The relationship between the nuclear mass and the nuclear binding energy is rooted in Einstein's mass-energy equivalence principle. To understand this, we examine the example of the helium-4 ($^4_2\text{He}$) nucleus.

Example: Helium-4 Nucleus

The mass of the individual nucleons are as follows:

• Mass of 2 protons: $2 \times 1.00728\, \text{u} = 2.01456\, \text{u}$
• Mass of 2 neutrons: $2 \times 1.00866\, \text{u} = 2.01732\, \text{u}$
• Measured mass of $^4_2\text{He}$ nucleus: $4.00151\, \text{u}$

The total mass of free nucleons is:

\Delta m = (2.01456 + 2.01732) - 4.00151 = 0.03037\, \text{u} \tag{1}

This difference in mass is called the mass defect. It represents the amount of mass converted into binding energy that holds the nucleus together.

Mass-Energy Equivalence

According to Einstein's equation:

E = \Delta m \cdot c^2 \tag{2}

Where:

• $E$ is the binding energy (in joules)
• $\Delta m$ is the mass defect (in kg)
• $c = 3.00 \times 10^8\, \text{m/s}$ is the speed of light

Since $1\, \text{u} = 1.6605 \times 10^{-27}\, \text{kg}$, we can compute the binding energy of helium-4:

$\Delta m = 0.03037\, \text{u} = 0.03037 \times 1.6605 \times 10^{-27} = 5.042 \times 10^{-29}\, \text{kg}$

BE = \Delta m \cdot c^2 = 5.042 \times 10^{-29} \cdot (3 \times 10^8)^2 = 4.54 \times 10^{-12}\, \text{J} \tag{3}

Conversion to Electronvolts

Using the conversion factor:

1\, \text{eV} = 1.6022 \times 10^{-19}\, \text{J} \Rightarrow 1\, \text{u} = 931.5\, \text{MeV} \tag{4}

We convert the binding energy into MeV:

BE = \frac{4.54 \times 10^{-12}}{1.6022 \times 10^{-13}} \approx 28.29\, \text{MeV} \tag{5}

Or directly from mass defect:

BE = 0.03037\, \text{u} \times 931.5\, \text{MeV/u} = 28.29\, \text{MeV} \tag{6}

General Formula for Binding Energy

For any nucleus with mass number $A$ and atomic number $Z$:

\Delta m = \left[ Z M_p + (A - Z) M_n \right] - M(A,Z) \tag{7}

BE = \Delta m \cdot c^2 = \left[ Z M_p + (A - Z) M_n - M(A,Z) \right] c^2 \tag{8}

Where:

• $M_p$ = mass of proton
• $M_n$ = mass of neutron
• $M(A,Z)$ = mass of nucleus

Binding Energy per Nucleon

To compare nuclear stability, we calculate the binding energy per nucleon:

\frac{BE}{A} \tag{9}

This ratio indicates the stability of a nucleus. A higher value signifies a more stable nucleus. The most stable nuclei have binding energy per nucleon around 8.7 MeV, peaking at iron-56 ($^{56}\text{Fe}$).

Graphical Representation

The variation of binding energy per nucleon versus mass number $A$ shows that:

• Light nuclei (e.g., H, He) have low binding energy per nucleon.
• Intermediate nuclei (e.g., $^{56}\text{Fe}$) are most stable.
• Heavy nuclei (e.g., U, Pu) have lower binding energy per nucleon, allowing for nuclear fission.

This trend explains nuclear fusion in light elements and fission in heavy elements as energy-releasing processes.

From experimental observations, some nuclides are found to be stable—meaning they do not undergo spontaneous decay—while others are unstable or radioactive, meaning they transform into different elements over time. Two principal factors govern nuclear stability:

1. Binding energy per nucleon
2. Neutron-to-proton (N/Z) ratio

Binding energy per nucleon

As discussed earlier, binding energy per nucleon reaches its maximum around $^{56}\text{Fe}$, which makes it the most stable nucleus in nature. Nuclei with intermediate mass numbers generally exhibit the highest binding energy per nucleon, leading to greater stability.

Notable stable nuclides include:

• $^{4}\text{He}$
• $^{12}\text{C}$
• $^{16}\text{O}$

These remain unchanged indefinitely under normal conditions due to their exceptionally stable nuclear configurations.

Neutron-to-proton ratio (N/Z)

The ratio of neutrons ($N$) to protons ($Z$) plays a vital role in nuclear stability. For light nuclei ($Z \leq 20$), stability is typically achieved when $N \approx Z$. However, as $Z$ increases, electrostatic repulsion among protons becomes more significant, requiring a greater number of neutrons to maintain nuclear cohesion. This causes the "band of stability" to curve upward in a plot of $N$ vs. $Z$.

Band of stability

Stable nuclei lie within the band of stability. Nuclei outside this region are unstable and undergo radioactive decay.

Radioactive decay and instability

Nuclei lying outside the band of stability are unstable and attempt to reach the stable zone by emitting particles. The three main types of nuclear instability are:

Beta-minus ($\beta^-$) decay

Occurs when a nucleus has too many neutrons compared to protons (above the band of stability at low mass number $A$). A neutron is converted into a proton, emitting a beta-minus particle and an antineutrino:

{}^A_Z X \rightarrow {}^A_{Z+1} Y + \beta^- + \bar{\nu} \tag{4.39}

Example:

${}^{14}_6\text{C} \rightarrow {}^{14}_7\text{N} + \beta^- + \bar{\nu}$

Where $\beta^-$ is an electron and $\bar{\nu}$ is an antineutrino with no charge and negligible rest mass.

Beta-plus ($\beta^+$) decay or electron capture

Occurs when a nucleus has too few neutrons (below the band of stability at low $A$). A proton is converted into a neutron, emitting a beta-plus particle and a neutrino:

{}^A_Z X \rightarrow {}^A_{Z-1} Y + \beta^+ + \nu \tag{4.40}

Example:

${}^8_5\text{O} \rightarrow {}^7_5\text{N} + \beta^+ + \nu$

Where $\beta^+$ is a positron and $\nu$ is a neutrino. This process is called an isobaric transformation since the mass number $A$ remains constant.

Alpha ($\alpha$) decay

Occurs in very heavy nuclei (typically $Z > 82$), where no neutron-to-proton ratio can offset the immense electrostatic repulsion. The nucleus emits an alpha particle ($^4_2\text{He}$) to become more stable:

{}^A_Z X \rightarrow {}^{A-4}_{Z-2} Y + {}^4_2\text{He} \tag{4.41}

Example:

${}^{208}_{84}\text{Po} \rightarrow {}^{204}_{82}\text{Pb} + {}^4_2\text{He}$

Radioactivity is an intrinsic quantum mechanical phenomenon whereby the nucleus of an unstable atom undergoes spontaneous transformation to achieve a more stable configuration. This process results in the emission of ionizing radiations such as:

• Alpha particles ($\alpha$) — helium nuclei consisting of 2 protons and 2 neutrons
• Beta particles ($\beta^-$ or $\beta^+$) — high-energy electrons or positrons emitted during nuclear beta decay
• Gamma rays ($\gamma$) — high-frequency electromagnetic radiation emitted during nuclear de-excitation

Materials that contain atoms undergoing such spontaneous emissions are termed radioactive.

Radioactive decay is fundamentally random and probabilistic at the level of individual nuclei, governed by quantum tunneling effects or weak nuclear interactions, depending on the decay type. It is impossible to predict the exact time when a specific atom will decay.

However, when considering a macroscopic sample with a very large number of radioactive atoms, the decay process becomes statistically predictable and the decay rate is proportional to the instantaneous number of undecayed nuclei, $N(t)$, at time $t$:

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

where:

• $\lambda$ is the decay constant (units: s$^{-1}$), representing the probability per unit time that a nucleus will decay
• The negative sign indicates the decrease of radioactive nuclei over time

Rearranging the differential equation:

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

Integrating from $N_0$ nuclei at time $t=0$ to $N$ nuclei at time $t$:

$\int_{N_0}^N \frac{dN'}{N'} = -\lambda \int_0^t dt' \quad \Rightarrow \quad \ln\frac{N}{N_0} = -\lambda t$

Exponentiating both sides:

$N(t) = N_0 e^{-\lambda t}$

This exponential decay law describes how the quantity of radioactive material decreases over time.

The activity, $A$, of a radioactive sample is defined as the number of decays per unit time:

$A(t) = -\frac{dN}{dt} = \lambda N(t) = \lambda N_0 e^{-\lambda t}$

The SI unit of activity is the Becquerel (Bq), which corresponds to one decay per second. Another traditional unit is the Curie (Ci), where $1\, \text{Ci} = 3.7 \times 10^{10} \, \text{Bq}$.

The half-life, $t_{1/2}$, of a radioactive isotope is the time required for half of the initial radioactive nuclei to decay, i.e., when $N(t_{1/2}) = \frac{N_0}{2}$. From the decay law:

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

Taking natural logarithms:

$-\lambda t_{1/2} = \ln \left( \frac{1}{2} \right) = -\ln 2 \quad \Rightarrow \quad t_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$

This fundamental relation allows us to easily convert between half-life and decay constant.

Example 1

A sample contains $N_0 = 10^{18}$ radioactive atoms with a half-life $t_{1/2} = 2000$ days. Calculate:

1. The fraction of atoms remaining after 5000 days
2. The activity of the sample after 5000 days

Solution

Using the decay law:

$\text{Fraction remaining} = \frac{N}{N_0} = e^{-\lambda t}$

First, find the decay constant:

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{2000\, \text{days}} = 3.465 \times 10^{-4} \, \text{day}^{-1}$

Calculate the fraction after $t = 5000$ days:

$\frac{N}{N_0} = e^{-\lambda t} = e^{-3.465 \times 10^{-4} \times 5000} = e^{-1.732} \approx 0.177$

Thus, about 17.7% of the original atoms remain.

To find activity, first convert half-life to seconds for SI consistency:

$t_{1/2} = 2000 \times 24 \times 3600 = 1.728 \times 10^{8} \, s \quad \Rightarrow \quad \lambda = \frac{0.693}{1.728 \times 10^{8}} = 4.01 \times 10^{-9} \, s^{-1}$

Activity at time $t$ is:

$A = \lambda N = 4.01 \times 10^{-9} \times 0.177 \times 10^{18} = 7.1 \times 10^{8} \, \text{Bq} = 710 \, \text{MBq}$

This shows the sample's activity has decreased to 710 megabecquerels after 5000 days.

Example 2

A radioactive sample initially has an activity $A_0 = 9 \times 10^{12} \, \text{Bq}$ and a half-life of 80 seconds. Calculate the time needed for its activity to reduce to $2 \times 10^{12} \, \text{Bq}$.

Solution

From the activity decay formula:

$A = A_0 e^{-\lambda t}$

Where:

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{80\, s} = 8.66 \times 10^{-3} \, s^{-1}$

Rearranging for $t$:

$t = \frac{1}{\lambda} \ln \left( \frac{A_0}{A} \right) = \frac{1}{8.66 \times 10^{-3}} \ln \left( \frac{9 \times 10^{12}}{2 \times 10^{12}} \right)$

$t = 115.5 \times \ln(4.5) \approx 115.5 \times 1.504 = 174 \, \text{seconds}$

Therefore, it takes approximately 174 seconds for the activity to decay to $2 \times 10^{12} \, \text{Bq}$.

Example 3

Strontium-90 has a half-life of 28.8 years. Calculate:

1. Its decay constant in seconds
2. The initial activity of 4 grams of Strontium-90
3. The remaining activity after 4 half-lives

Solution

a. Decay constant $\lambda$:

$t_{1/2} = 28.8 \times 365.25 \times 24 \times 3600 = 9.08 \times 10^8 \, s$

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{9.08 \times 10^8} = 7.63 \times 10^{-10} \, s^{-1}$

b. Initial activity $A_0$:

Calculate number of atoms in 4 g of Strontium-90 (atomic mass = 90 g/mol):

$n = \frac{4\, g}{90\, g/mol} = 0.0444\, mol$

Using Avogadro's number $N_A = 6.022 \times 10^{23} \, \text{atoms/mol}$:

$N_0 = n \times N_A = 0.0444 \times 6.022 \times 10^{23} = 2.67 \times 10^{22} \, \text{atoms}$

Then initial activity:

$A_0 = \lambda N_0 = 7.63 \times 10^{-10} \times 2.67 \times 10^{22} = 2.04 \times 10^{13} \, \text{Bq} = 20.4 \, \text{TBq}$

c. Activity after 4 half-lives:

$A = A_0 e^{-\lambda (4 t_{1/2})} = A_0 e^{-4 \ln 2} = A_0 \left( \frac{1}{2} \right)^4 = \frac{A_0}{16}$

Numerically:

$A = \frac{20.4 \, \text{TBq}}{16} = 1.275 \, \text{TBq}$

Radioisotopes have diverse applications across multiple disciplines:

1. Medicine: Radioisotopes like Iodine-131 are used in cancer treatment (e.g., thyroid cancer), while Technetium-99m (Tc-99m) is widely used in diagnostic imaging due to its ideal half-life and gamma emission.
2. Industry: Radiotracers detect leaks in pipelines; thickness gauges monitor manufacturing quality; smoke detectors employ Americium-241 as a radiation source.
3. Agriculture: Radioisotopes help study plant nutrient uptake, soil erosion, and pest control strategies.
4. Research: Used for studying material properties, tracing biochemical pathways, and dating archaeological samples (radiocarbon dating).

Although radioisotopes are invaluable tools, their ionizing radiation poses serious health risks:

1. Cellular damage and mutation: Radiation can break DNA strands causing mutations that may lead to cancer.
2. Radiation sickness: Acute exposure can cause cell death, especially in rapidly dividing cells like those in bone marrow and intestinal lining.
3. Cataracts: Eye lens exposure to radiation can result in opacification.
4. Bone marrow suppression: Leads to immunosuppression due to reduced white blood cell production.
5. Leukemia: Prolonged radiation exposure increases risk of blood cancers.

This text covers nuclear fusion, nuclear fission, nuclear reactors, and related calculations for energy released during these processes, all within the context of advanced physics (likely A-level or Form 6).

Nuclear fusion

Fusion is the process of combining light nuclei (small nucleons) to form a heavier nucleus. From the binding energy curve, light elements have lower binding energy per nucleon; fusion increases the binding energy per nucleon, releasing energy.

Example:

1. Fusion of tritium ($^3$H) and
2. deuterium ($^2$H) releases energy, which powers the Sun.

Energy from fusion can be used to generate electricity.

Nuclear fission

Fission is the process of splitting a heavy nucleus into two lighter nuclei. Heavy nuclei have lower binding energy per nucleon compared to medium-weight nuclei.

Splitting heavy nuclei (e.g., Uranium-235) into medium nuclei increases binding energy per nucleon, releasing energy.

Fission reactions produce neutrons and fragments that carry kinetic energy.

Energy released from U-235 fission is about 173.29 MeV per fission event.

Neutrons released are fast (~2 MeV) but thermal (slow) neutrons are more effective in sustaining fission chain reactions.

Nuclear reactor operation

Chain reaction: Neutrons from one fission induce further fission, producing more neutrons, sustaining the reaction.

Reactor design includes:

1. Chain reaction
   Neutrons released from the fission of one nucleus induce fission in other nuclei, releasing more neutrons and sustaining a continuous, controlled reaction.

2. Moderator
   The moderator (commonly water) slows down fast neutrons to thermal speeds, increasing the likelihood of further fission events.

3. Control rods
   Control rods are made of materials that absorb excess neutrons, allowing operators to regulate or halt the chain reaction safely.

4. Cooling system
   The cooling system removes heat generated by fission, often converting water to steam, which drives turbines to generate electricity.

5. Containment
   A containment structure is designed to prevent the release of radioactive materials, ensuring safety during reactor operation.

Fuel enrichment increases percentage of U-235 from natural uranium (~0.7%) to reactor fuel (~4-5%).

Sample calculations

1. How much deuterium ($^2$H) in kg/s is needed to generate 100 MW by fusion?
2. Energy in joules released from 50 kg of U-235.

a. How much $^2$H is required for 100 MW fusion?

Given:

• Energy per fusion reaction $E = 2.8179 \times 10^{-12} \text{ J}$ (from fusion of tritium and deuterium)
• Mass of $^2$H $m = 2.014102 \, u \times 1.67 \times 10^{-27} \text{ kg/u} = 3.3636 \times 10^{-27} \text{ kg}$
• Power required $P = 100 \times 10^{6} \text{ W}$

Formula:

$\text{mass flow rate } m_s = \frac{P}{E} \times m$

b. Energy released from 50 kg of U-235

Given:

• Energy per fission $E = 173.29 \, \text{MeV} = 173.29 \times 1.602 \times 10^{-13} \text{ J}$
• Mass of one U-235 atom $m_{\text{atom}} = 235.043929 \, u = 235.043929 \times 1.67 \times 10^{-27} \text{ kg}$
• Total mass $= 50 \text{ kg}$

Calculate number of atoms: $N = \frac{50}{m_{\text{atom}}}$

then total energy: $E_{\text{total}} = N \times E$