1. Nuclear Instability:

• Nuclear instability refers to the tendency of an atomic nucleus to undergo changes in its composition or structure, leading to the emission of radiation or particles. This can occur due to various factors, such as:
• Excess energy: Nuclei with high energy levels may release energy through radiation or particle emission.
• Unbalanced proton-neutron ratio: Nuclei with an unstable ratio of protons to neutrons may undergo changes to achieve a more stable configuration.
• Weak nuclear forces: The strong nuclear force may not be enough to hold the nucleus together, leading to instability.
• Types of nuclear instability:
  Radioactive decay: Spontaneous emission of radiation or particles.
  Nuclear fission: Splitting of a heavy nucleus into lighter nuclei.
  Nuclear fusion: Combination of two light nuclei to form a heavier nucleus.
• Causes of nuclear instability:
  High proton-neutron ratio
  Odd number of protons or neutrons
  High energy levels
  Weak nuclear forces
  External factors (e.g., radiation, temperature, pressure)
• Effects of nuclear instability:
  Radiation emission
  Particle emission (e.g., alpha, beta, gamma)
  Nuclear reactions (e.g., fission, fusion)
  Changes in nuclear composition
  Energy release or absorption
• There are several distinct isotopes for each element in the periodic table. Several thousand distinct nuclei may be produced by adding all of these isotopes together.
• Nevertheless, the majority of these isotopes decay to become more stable through the release of radiation. There are only 253 stable nuclides in all.
• The half-lives of the remaining nuclides range from billions of years to a fraction of a microsecond, and they all decay.

  
  Figure 1 A plot of neutron number, N, against atomic number, Z. The stable nuclei form the black line down the centre of the nuclides. Nuclei above the stable line decay by [math] β^- [/math] decay; nuclei below the stable line can decay by [math] α,β^+ [/math] decay or [math] K^- [/math] capture.

• Figure 1 shows a chart of nuclei and their stability. The neutron number N is plotted on the y-axis, against atomic (or proton) number Z on the x-axis.
• The least stable nuclei, with very short half-lives, are shown in blue, then the colours green, yellow and red show increasingly stable nuclei.
• The stable nuclei form the black line down the centre of the chart.
• The highest atomic number for a stable nucleus is 82 – this is the element lead.
• The element above lead in the periodic table is bismuth. Its isotope bismuth-209 is nearly stable, but it decays by alpha emission with a very long half-life of [math] 2 \times 10^{19} [/math] years- the age of the Universe is [math] 1.37 \times 10^{10} [/math]
• Uranium-238 has the highest naturally occurring atomic number of 92.
• The isotope uranium-238 decays with a half- life of 4.5 billion years, which is about the same as the age of our Solar System.
• The chart in Figure 1 shows that, for small values of N and Z, stable nuclei have roughly equal numbers of protons and neutrons.
• Examples of nuclei with equal numbers of protons and neutrons include [math] _2^4 \text{He}, _6^{12} \text{C}, _7^{14} \text{N}, _8^{16} \text{O},   _{14}^{28} \text{Si}, \text{and } _{20}^{40} \text{Ca} [/math].
• However, as Z increases, the chart shows that the of neutrons becomes higher than the number of protons.
• A uranium 238 nucleus has 92 protons and 146 neutrons. The physical reason for this is that the electrostatic repulsion of the protons becomes more significant as the nucleus grows. This repulsive effect is balanced in a stable nucleus by extra neutrons, which provide extra attractive nuclear interactions.

• ⇒ Unstable nuclei:

• Unstable nuclei are atomic nuclei that have an unstable combination of protons and neutrons, leading to radioactivity. They can undergo various types of decay, such as:
• Alpha decay (α-decay): Emission of an alpha particle (2 protons + 2 neutrons)
• 
  Figure 2 Alpha decay
• Beta decay (β-decay): Emission of a beta particle (electron or positron)
  Gamma decay (γ-decay): Emission of a gamma ray (high-energy electromagnetic radiation)
  Electron capture: Capture of an electron by the nucleus
  Spontaneous fission: Splitting of the nucleus into two or more smaller nuclei
• Unstable nuclei can be formed in various ways, such as:
  Radioactive decay of stable nuclei
  Nuclear reactions (e.g., particle collisions)
  Cosmic ray interactions
  Artificial nuclear reactions (e.g., particle accelerators)

2. Decay modes of unstable nuclei:

• A nucleus in terms of its atomic number Z, which is the number of protons in it, and its mass number A, which is the sum of the protons and neutrons inside the nucleus.
• This is written as follows:
• 
• When an unstable nucleus decays, it emits radiations, which change the nucleus. So, the numbers A and Z may change.
• When Z changes, the symbol X changes too, as a new element has been formed. This product of the decay is called a daughter nucleus.

2. Radioactive decay and representation in simple decay equations:

• ⇒Alpha decay:

  
  Figure 3 A graphical representation of alpha decay: A decreases by 4 and Z decreases by 2

• An alpha particle, α , is a helium nucleus, so it has the symbol [math] _2^4\text{He} [/math]. A typical alpha decay is described below for an emitter americium-241:
• [math] _{95}^{241}\text{Am} \rightarrow _{93}^{237}\text{Np} + _2^4\text{He} [/math]
• Both the mass number and the atomic number must be conserved. The new element formed has 93 protons, which is neptunium.
• Alpha decay is very rare for elements with Z less than 82.
• The process of alpha emission may be represented on a plot of nucleon number, A, against atomic number, Z (see Figure 3).

• ⇒Beta emission:

• A beta particle, β, is a fast-moving electron. It has a charge of -1, and it is very small in comparison with a proton or neutron. A beta particle is described by the symbol [math] _{-1}^{0}e  [/math]. (The electron mass is about [math] \frac{1}{2000} [/math]  times that of a proton or neutron, so this has a negligible effect on the mass of an atom.)
• The isotope nickel-63 is a beta emitter and its decay is described below. You will recall from last year’s work that an antineutrino, , is also emitted with the beta particle.

  
  Figure 4 A graphical representation of β decay: A remains the same and Z increases by 1

• This particle has virtually no mass but it does have energy:
• [math] _{28}^{63}\text{Ni} \rightarrow _{29}^{63}\text{Cu} + _{-1}^{0}e + \bar{v} [/math]
• When the electron leaves the nucleus, its atomic number increases by 1, and copper-63 is formed. Figure 4 shows the change graphically.

• ⇒ Positron emission and K-capture:

• Nuclei that are rich in protons, with an atomic number less than 82, tend to decay by positron, [math] \beta^+ [/math] , emission or by K-capture.
• A positron is the antiparticle of an electron. A positron has the same mass as an electron, but has a positive charge.
• A nucleus that decays by positron decay is iodine-124
• [math] _{53}^{124}\text{I} \rightarrow _{52}^{124}\text{Te} + _1^0e + v [/math]
• In positron decay, the atomic number decreases by 1. This is shown graphically in Figure 5.

  
  Figure 5 A graphical representation of [math] β^+ [/math]decay

• A neutrino, which is the antiparticle of the antineutrino, is emitted together with the positron.
  However, iodine-124 also decays by K-capture.
  In an atom, electrons that orbit the nucleus in the lowest level are in the ‘K shell.
  These electrons are very tightly bound to the nucleus and actually spend some time inside the nucleus itself.
• The nucleus can capture such an electron, so that a proton is turned into a neutron. K-capture in iodine-124 can be described as follows:
• [math] _{53}^{124}\text{I} + _{-1}^0e \rightarrow _{52}^{124}\text{Te} + v [/math]

4. Gamma emission:

• When a nucleus decays by emission or by [math] α , β^+,\, \text{or} \, β^- [/math] emission or by K-capture, it is often left in an excited state. This is similar to an atom being in an excited state, when an electron is in a higher energy level.
• When an electron drops to a lower energy level, a photon is emitted. Such a photon has an energy of a few eV or perhaps keV in heavy atoms.
• Protons and neutrons can be left in a higher energy level after a nucleus emits a radioactive particle.
• When the nucleon drops back to a lower level, a photon is emitted.
• The energy of these photons is often measured in MeV, and these high-energy photons are the gamma rays that we detect.
• Figure 6 emitted rays are gamma rays
• In most examples of radioactive decay, the excited nucleus releases its energy very quickly, and gamma rays are emitted shortly after an [math] α , β^+ , \, \text{or} \, β^- [/math] particle (with a half-life shorter than [math] 10^9 [/math]s).
• When the half-life for gamma emission is much more than [math] 10^9 [/math]s, we say that the nucleus is left in a metastable state. The half-life for metastable states varies from seconds to many years.
• To distinguish a metastable state from a stable state, we use the letter ‘m’.
• For example, silver-107m is the metastable state of the common isotope silver-107. These states are also distinguished as shown below, when we use atomic and mass numbers:
  – Silver-107 in its ground (stable) state [math] _{47}^{107}\text{Ag} [/math]
  – Silver-107 in its metastable state [math] _{47}^{107m}\text{Ag} [/math]

• Metastable technetium-99m:

• One metastable radioisotope is worth a separate comment. Technetium-99m is a decay product of molybdenum-99, which can be produced in nuclear reactors.
• The relevant nuclear equations are as follows:
• [math] _{42}^{99}\text{Mo} \rightarrow _{43}^{99m}\text{Tc} + _{-1}^0e + \bar{v} \quad \text{(half-life 66 hours)} \\ _{43}^{99m}\text{Tc} \rightarrow _{43}^{99}\text{Tc} + \gamma \quad \text{(half-life 6 hours)} [/math]
• The half-life of molybdenum-99 is long enough for it to be transported to hospitals, where it is then put to good use.
• Figure 7 Energy level diagram for molybdenum-99 and technetium-99
• The technetium-99m is chemically separated from the molybdenum and it is used as a diagnostic tracer. The short half-life of 6 hours makes technetium-99m ideal: it produces a relatively high activity but for a short time.
• Nuclear energy level diagram (figure 7):
  1. Energy axis: The vertical axis represents the energy of the nucleus.
  2. Ground state: The lowest energy level, representing the most stable state of the nucleus.
  3. Excited states: Higher energy levels, representing less stable states of the nucleus.
  4. Nuclear shells: Horizontal lines representing energy levels where protons or neutrons occupy specific nuclear shells (similar to electron shells in atomic physics).
  5. Spin and parity: Labels indicating the spin (angular momentum) and parity (symmetry) of each energy level.
• These diagrams help physicists:
  1. Understand nuclear structure and stability
  2. Predict nuclear reactions and decays
  3. Analyze experimental data from nuclear physics experiments
• Some common features in nuclear energy level diagrams include:
  – Shell gaps: Energy gaps between nuclear shells, indicating increased stability
  – Pairing energies: Energy reductions due to proton-proton or neutron-neutron pairing
  – Isomeric states: Long-lived excited states with different spin or parity

Nuclear Radius

5. The radius of atomic nuclei:

• ⇒Closest approach of alpha particles:

• In the early 20th century, Rutherford’s gold foil experiment involved scattering alpha particles (helium nuclei) off atomic nuclei. By analyzing the scattering angles and energies, researchers could estimate the radius of the nucleus. The closest approach of alpha particles to the nucleus is related to the nuclear radius.

• ⇒Electron diffraction:

• Electron diffraction involves scattering electrons off atomic nuclei. By analyzing the diffraction patterns, researchers can determine the nuclear radius. This method is similar to X-ray diffraction, but uses electrons instead.
• Both methods provide estimates of the nuclear radius, but with some limitations:
  – Alpha particle scattering: Assumes a sharp nuclear surface and doesn’t account for nuclear deformations.
  – Electron diffraction: Assumes a spherical nucleus and may be affected by electron-nucleus interactions.
• Despite these limitations, these methods have contributed significantly to our understanding of nuclear sizes and shapes.
• Keep in mind that modern methods, like electron microscopy and advanced scattering techniques, provide more precise measurements of nuclear radii.
• Typical values for nuclear radius are:
  – Atomic number (Z): 1-100
  – Mass number (A): 1-250
  – Nuclear radius (R): 1-10 femtometers (fm)
• Some examples of nuclear radii are:
  – Hydrogen (Z=1, A=1): 0.7-1.1 fm
  – Helium (Z=2, A=4): 1.5-2.5 fm
  – Oxygen (Z=8, A=16): 3.0-4.0 fm
  – Uranium (Z=92, A=238): 6.5-7.5 fm
• Keep in mind that nuclear radii can vary slightly depending on the specific isotope and nuclear model used. These values are approximate and based on empirical formulas like the “liquid drop model”.
• Remember, nuclear radii are incredibly small, and even the largest nuclei are only about [math] 10^{-14} [/math] meters in radius.

6. The Coulomb equation for the closest approach estimate:

• The Coulomb equation for the closest approach estimate is a fundamental concept in nuclear physics. Here’s a detailed explanation:

• ⇒Coulomb Equation:

• The Coulomb equation describes the interaction between two charged particles, such as an alpha particle (helium nucleus) and a target nucleus. The equation is:
• [math] V(r) = k \frac{Z_1 Z_2}{r} [/math]
• where:- V(r) is the Coulomb potential energy
  – k is Coulomb’s constant (approximately)
  –  and  are the atomic numbers (number of protons) of the two nuclei
  – r is the distance between the centers of the two nuclei

• ⇒ Closest Approach Estimate:

• To estimate the closest approach distance () of an alpha particle to a target nucleus, we set the kinetic energy of the alpha particle equal to the Coulomb potential energy at the distance of closest approach:
• Kinetic Energy = Coulomb Potential Energy
• [math] \frac{1}{1} mv^2 = k \frac{Z_1 Z_2}{r_0} [/math]
• Where m is the mass of the alpha particle and v is its velocity.
  Rearranging the equation to solve for [math]r_0[/math], we get:
• [math] r_0 = k \frac{Z_1 Z_2}{\frac{1}{2} mv^2} \\ r_0 = k \frac{2 \, \cdot  Z_1 Z_2}{mv^2} [/math]
• This equation gives us the closest approach distance, which is related to the nuclear radius. By analyzing the scattering data, we can estimate the nuclear radius.
  – This equation assumes a head-on collision between the alpha particle and the target nucleus.
  – The Coulomb potential energy is a simplification, as it doesn’t account for the strong nuclear force or other effects.
  – This estimate provides a rough value for the nuclear radius, which can be refined with more advanced models and experiments.
• By understanding the Coulomb equation and closest approach estimate, you’ll be well-prepared to explore nuclear physics and estimate nuclear radii:

7. Dependence of radius on nucleon number:

• The dependence of radius on nucleon number is a fundamental concept in nuclear physics. The radius of a nucleus (R) is found to be proportional to the cube root of the nucleon number (A):
• [math] R ∝ A^{\frac{1}{3}}[/math]
• This means that as the number of nucleons (protons and neutrons) in the nucleus increases, the radius of the nucleus grows, but at a slower rate. This relationship is a consequence of the following factors:
  1. Strong nuclear force: The strong nuclear force, which holds the nucleus together, becomes weaker as the distance between nucleons increases.
  2. Constant density: The nuclear density (number of nucleons per unit volume) remains relatively constant across all nuclei.
  3. Spherical nucleus: The nucleus is approximately spherical in shape, so the volume (V) is related to the radius (R) by
• [math] \frac{4}{3} \pi R^3 [/math]
• By combining these factors, we can derive the [math] R ∝ A^{\frac{1}{3}}[/math] relationship, which indicates that the radius grows slower than the volume as the nucleus gets larger.
• – This relationship is an approximation, and actual nuclear radii may vary slightly due to other factors like nuclear shape and deformation.
  – The constant [math] R_0 [/math] in the equation
• [math] R = R_0 A^{\frac{1}{3}}[/math]
• is approximately 1.2 femtometers (fm).
  – This relationship holds well for most nuclei, but there are some deviations for certain regions of the periodic table.
• The equation [math] R ∝ A^{\frac{1}{3}} [/math] can be interpreted as evidence for the constant density of nuclear material because:
  – Volume proportional to cube of radius: The volume of the nucleus (V) is proportional to the cube of the radius ([math] R^3 [/math]).
  – Volume proportional to mass number: The volume (V) is also proportional to the mass number (A).
  – Constant density:
• Since
• [math] V ∝ R^3 [/math]
• And
• [math] V ∝ A [/math]
• We can combine these relationships to get
• [math] V ∝ A^{\frac{1}{3}} [/math]
• This means that as the mass number (A) increases, the radius (R) grows, but at a slower rate, indicating that the density of the nuclear material remains constant.
• Think of it like a sphere: if you increase the radius, the volume grows faster than the radius. But in the case of nuclei, the radius grows slower than the volume, suggesting that the density remains constant.
• This constant density is a fundamental property of nuclear material, indicating that the strong nuclear force holds the nucleus together in a way that maintains a consistent density across all nuclei.

8. Calculation of nuclear density:

• To calculate the nuclear density, we can use the following steps:
  1. Determine the mass number (A) of the nucleus.
  2. Use the equation

     [math] R = R_0 A^{\frac{1}{3}} [/math]

     to estimate the radius (R) of the nucleus.

  3. Calculate the volume (V) of the nucleus using the formula

     [math] V = \frac{4}{3} \pi R^3 [/math]

  4. Use the mass number (A) and the atomic mass unit (amu) to calculate the mass (m) of the nucleus.
  5. Divide the mass (m) by the volume (V) to get the nuclear density (ρ).
• The nuclear density (ρ) is typically expressed in units of grams per cubic centimeter (g/cm³) or grams per femtometer cubed (g/fm³).
• Let’s say we want to calculate the nuclear density of a nucleus with a mass number (A) of 56.
  1. Estimate the radius (R) using

     [math]R = R_0 A^{\frac{1}{3}} [/math]

     Where  is approximately 1.2 fm.

  2. Calculate the volume (V) using

     [math] V = \frac{4}{3} π R^3[/math]

  3. Calculate the mass (m) using the mass number (A) and the atomic mass unit (amu).
  4. Divide the mass (m) by the volume (V) to get the nuclear density (ρ).
     Plugging in the values, we get:

      

     [math] R ≈ 4.8 fm \\
     V ≈ 2.4 × 10^{-45} m^3 \\
     m ≈ 1.4 × 10^{-25} g \\
     ρ ≈ 2.3 × 10^{17} g/cm^3 [/math]

• So, the nuclear density of this nucleus is approximately 3 × 10¹⁷ g/cm³.
• Keep in mind that this is a rough estimate, as actual nuclear densities may vary slightly depending on the specific nucleus and its properties.

Example 1:

Given data:
Nuclear mass [math] = m =12.000[/math] u (carbon-12)
Nuclear radius [math] = R = 2.7 fm = 2.7 * 10^{-15} m [/math]

Find data:
Density = ρ =?
Volume= V=?

Formula:

[math] \text{Density} = \frac{\text{mass}}{\text{volume}} \\
ρ = \frac{m}{V} \\
V = \frac{4}{3} \pi R^3 [/math]

Solution:

[math] V = \frac{4}{3} \pi R^3 \\
V = \frac{4}{3} (3.14)(2.7 * 10^{-15})^3 \\
V = 82.4 * 10^{-45} m^3 \\
ρ = \frac{m}{V} \\
ρ = \frac{12 * 1.66 * 10^{-27}}{82.4 * 10^{-45}} \\
ρ = 2.4 * 10^{17} kg/m^3 [/math]

“The nuclear density of carbon-12 is approximately , indicating a highly dense nucleus.”

Example 2:

Given data:
Nuclear mass = 20.000 u (neon-20)
Nuclear radius [math] = 3.5 fm = 3.5 * 10^{-15} m [/math]

Find data:
Density = ρ =?
Volume= V=?

Formula:

[math] \text{Density} = \frac{\text{mass}}{\text{volume}} \\
ρ = \frac{m}{V} \\
V = \frac{4}{3} \pi R^3 [/math]

Solution:

[math] V = \frac{4}{3} \pi R^3 \\
V = \frac{4}{3} (3.14)(3.5 * 10^{-15})^3 \\
V = 179.5 * 10^{-45} m^3 \\
ρ = \frac{m}{V} \\
ρ = \frac{20 * 1.66 * 10^{-27}}{179.5 * 10^{-45}} \\
ρ = 1.85 * 10^{17} kg/m^3 [/math]

“Neon-20 has a nuclear density of approximately , slightly lower than carbon-12 due to its larger radius.”