In every atom, the positive charge and mass are densely concentrated at the center, forming the nucleus . The volume of a nucleus is about 10 -12 \times the volume of the atom, meaning an atom is almost empty space. Nevertheless, the nucleus contains more than 99.9% of the mass of an atom.
The mass of an atom is extremely small. To measure atomic masses, the atomic mass unit ($u$) is defined as 1/12th of the mass of a carbon-12 ($^{12}\text{C}$) atom.
$$1\text{ u} = \frac{\text{mass of one }^{12}\text{C atom}}{12} = \frac{1.992647 \times 10^{-26}\text{ kg}}{12} \approx 1.660539 \times 10^{-27}\text{ kg}$$
Historical Chronicle: Discovery of the Neutron
A nuclear species or nuclide is represented as $_Z^A\text{X}$, where $X$ is the chemical symbol, $Z$ is the atomic number (number of protons), and $A$ is the mass number (total number of protons and neutrons, also called nucleons ). The neutron number is $N = A - Z$.
| Classification Type | Definition & Properties | Examples |
|---|---|---|
| Isotopes | Nuclides with the same atomic number $Z$ but different mass numbers $A$ (same proton count, different neutron count). They exhibit identical chemical properties. | $_1^1\text{H}$, $_1^2\text{H}$ (Deuterium), $_1^3\text{H}$ (Tritium); $_{17}^{35}\text{Cl}$, $_{17}^{37}\text{Cl}$ |
| Isobars | Nuclides with the same mass number $A$ but different atomic numbers $Z$. | $_1^3\text{H}$ and $_2^3\text{He}$ |
| Isotones | Nuclides with the same neutron number $N = A - Z$ but different atomic numbers $Z$. | $_{80}^{198}\text{Hg}$ and $_{79}^{197}\text{Au}$ ($N = 118$) |
By performing scattering experiments using fast electrons as projectiles, the sizes of nuclei of various elements have been accurately measured.
The radius $R$ of a nucleus of mass number $A$ is given by the empirical relation:
$$R = R_0 A^{1/3}$$
where $R_0 \approx 1.2 \times 10^{-15}\text{ m} = 1.2\text{ fm}$ (femtometer).
Since the volume of a sphere is $V = \frac{4}{3}\pi R^3$, substituting the radius formula gives:
$$V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$$
This shows that the volume $V$ is directly proportional to the mass number $A$. Consequently, the nuclear density ($\rho$), which is mass divided by volume, is constant and independent of $A$ for all nuclei:
$$\rho = \frac{A \times m_u}{\frac{4}{3}\pi R_0^3 A} = \frac{3 m_u}{4 \pi R_0^3} \approx 2.3 \times 10^{17}\text{ kg m}^{-3}$$
This density is extremely large compared to ordinary matter (e.g., water is 10 3 kg m -3 ) and is comparable to the density of matter in neutron stars.
Mathematical Problem Statement: Given the mass of an iron nucleus as 55.85 u and $A = 56$, find the nuclear density.
Step-by-Step Algorithmic Solution Trace:
- Identify given values: Mass $m = 55.85\text{ u} = 55.85 \times 1.66 \times 10^{-27}\text{ kg} \approx 9.27 \times 10^{-26}\text{ kg}$, and $A = 56$.
- Calculate nuclear volume: $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi (1.2 \times 10^{-15}\text{ m})^3 \times 56 \approx 4.05 \times 10^{-43} ext{ m}^3$.
- Compute density: $\rho = \frac{\text{mass}}{\text{volume}} = \frac{9.27 \times 10^{-26}\text{ kg}}{4.05 \times 10^{-43}\text{ m}^3} \approx 2.29 \times 10^{17}\text{ kg m}^{-3}$.
Final Evaluation Output: $\rho \approx 2.29 \times 10^{17}\text{ kg m}^{-3}$
Einstein showed that mass is another form of energy, related by the famous equivalence relation:
$$E = m c^2$$
where $c \approx 3 \times 10^8\text{ m s}^{-1}$ is the velocity of light in vacuum.
Mathematical Problem Statement: Calculate the energy equivalent of 1 g of substance.
Step-by-Step Algorithmic Solution Trace:
- Convert mass to SI units: $m = 1\text{ g} = 10^{-3}\text{ kg}$.
- Apply Einstein's relation: $E = m c^2 = 10^{-3}\text{ kg} \times (3 \times 10^8\text{ m s}^{-1})^2$.
- Perform arithmetic: $E = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13}\text{ J}$.
Final Evaluation Output: $E = 9 \times 10^{13}\text{ J}$
The mass of a stable nucleus $M$ is always less than the sum of the individual masses of its constituent protons and neutrons. This difference is called the mass defect ($\Delta M$):
$$\Delta M = [Z m_p + (A - Z) m_n] - M$$
The energy equivalent of this mass defect is the nuclear binding energy ($E_b$), which is the energy required to separate a nucleus into its individual nucleons:
$$E_b = \Delta M c^2$$
A more useful measure of nuclear stability is the binding energy per nucleon ($E_{bn}$):
$$E_{bn} = \frac{E_b}{A}$$
The plot of $E_{bn}$ versus mass number $A$ reveals critical features of nuclear forces:
Mathematical Problem Statement: Find the energy equivalent of 1 u in Joules and MeV, and express the mass defect of $^{16}_8\text{O}$ ($\Delta M = 0.13691\text{ u}$) in $\text{MeV/c}^2$.
Step-by-Step Algorithmic Solution Trace:
- Convert $1\text{ u}$ to energy in Joules: $E = 1.6605 \times 10^{-27}\text{ kg} \times (2.9979 \times 10^8\text{ m/s})^2 \approx 1.4924 \times 10^{-10}\text{ J}$.
- Convert Joules to eV: $E = \frac{1.4924 \times 10^{-10}\text{ J}}{1.602 \times 10^{-19}\text{ J/eV}} \approx 9.315 \times 10^8\text{ eV} = 931.5\text{ MeV}$. Thus, $1\text{ u} = 931.5\text{ MeV/c}^2$.
- Calculate binding energy of $^{16}_8\text{O}$: $E_b = 0.13691\text{ u} \times 931.5\text{ MeV/u} \approx 127.5\text{ MeV}$.
Final Evaluation Output: $1\text{ u} = 931.5\text{ MeV/c}^2$, and $E_b(^{16}_8\text{O}) = 127.5\text{ MeV}$
The nuclear force is the strong attractive force that binds protons and neutrons together inside the tiny volume of a nucleus, overcoming the strong electrostatic Coulomb repulsion between protons.
The potential energy between two nucleons as a function of their separation distance $r$ shows:
Radioactivity is a nuclear phenomenon in which an unstable nucleus spontaneously undergoes decay to achieve stability.
Historical Chronicle: Discovery of Radioactivity
Nuclear reactions release energy on the order of millions of electron volts ($\text{MeV}$), which is about a million \times greater than conventional chemical reactions (which release energy on the order of $\text{eV}$).
Nuclear fission is the splitting of a heavy nucleus into two lighter nuclei of intermediate mass when bombarded with neutrons.
Example reaction for Uranium-235:
$$_0^1\text{n} + _{92}^{235}\text{U} \rightarrow _{92}^{236}\text{U} \rightarrow _{56}^{144}\text{Ba} + _{36}^{89}\text{Kr} + 3 _0^1\text{n}$$
The average energy released per fission event is approximately 200 MeV , primarily appearing as the kinetic energy of the fragments and neutrons, which is subsequently converted into heat.
Nuclear fusion is the process where two or more light nuclei combine to form a single heavier nucleus, releasing enormous energy.
To overcome the electrostatic Coulomb barrier ($\approx 400\text{ keV}$ for two protons), the nuclei must have extremely high kinetic energy, requiring temperatures of about 3 &× 10 9 K (thermonuclear fusion).
In the Sun, fusion occurs via the multi-step proton-proton cycle at a core temperature of 1.5 &× 10 7 K due to high-energy tail distributions and quantum tunneling:
$$4 _1^1\text{H} + 2e^- \rightarrow _2^4\text{He} + 2\nu + 6\gamma + 26.7\text{ MeV}$$
Replicating stellar fusion on Earth requires confining a high-temperature mixture of positive ions and electrons ( plasma ) at temperatures around 10 8 K . Since no physical container can withstand such temperatures, magnetic confinement techniques (such as Tokamaks) are being developed globally.
Mathematical Problem Statement: If both the number of protons and neutrons are conserved in a nuclear reaction, how is mass converted into energy?
Step-by-Step Algorithmic Solution Trace:
- Note that the total number of protons and neutrons is conserved, meaning the sum of their individual rest masses remains constant.
- However, the total binding energy of the reactant nuclei is different from that of the product nuclei.
- The difference in total binding energy ($\Delta E_b$) corresponds to a difference in the actual nuclear masses (mass defect). This mass difference is released as kinetic energy or radiation according to $\Delta E = \Delta M c^2$.
Final Evaluation Output: Energy is released due to the difference in nuclear binding energies, which manifests as a change in the total rest mass of the system.