Atomic structure and constituent particles

An atom has a small, positively charged nucleus surrounded by electrons, which occupy the space around it. The nucleus contains protons and neutrons; these particles are collectively called nucleons.

Each proton has a positive charge, while each electron has an equal-sized negative charge. Neutrons have no charge. A neutral atom has equal numbers of protons and electrons, so its positive and negative charges cancel.

Charges and masses of atomic particles

The particles’ charges and masses can be given in standard units or compared using relative values:

• Proton: charge $$+1.602 \times 10^{-19},\mathrm{C}$$; mass approximately $$1.673 \times 10^{-27},\mathrm{kg}$$. Its relative charge is $$+1$$ and its relative mass is approximately $$1$$.
• Neutron: charge $$0,\mathrm{C}$$; mass approximately $$1.675 \times 10^{-27},\mathrm{kg}$$. Its relative charge is $$0$$ and its relative mass is approximately $$1$$.
• Electron: charge $$-1.602 \times 10^{-19},\mathrm{C}$$; mass approximately $$9.11 \times 10^{-31},\mathrm{kg}$$. Its relative charge is $$-1$$ and its relative mass is approximately $$1/1836$$.

A proton and a neutron have similar masses, but an electron is much lighter. Almost all of an atom’s mass is therefore concentrated in its nucleus.

The atomic mass unit

The atomic mass unit, written $$\mathrm{u}$$ or $$\mathrm{amu}$$, is defined as one twelfth of the mass of a carbon-12 atom. It is approximately $$1.661 \times 10^{-27},\mathrm{kg}$$.

Specific charge

An object's specific charge is its electric charge per unit mass. It is found by dividing the object's charge by its mass:

Here, $$q$$ is the charge in coulombs and $$m$$ is the mass in kilograms. Specific charge is measured in $$\mathrm{C,kg^{-1}}$$. Since mass is positive, the sign of the specific charge is the sign of the object's charge.

A proton has a specific charge of approximately $$+9.58 \times 10^{7},\mathrm{C,kg^{-1}}$$. An electron has a specific charge of approximately $$-1.76 \times 10^{11},\mathrm{C,kg^{-1}}$$. The electron's specific charge is negative because its charge is negative.

Specific charge of a nucleus

A nucleus contains $$Z$$ protons, where $$Z$$ is its proton number. Each proton has charge $$+e$$, so the nucleus has charge $$+Ze$$. Its specific charge is therefore $$+Ze/m_{\text{nucleus}}$$, where $$m_{\text{nucleus}}$$ is the mass of the nucleus.

Specific charge of an ion

An ion has a net charge because it has lost or gained electrons. Losing electrons produces a positive ion; gaining electrons produces a negative ion. To find its specific charge, divide its net charge by its mass.

For an atomic ion, the mass of the nucleus is often used as an approximation to the mass of the whole ion. This is appropriate when the electron masses are negligible for the precision required.

Nuclide notation

A nucleus contains protons and neutrons. Its proton number, $$Z$$, is the number of protons it contains. This number identifies the chemical element.

Its nucleon number, $$A$$, is the total number of protons and neutrons. The number of neutrons is therefore $$A-Z$$.

Reading nuclide notation

A nuclide of element $$X$$ is written as:

The nucleon number is at the upper left of the element symbol, and the proton number is at the lower left.

For example, $$\ce{^{14}_{6}C}$$ represents a carbon nucleus with $$6$$ protons and $$14-6=8$$ neutrons. Nuclide notation specifies the composition of the nucleus; it does not specify the number of electrons.

Isotopes and isotopic abundance

Atoms of the same element always have the same number of protons. They can, however, have different numbers of neutrons. Atoms of the same element with different neutron numbers are called isotopes.

The nucleon number is the total number of protons and neutrons in an atom. Isotopes of an element therefore have the same proton number but different nucleon numbers. For example, $$^{10}{5}\mathrm{B}$$ and $$^{11}{5}\mathrm{B}$$ both have 5 protons, but they have 5 and 6 neutrons respectively.

A sample of an element can contain a mixture of isotopes. An isotope’s fractional abundance is the fraction of the atoms in the sample that are that isotope. The fractional abundances of all the isotopes in the sample add up to 1.

Mean mass of an isotopic mixture

The mean atomic mass is a weighted mean: each isotopic mass contributes according to how abundant that isotope is. An isotope present in a larger fraction of the atoms has a greater effect on the mean.

Here, $$\bar{m}$$ is the mean atomic mass, and $$f_i$$ and $$m_i$$ are the fractional abundance and mass of isotope $$i$$. To calculate the mean, multiply each isotopic mass by its fractional abundance, then add the results. Convert a percentage abundance to a fraction by dividing it by 100.

For example, suppose a boron sample contains 20% $$^{10}{5}\mathrm{B}$$ and 80% $$^{11}{5}\mathrm{B}$$. Using approximate isotopic masses of $$10.0,\mathrm{u}$$ and $$11.0,\mathrm{u}$$, its mean atomic mass is:

The unit $$\mathrm{u}$$ is the atomic mass unit. The mean is closer to $$11.0,\mathrm{u}$$ because the heavier isotope is more abundant. A nucleon number is a count of particles; an isotopic mass is a mass and need not equal the nucleon number exactly.

Finding a missing isotopic abundance

The weighted-mean relationship can also be used in reverse. For a mixture of two isotopes, if one has fractional abundance $$f$$, the other has fractional abundance $$1-f$$.

Suppose another sample contains the same two boron isotopes, with the same approximate masses, and has a mean atomic mass of $$10.3,\mathrm{u}$$. Let $$f$$ be the fractional abundance of the lighter isotope. Then:

The masses in this equation are in $$\mathrm{u}$$. Solving gives $$f=0.70$$. The lighter isotope therefore has an abundance of 70%, and the heavier isotope has an abundance of 30%.