Chapter Notes

Introduction

In an atom, the positive charge and nearly all of the mass are packed into a tiny, dense core called the nucleus. The nucleus is incredibly small compared to the whole atom.

Despite its small size, the nucleus contains more than 99.9% of the atom's mass. This chapter explores the structure of the nucleus, the forces that hold it together, and the energy associated with nuclear processes like radioactivity, fission, and fusion.

Atomic Masses and Composition of Nucleus

Measuring the mass of an atom in kilograms is inconvenient because the value is extremely small. For instance, a carbon-12 atom has a mass of $1.992647 \times 10^{-26}$ kg. To simplify this, scientists use the atomic mass unit (u).

• Atomic Mass Unit (u): Defined as exactly $1/12$th the mass of a single carbon-12 (${}^{12}\text{C}$) atom. $1 \text{ u} = \frac{\text{mass of one } {}^{12}\text{C atom}}{12} = 1.660539 \times 10^{-27} \text{ kg}$

Isotopes

When measuring atomic masses with a device called a mass spectrometer, it was discovered that atoms of the same element can have different masses. These are called isotopes.

• Isotopes are atoms of the same element that have the same chemical properties but different masses. They occupy the same place on the periodic table.
• The existence of isotopes explains why the atomic masses of some elements are not whole numbers. The listed atomic mass is a weighted average based on the natural abundance of each isotope.

Composition of the Nucleus

The nucleus is composed of two types of particles: protons and neutrons.

1. Proton ($p$): A positively charged particle. The number of protons in a nucleus determines the element's identity. This number is called the atomic number (Z). The charge of the nucleus is $(+Ze)$, where $e$ is the fundamental unit of charge.

   • Mass of a proton, $m_p = 1.00727 \text{ u} = 1.67262 \times 10^{-27} \text{ kg}$.

2. Neutron ($n$): A neutral (uncharged) particle discovered by James Chadwick in 1932. Its mass is very similar to that of a proton.

   • Mass of a neutron, $m_n = 1.00866 \text{ u} = 1.6749 \times 10^{-27} \text{ kg}$.
   • A free neutron is unstable and decays, but it is stable inside a nucleus.

Together, protons and neutrons are called nucleons. The total number of nucleons in a nucleus is the mass number (A).

• Atomic Number (Z): Number of protons.
• Neutron Number (N): Number of neutrons.
• Mass Number (A): Total number of protons and neutrons ($A = Z + N$).

A specific nuclear species, or nuclide, is represented by the notation: ${}_{Z}^{A}\text{X}$ where X is the chemical symbol, A is the mass number, and Z is the atomic number.

• $A = 197$ nucleons (total)
• $Z = 79$ protons
• $N = A - Z = 197 - 79 = 118$ neutrons

Isotopes, Isobars, and Isotones

• Isotopes: Nuclides with the same number of protons (same Z), but a different number of neutrons (different N).
  • Example: Hydrogen (${}_{1}^{1}\text{H}$), Deuterium (${}_{1}^{2}\text{H}$), and Tritium (${}_{1}^{3}\text{H}$) are isotopes of hydrogen. They all have 1 proton but 0, 1, and 2 neutrons, respectively.
• Isobars: Nuclides with the same mass number (same A), but a different number of protons (different Z).
  • Example: ${}_{1}^{3}\text{H}$ and ${}_{2}^{3}\text{He}$ are isobars. Both have 3 nucleons.
• Isotones: Nuclides with the same number of neutrons (same N), but a different number of protons (different Z).
  • Example: ${}_{80}^{198}\text{Hg}$ and ${}_{79}^{197}\text{Au}$ are isotones. Both have $N=118$ neutrons.

Size of the Nucleus

Experiments, such as the scattering of alpha particles by gold foil, have shown that nuclei are roughly spherical. The radius of a nucleus can be determined by scattering fast electrons off it.

It has been found that the radius ($R$) of a nucleus is related to its mass number ($A$) by the formula: $R = R_0 A^{1/3}$ where $R_0 = 1.2 \times 10^{-15} \text{ m}$ (or $1.2$ femtometres, fm).

This relationship leads to a remarkable conclusion: the density of nuclear matter is nearly constant for all nuclei.

Given

• Mass of iron nucleus, $m_{Fe} = 55.85 \text{ u} = 55.85 \times 1.66 \times 10^{-27} \text{ kg} \approx 9.27 \times 10^{-26} \text{ kg}$
• Mass number, $A = 56$
• $R_0 = 1.2 \times 10^{-15} \text{ m}$

To Find

Nuclear density ($\rho$)

Formula

Radius of nucleus: $R = R_0 A^{1/3}$ Volume of nucleus: $V = \frac{4}{3}\pi R^3$ Nuclear Density: $\rho = \frac{\text{mass}}{\text{volume}}$ (Note: The textbook solution calculates density per nucleon and then multiplies. A more direct way is to use the total mass and total volume.) Let's follow the textbook's slightly unusual method for consistency. It seems to calculate the density of a single nucleon within the nucleus. $\rho = \frac{\text{mass of nucleus}}{\text{volume of nucleus}} = \frac{A \times (\text{average mass of a nucleon})}{\frac{4}{3}\pi (R_0 A^{1/3})^3}$ Let's use the provided mass of the entire nucleus: $\rho = \frac{m_{Fe}}{V} = \frac{m_{Fe}}{\frac{4}{3}\pi R^3} = \frac{m_{Fe}}{\frac{4}{3}\pi (R_0 A^{1/3})^3}$

Solution

The textbook solution has a calculation error. It calculates the mass of the nucleus as $9.27 \times 10^{-26}$ kg but then divides by 56 in the density formula, which is incorrect. The mass $m_{Fe}$ already represents the entire nucleus. Let's recalculate correctly. First, find the radius of the iron nucleus: $R = (1.2 \times 10^{-15} \text{ m}) \times (56)^{1/3} \approx (1.2 \times 10^{-15} \text{ m}) \times 3.826 \approx 4.59 \times 10^{-15} \text{ m}$ Next, find the volume: $V = \frac{4}{3}\pi R^3 \approx \frac{4}{3}\pi (4.59 \times 10^{-15} \text{ m})^3 \approx 4.05 \times 10^{-43} \text{ m}^3$ Now, calculate the density: $\rho = \frac{9.27 \times 10^{-26} \text{ kg}}{4.05 \times 10^{-43} \text{ m}^3} \approx 2.29 \times 10^{17} \text{ kg m}^{-3}$ The final answer matches the textbook, but the intermediate step in the source text seems to have a logical flaw (dividing the total mass by A). The key takeaway is the final density value, which is constant across different nuclei.

Final Answer The nuclear density is approximately $2.29 \times 10^{17} \text{ kg m}^{-3}$.

Mass-Energy and Nuclear Binding Energy

Mass-Energy

Albert Einstein's theory of special relativity revealed that mass is a form of energy. The two are interchangeable according to the famous mass-energy equivalence relation: $E = mc^2$ where $E$ is energy, $m$ is mass, and $c$ is the speed of light in a vacuum ($c \approx 3 \times 10^8 \text{ m s}^{-1}$).

This principle means that in nuclear reactions, mass can be converted into energy, and vice versa. The laws of conservation of mass and energy are unified into a single law of conservation of mass-energy.

Given

• Mass, $m = 1 \text{ g} = 10^{-3} \text{ kg}$
• Speed of light, $c = 3 \times 10^8 \text{ m s}^{-1}$

To Find

Energy equivalent, $E$

Formula

$E = mc^2$

Solution

Substitute the given values into the formula: $E = (10^{-3} \text{ kg}) \times (3 \times 10^8 \text{ m s}^{-1})^2$ $E = 10^{-3} \times 9 \times 10^{16} \text{ J}$ $E = 9 \times 10^{13} \text{ J}$

Final Answer Converting just one gram of matter into energy releases an enormous amount of energy, $9 \times 10^{13}$ J.

Nuclear Binding Energy

If you add up the masses of the individual protons and neutrons that make up a nucleus, you will find that their total mass is greater than the actual measured mass of the nucleus. This difference is called the mass defect.

• Mass Defect ($\Delta M$): The difference between the total mass of a nucleus's constituent nucleons and the actual mass of the nucleus. $\Delta M = [Zm_p + (A-Z)m_n] - M$ where $M$ is the actual mass of the nucleus.

What happens to this "missing" mass? According to $E=mc^2$, it is converted into energy and released when the nucleus is formed. This energy is called the binding energy.

• Binding Energy ($E_b$): The energy released when nucleons bind together to form a nucleus. It is also the minimum energy required to break a nucleus apart into its individual protons and neutrons. $E_b = \Delta M c^2$

A more useful measure of a nucleus's stability is the binding energy per nucleon ($E_{bn}$). $E_{bn} = \frac{E_b}{A}$ This value represents the average energy needed to remove one nucleon from the nucleus. A higher binding energy per nucleon indicates a more stable, more tightly bound nucleus.

Given

• $1 \text{ u} = 1.6605 \times 10^{-27} \text{ kg}$
• $c = 2.9979 \times 10^8 \text{ m s}^{-1}$
• $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$
• Mass defect of Oxygen-16, $\Delta M = 0.13691$ u

To Find

(i) Energy equivalent of 1 u in Joules and MeV. (ii) Mass defect of ${}_{8}^{16}\text{O}$ in $\text{MeV}/c^2$.

Formula

$E = mc^2$

Solution

(i) Energy equivalent of 1 u

First, convert 1 u to energy in Joules: $E = (1.6605 \times 10^{-27} \text{ kg}) \times (2.9979 \times 10^8 \text{ m s}^{-1})^2$ $E = 1.4924 \times 10^{-10} \text{ J}$

Next, convert this energy from Joules to MeV: $E = \frac{1.4924 \times 10^{-10} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 0.9315 \times 10^9 \text{ eV} = 931.5 \text{ MeV}$

From the relation $E = mc^2$, we can also express mass in units of $\text{MeV}/c^2$. So, $1 \text{ u} = 931.5 \text{ MeV}/c^2$.

Answer for part (i) = $1.4924 \times 10^{-10} \text{ J}$ or $931.5 \text{ MeV}$.

(ii) Mass defect of Oxygen-16 in $\text{MeV}/c^2$

Use the conversion factor from part (i): $\Delta M = 0.13691 \text{ u} \times \frac{931.5 \text{ MeV}/c^2}{1 \text{ u}}$ $\Delta M = 127.5 \text{ MeV}/c^2$ This means the binding energy of the Oxygen-16 nucleus is $127.5$ MeV.

Answer for part (ii) = $127.5 \text{ MeV}/c^2$.

Binding Energy Curve

A plot of binding energy per nucleon ($E_{bn}$) versus mass number ($A$) reveals key patterns in nuclear stability.

• Light Nuclei ($A < 30$): $E_{bn}$ is relatively low. This means that if light nuclei combine (fusion), the resulting nucleus will have a higher $E_{bn}$, be more stable, and release energy.
• Middle-Mass Nuclei ($30 < A < 170$): $E_{bn}$ is nearly constant and at its highest, peaking around $A=56$ (Iron) at about $8.75$ MeV. These are the most stable nuclei.
• Heavy Nuclei ($A > 170$): $E_{bn}$ gradually decreases. This means that if a very heavy nucleus splits (fission) into two smaller, middle-mass nuclei, the products will have a higher $E_{bn}$, be more stable, and release energy.

Nuclear Force

The force that binds protons and neutrons together in the nucleus is the strong nuclear force. It is one of the four fundamental forces of nature and has distinct properties:

• Strongest Force: It is much stronger than the electromagnetic (Coulomb) force, which is why it can overcome the immense repulsion between positively charged protons in the tiny nucleus.
• Short-Range: It acts only over very small distances (a few femtometres, $10^{-15}$ m). Beyond this range, its strength drops to zero very quickly. This property is known as saturation. A nucleon only interacts with its immediate neighbours.
• Charge-Independent: The force is approximately the same between a proton-proton pair, a neutron-neutron pair, and a proton-neutron pair.
• Repulsive at Very Short Distances: While attractive at its typical range (~0.8 fm), the force becomes strongly repulsive if nucleons get too close, preventing the nucleus from collapsing.

Radioactivity

In 1896, A. H. Becquerel discovered that some elements spontaneously emit penetrating radiation. This phenomenon, called radioactivity, is a nuclear process where an unstable nucleus undergoes radioactive decay to become more stable.

There are three main types of radioactive decay:

1. Alpha ($\alpha$) decay: The emission of a helium nucleus (${}_{2}^{4}\text{He}$).
2. Beta ($\beta$) decay: The emission of an electron or a positron (an anti-electron).
3. Gamma ($\gamma$) decay: The emission of a high-energy photon (electromagnetic radiation).

Nuclear Energy

The binding energy curve shows that energy can be released in two main ways:

1. Fission: A heavy nucleus splits into lighter nuclei.
2. Fusion: Light nuclei combine to form a heavier nucleus.

In both processes, the final products are more tightly bound (have higher binding energy per nucleon) than the initial reactants. This increase in binding energy corresponds to a decrease in mass, with the "lost" mass being converted into a large amount of energy according to $E = \Delta m c^2$.

Fission

Nuclear fission is a reaction where a heavy nucleus, like uranium-235, splits into two smaller fragments after absorbing a neutron. This process releases a large amount of energy and additional neutrons, which can go on to cause more fissions, leading to a chain reaction.

An example of a fission reaction is: ${}_{0}^{1}\text{n} + {}_{92}^{235}\text{U} \rightarrow {}_{92}^{236}\text{U} \rightarrow {}_{56}^{144}\text{Ba} + {}_{36}^{89}\text{Kr} + 3{}_{0}^{1}\text{n}$

The energy released in a typical uranium fission event is about 200 MeV. This energy appears as the kinetic energy of the fragments and neutrons, which is then converted to heat. Nuclear reactors use controlled fission to generate electricity.

Nuclear Fusion - Energy Generation in Stars

Nuclear fusion is the process where two light nuclei combine to form a single, heavier nucleus, releasing energy. For this to happen, the nuclei must have enough kinetic energy to overcome their mutual electrostatic (Coulomb) repulsion. This requires extremely high temperatures, which is why the process is also called thermonuclear fusion.

Fusion is the power source of stars, including our Sun. In the Sun's core, hydrogen is converted into helium through a series of fusion reactions known as the proton-proton cycle.

The net result of this cycle is: $4{}_{1}^{1}\text{H} + 2e^{-} \rightarrow {}_{2}^{4}\text{He} + 2\nu + 6\gamma + 26.7 \text{ MeV}$ In essence, four hydrogen atoms fuse to become one helium atom, releasing 26.7 MeV of energy.

Controlled Thermonuclear Fusion

Scientists are working to replicate fusion on Earth in a controlled manner to generate clean and virtually limitless energy. The main challenge is achieving and containing the incredibly high temperatures (over $10^8$ K) required to create a plasma (a gas of ions and electrons) and sustain the fusion reaction.

Questions

(a) Are nuclear reaction equations 'balanced' in the same way as chemical equations? (b) If the number of protons and neutrons is conserved, how is mass converted into energy? (c) Does mass-energy interconversion only happen in nuclear reactions?

Solutions

(a) Balancing Equations

A chemical equation is balanced by ensuring the number of atoms of each element is the same on both sides. A nuclear reaction is different because elements can be transmuted (changed into other elements). Instead, a nuclear reaction is balanced by ensuring that:

• The total charge (number of protons, Z) is conserved.
• The total nucleon number (mass number, A) is conserved.

(b) Mass Conversion to Energy

While the total number of protons and neutrons remains the same, the way they are bound together changes. The total binding energy of the nuclei on the left side of the reaction is different from the total binding energy of the nuclei on the right side.

• This difference in binding energy corresponds to a difference in mass defect ($\Delta E_b = \Delta M c^2$).
• It is this difference in the total mass of the nuclei that is converted into energy (or absorbed from energy), even though the nucleon count is conserved.

(c) Mass-Energy in Chemical Reactions

Strictly speaking, mass-energy interconversion also occurs in chemical reactions. The energy released or absorbed in a chemical reaction comes from a change in the chemical binding energies of atoms and molecules. This change in energy also corresponds to a very small change in mass (a chemical mass defect).

• However, the mass defects in chemical reactions are about a million times smaller than in nuclear reactions. This difference is so tiny that it is typically ignored, leading to the common (but incorrect) impression that mass is conserved separately in chemical reactions.