Atomic Masses and Composition of Nucleus

Why a special unit of mass

• Atomic and nuclear masses are absurdly small in kilograms: the mass of a ${}^{12}\text{C}$ atom is $1.992647\times10^{-26}$ kg. So a dedicated unit is used.
• Definition to quote: one atomic mass unit (1 u) is one-twelfth of the mass of one neutral carbon-12 atom.

$$1\ \text{u} = \dfrac{\text{mass of one } {}^{12}\text{C atom}}{12} = \dfrac{1.992647\times10^{-26}}{12}\ \text{kg} = 1.660539\times10^{-27}\ \text{kg}$$

• Atomic masses expressed in u come out close to integral multiples of the mass of a hydrogen atom, but with striking exceptions — chlorine is $35.46$ u, nowhere near an integer.
• Accurate atomic masses are measured with a MASS SPECTROMETER.

Isotopes and the weighted average

• The mass spectrometer reveals that a single element has atomic species of different mass but identical chemical properties. These are ISOTOPES — literally "the same place", since they occupy the same slot in the periodic table.
• Practically every element is a mixture of several isotopes, in fixed relative abundances that differ from element to element.
• The quoted atomic mass of an element is the weighted average of its isotopic masses:

$$\bar m = \dfrac{f_1 m_1 + f_2 m_2 + \ldots}{100}$$

• Standard example — chlorine has isotopes of mass $34.98$ u ($75.4\%$) and $36.98$ u ($24.6\%$):

$$\bar m = \dfrac{75.4\times34.98 + 24.6\times36.98}{100} = 35.47\ \text{u}$$

which agrees with the tabulated atomic mass of chlorine.

• Hydrogen, the lightest element, has three isotopes: $1.0078$ u (protium, abundance $99.985\%$), $2.0141$ u (deuterium) and $3.0160$ u (tritium). Tritium nuclei are unstable, do not occur naturally, and are produced artificially in laboratories.

The proton

• The nucleus of the commonest hydrogen atom is the PROTON.

$$m_p = 1.00727\ \text{u} = 1.67262\times10^{-27}\ \text{kg}$$

• This equals the mass of the hydrogen atom ($1.00783$ u) minus the mass of one electron ($m_e = 0.00055$ u) — a useful check, and the reason atomic-mass data can be used directly in binding-energy sums.
• A proton carries one unit of fundamental positive charge, $+e$, and is stable.
• The positive charge of a nucleus is entirely due to its protons. Since the atom is neutral and has $Z$ electrons of total charge $-Ze$ outside, the nuclear charge is $+Ze$ and the number of protons is exactly $Z$.
• It was once thought that the nucleus might contain electrons; quantum-theoretical arguments ruled this out. ALL the electrons of an atom are outside the nucleus.

Discovery of the neutron

• The argument that forced a second constituent: deuterium and tritium are isotopes of hydrogen, so each has just ONE proton; yet the masses of the hydrogen, deuterium and tritium nuclei are in the ratio $1:2:3$. The extra mass must be neutral matter, present in amounts of about one and two proton-masses respectively — i.e. in multiples of a basic unit.
• James Chadwick verified this in 1932. Bombarding beryllium nuclei with $\alpha$-particles produced a neutral radiation that could knock protons out of light nuclei such as helium, carbon and nitrogen.
• The radiation could not be photons: applying conservation of energy and momentum showed that photons able to do this would need far more energy than the $\alpha$-bombardment of beryllium could supply.
• Chadwick's resolution: the radiation consists of a new neutral particle, the NEUTRON, whose mass conservation of energy and momentum showed to be very nearly that of the proton. He received the 1935 Nobel Prize in Physics for the discovery.

$$m_n = 1.00866\ \text{u} = 1.6749\times10^{-27}\ \text{kg}$$

• A FREE neutron is unstable, unlike a free proton: it decays into a proton, an electron and an antineutrino, with a mean life of about $1000$ s. Inside a nucleus, however, the neutron is stable.

Describing the composition of a nucleus

• $Z$ = atomic number = number of protons.
• $N$ = neutron number = number of neutrons.
• $A$ = mass number $= Z + N$ = total number of protons and neutrons.
• A proton or a neutron is called a NUCLEON, so $A$ is the number of nucleons.
• Nuclear species (nuclides) are written ${}^{A}_{Z}\text{X}$, with $X$ the chemical symbol. Example: gold is ${}^{197}_{79}\text{Au}$ — 197 nucleons, of which 79 are protons and the remaining 118 are neutrons.
• The nuclear mass is close to $A$ times the average nucleon mass, $m \approx 1.66\times10^{-27}$ kg; the nuclear charge is $+Ze$.

Isotopes, isobars, isotones

• Isotopes of a given element have the same number of protons but different numbers of neutrons. Deuterium ${}^{2}_{1}\text{H}$ has one proton and one neutron; tritium ${}^{3}_{1}\text{H}$ has one proton and two neutrons. Gold has 32 isotopes, from $A = 173$ to $A = 204$.
• Because chemical properties depend on electronic structure, and isotopes have identical electronic structure, they behave identically chemically and sit in the same place in the periodic table.

Balancing a nuclear reaction — the 2-mark question type

• In a nuclear reaction the mass number $A$ and the charge (atomic) number $Z$ are each conserved separately, summed over both sides. The number of atoms of each element is NOT conserved, because elements are transmuted.
• Method: write the unknown as ${}^{A}_{Z}\text{Q}$, equate the sums of the superscripts and the sums of the subscripts, and read off the particle.
• Worked: ${}^{10}_{5}\text{B} + {}^{1}_{0}\text{n} \to {}^{4}_{2}\text{He} + {}^{A}_{Z}\text{Q}$. Mass numbers: $10 + 1 = 4 + A \Rightarrow A = 7$. Charge numbers: $5 + 0 = 2 + Z \Rightarrow Z = 3$. So the missing nucleus is ${}^{7}_{3}\text{Li}$.
• Worked: ${}^{94}_{42}\text{Mo} + {}^{2}_{1}\text{H} \to {}^{95}_{43}\text{Te} + {}^{A}_{Z}\text{Q}$. Mass numbers: $94 + 2 = 95 + A \Rightarrow A = 1$. Charge numbers: $42 + 1 = 43 + Z \Rightarrow Z = 0$. So the missing particle is a neutron, ${}^{1}_{0}\text{n}$.
• Worked: ${}^{22}_{11}\text{Na} + X \to {}^{22}_{10}\text{Ne} + \nu_e$. Mass number is 22 on both sides, so $X$ has $A = 0$; charge: $11 + q = 10 \Rightarrow q = -1$. So $X$ is an electron, ${}^{0}_{-1}e$, and the process is ELECTRON CAPTURE: ${}^{22}_{11}\text{Na} + e^{-} \to {}^{22}_{10}\text{Ne} + \nu_e$.
• Worked: ${}^{238}_{92}\text{U} + n \to {}^{Y}_{X}\text{Np} + e^{-} + \bar\nu + Q$. Mass numbers: $238 + 1 = Y \Rightarrow Y = 239$. Charge numbers: $92 + 0 = X + (-1) \Rightarrow X = 93$, which is indeed neptunium.

Standard results

• $1\ \text{u} = 1.660539\times10^{-27}$ kg $= 931.5\ \text{MeV}/c^2$.
• $m_p = 1.00727$ u, $m_n = 1.00866$ u, $m_e = 0.00055$ u, $m_H(\text{atom}) = 1.007825$ u.
• $A = Z + N$; nuclear charge $= +Ze$; nuclide symbol ${}^{A}_{Z}\text{X}$.
• Same $Z$: isotopes. Same $A$: isobars. Same $N$: isotones.
• Free neutron: unstable, mean life $\approx 1000$ s. Bound neutron: stable. Free proton: stable.
• In every nuclear reaction, $\sum A$ and $\sum Z$ are separately conserved.