, use the conversion \(1 u = 931.5 MeV\) (or the approximation 930 MeV as specified in the question paper).

在 CIE 考试中，注意检查质量是以千克还是原子质量单位给出的。当能量要求以焦耳为单位时，使用 \(E = \Delta m c²\)，质量以千克为单位。当能量要求以 MeV 为单位时，使用换算关系 \(1 u = 931.5 MeV\)（或按题目指定的近似值 930 MeV）。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Calculating Energy Released in Radioactive Decay | 放射性衰变释放能量的计算