Atomic Masses and Nuclear Composition
Introduction
Our understanding of matter at the atomic level has evolved significantly. While the atom was once considered the smallest indivisible particle, it is now known to be composed of even smaller constituents: electrons, protons, and neutrons. We have discussed the nuclear model of the atom, where a tiny, dense nucleus contains protons and neutrons, and electrons orbit this nucleus. Nuclear physics is the branch of physics dedicated to the study of the nucleus itself – its composition, structure, properties, and the forces that hold it together.
The nucleus is located at the center of the atom and contains most of the atom's mass. It is held together by the strong nuclear force, one of the fundamental forces in nature, which is much stronger than the electrostatic repulsion between the protons within the nucleus.
To understand the nucleus, we first need to accurately measure the masses of atoms and understand the composition of the nucleus in terms of its constituent particles.
Atomic Masses And Composition Of Nucleus (Isotopes, Isobars, Isotones)
Atoms are characterised by their atomic number ($Z$) and mass number ($A$). The nucleus of an atom contains protons and neutrons, collectively called nucleons.
Composition of the Nucleus
- Protons: Positively charged particles ($+e$) with a mass approximately 1836 times the mass of an electron. The number of protons in the nucleus of an atom is equal to the atomic number ($Z$). The atomic number determines the chemical element to which the atom belongs.
- Neutrons: Electrically neutral particles with a mass slightly greater than that of a proton.
The total number of protons and neutrons in a nucleus is called the mass number ($A$).
$ A = \text{Number of protons} + \text{Number of neutrons} = Z + N $
Where $N$ is the number of neutrons. Thus, the number of neutrons is $N = A - Z$.
A nucleus is often represented by the notation $_Z^A X$, where X is the chemical symbol of the element, A is the mass number, and Z is the atomic number. For example, $_6^{12}C$ represents a carbon nucleus with 6 protons and 12 nucleons (6 protons + 6 neutrons). $_8^{16}O$ represents an oxygen nucleus with 8 protons and 16 nucleons (8 protons + 8 neutrons).
Atomic Mass Unit (amu or u)
The masses of atoms are very small, making it inconvenient to express them in kilograms. A special unit called the atomic mass unit (amu) or unified atomic mass unit (u) is used.
One atomic mass unit is defined as one-twelfth (1/12th) of the mass of an unbound atom of carbon-12 ($^{12}C$) in its nuclear and electronic ground state.
$ 1 \, u = \frac{\text{Mass of one } ^{12}C \text{ atom}}{12} \approx 1.6605 \times 10^{-27} \, kg $
The approximate masses of the fundamental particles in atomic mass units are:
- Proton mass: $m_p \approx 1.007276 \, u$
- Neutron mass: $m_n \approx 1.008665 \, u$
- Electron mass: $m_e \approx 0.00054858 \, u$
Isotopes, Isobars, and Isotones
Nuclei can be classified into different types based on their number of protons and neutrons:
- Isotopes: Atoms of the same element that have the same atomic number ($Z$) but different mass numbers ($A$). This means they have the same number of protons but different numbers of neutrons. Isotopes have the same chemical properties because they have the same number of electrons (in a neutral atom), but they may have different physical properties (like mass, density, nuclear stability). Examples: Hydrogen has three common isotopes: Protium ($_1^1H$, Z=1, A=1, N=0), Deuterium ($_1^2H$, Z=1, A=2, N=1), and Tritium ($_1^3H$, Z=1, A=3, N=2). Carbon has isotopes like $^{12}C$ (6 protons, 6 neutrons) and $^{14}C$ (6 protons, 8 neutrons).
- Isobars: Atoms of different elements that have the same mass number ($A$) but different atomic numbers ($Z$). This means they have the same total number of nucleons but different numbers of protons and neutrons. Isobars have different chemical properties because they are different elements. Examples: Argon ($_{18}^{40}Ar$), Potassium ($_{19}^{40}K$), and Calcium ($_{20}^{40}Ca$) are isobars, as they all have a mass number of 40.
- Isotones: Nuclei that have the same number of neutrons ($N$) but different atomic numbers ($Z$) and different mass numbers ($A$). Examples: Chlorine-37 ($_{17}^{37}Cl$, N=37-17=20) and Potassium-39 ($_{19}^{39}K$, N=39-19=20) are isotones because they both have 20 neutrons.
Discovery Of Neutron
The discovery of the neutron was a crucial step in understanding the composition of the nucleus. Initially, it was thought that the nucleus contained only protons and electrons. This "proton-electron model" of the nucleus explained the atomic number (number of protons) and mass number (protons + nuclear electrons) for some light nuclei, but it had several inconsistencies (e.g., incorrect nuclear spin, inability to explain beta decay properly without hypothesising electron creation).
In 1932, James Chadwick performed experiments involving bombarding Beryllium with alpha particles. He observed that a highly penetrating, electrically neutral radiation was emitted. This radiation could eject protons from paraffin wax, and by measuring the energy of these ejected protons, Chadwick was able to determine the mass of the neutral particles. He found that these particles had a mass slightly greater than that of a proton and were electrically neutral. He called these particles neutrons.
The discovery of the neutron led to the modern understanding of the nucleus as being composed of protons and neutrons. This proton-neutron model of the nucleus resolved the inconsistencies of the earlier model and correctly explained the atomic mass and charge of nuclei.
Chadwick's discovery was a major advancement and earned him the Nobel Prize in Physics in 1935.
Size Of The Nucleus (Radius Formula $ R = R_0 A^{1/3} $)
The experiments involving the scattering of alpha particles by atomic nuclei (Rutherford scattering) provided the first estimates of the size of the nucleus. These experiments showed that the nucleus is extremely small, of the order of $10^{-14}$ to $10^{-15}$ meters (femtometers or fermis), which is about $10^5$ times smaller than the size of the atom ($\sim 10^{-10}$ m).
Experimental Determination of Nuclear Size
In Rutherford scattering, when an alpha particle is directed head-on ($b=0$) towards a nucleus, it approaches the nucleus until its kinetic energy is completely converted into electrostatic potential energy at the point of closest approach. At this point, the alpha particle momentarily stops before being repelled backwards. Assuming the electrostatic repulsion holds down to this distance, the distance of closest approach gives an upper limit to the size of the nucleus.
If $K_\alpha$ is the initial kinetic energy of the alpha particle (charge $2e$) and $Ze$ is the charge of the nucleus, and $r_0$ is the distance of closest approach, then:
$ K_\alpha = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0} $
$ r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K_\alpha} $
Since scattering experiments with very energetic alpha particles show deviations from Rutherford scattering predictions at very close distances, it indicates that the strong nuclear force starts to act, meaning the alpha particle has "touched" the nucleus. This provides a more accurate estimate of the nuclear radius.
More accurate measurements of nuclear radii are obtained from electron scattering experiments (since electrons are not subject to the strong nuclear force and probe the charge distribution) and studies of pionic atoms and mirror nuclei.
Relation between Nuclear Radius and Mass Number
Experimental studies show that the volume of a nucleus is approximately proportional to the number of nucleons ($A$) it contains. Since the volume of a sphere is proportional to the cube of its radius ($V \propto R^3$), this suggests that $R^3 \propto A$, or $R \propto A^{1/3}$.
The relationship between the nuclear radius ($R$) and the mass number ($A$) is empirically given by the formula:
$ R = R_0 A^{1/3} $
Where:
- $R$ is the radius of the nucleus.
- $A$ is the mass number (total number of nucleons).
- $R_0$ is an empirical constant called the Fermi radius or nuclear radius constant. Its value is approximately $R_0 \approx 1.2 \times 10^{-15} \, m = 1.2 \, fm$ (femtometer or fermi). This value is determined experimentally from scattering experiments.
This formula indicates that the density of nuclear matter is approximately constant for all nuclei, independent of the mass number.
Example:
- Radius of Carbon-12 nucleus ($A=12$): $R_{C} = R_0 (12)^{1/3} \approx 1.2 \times 2.289 \, fm \approx 2.75 \, fm$.
- Radius of Gold-197 nucleus ($A=197$): $R_{Au} = R_0 (197)^{1/3} \approx 1.2 \times 5.819 \, fm \approx 6.98 \, fm$.
Nuclear Density
The volume of a nucleus is $V \approx \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$.
The mass of the nucleus is approximately $M_{nucleus} \approx A \cdot u$ (where $u$ is the atomic mass unit, close to the average mass of a nucleon).
The nuclear density ($\rho_{nucleus}$) is mass per unit volume:
$ \rho_{nucleus} = \frac{M_{nucleus}}{V} \approx \frac{A \cdot u}{\frac{4}{3}\pi R_0^3 A} = \frac{u}{\frac{4}{3}\pi R_0^3} $
Substitute $u \approx 1.66 \times 10^{-27} \, kg$ and $R_0 \approx 1.2 \times 10^{-15} \, m$:
$ \rho_{nucleus} \approx \frac{1.66 \times 10^{-27} \, kg}{\frac{4}{3}\pi (1.2 \times 10^{-15} \, m)^3} \approx \frac{1.66 \times 10^{-27}}{4.189 \times 1.728 \times 10^{-45}} \, kg/m^3 $
$ \rho_{nucleus} \approx \frac{1.66}{7.238} \times 10^{18} \, kg/m^3 \approx 0.23 \times 10^{18} \, kg/m^3 = 2.3 \times 10^{17} \, kg/m^3 $
This density is incredibly high, about $2.3 \times 10^{14}$ times the density of water ($10^3 \, kg/m^3$). This constant and extremely high density of nuclear matter is a characteristic property of nuclei, supporting the idea that nucleons are tightly packed within the nucleus.
Example 1. The radius of the nucleus of Germanium ($_{32}^{70}Ge$) is approximately 4.8 fm. Using the formula $R = R_0 A^{1/3}$, estimate the value of the nuclear radius constant $R_0$ from this data. (1 fm = $10^{-15}$ m).
Answer:
Given:
Radius of Germanium-70 nucleus, $R = 4.8 \, fm = 4.8 \times 10^{-15} \, m$
Mass number of Germanium-70, $A = 70$
We use the formula $R = R_0 A^{1/3}$ and rearrange to find $R_0$:
$ R_0 = \frac{R}{A^{1/3}} $
Calculate $A^{1/3}$: $(70)^{1/3} \approx 4.121$
Substitute the values:
$ R_0 = \frac{4.8 \, fm}{4.121} $
$ R_0 \approx 1.165 \, fm $
The estimated value of the nuclear radius constant $R_0$ from this data is approximately 1.165 fm, which is close to the standard value of 1.2 fm.