S1.2 The nuclear atomIB Chemistry SL: Revision notes
Section 1
Structure of the atom
An atom has a tiny, dense, positively charged nucleus containing protons and neutrons (together called nucleons). Negatively charged electrons occupy the space outside the nucleus. The nucleus is about 10⁻¹⁵ m across while the atom is about 10⁻¹⁰ m, so most of an atom is empty space but almost all of its mass is in the nucleus.
| Particle | Relative mass | Relative charge |
|---|---|---|
| proton | 1 | +1 |
| neutron | 1 | 0 |
| electron | 0.0005 | −1 |
Section 2
Evidence for the nuclear model
In the gold-foil experiment almost all alpha particles passed straight through (so the atom is mostly empty space), a few were deflected strongly and about 1 in 8000 bounced back (so the positive charge and mass are concentrated in a very small nucleus). This replaced the earlier model of positive charge spread through the whole atom.
Section 3
Nuclear symbols
The nuclear symbol shows the mass number A (protons + neutrons) and the atomic number Z (protons).
- protons = Z
- neutrons = A − Z
- electrons = Z − charge (a 2+ ion has lost 2 electrons; a 2− ion has gained 2)
Example: has 16 protons, 16 neutrons and 18 electrons.
Forming an ion changes only the number of electrons - never the protons or neutrons.
Section 4
Isotopes
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have the same chemical properties, because chemistry depends on electrons, but slightly different physical properties that depend on mass, such as density, rate of diffusion and melting or boiling point. Example: ³⁵Cl and ³⁷Cl; ¹H, ²H and ³H.
Section 5
Relative atomic mass from isotopic abundance
The relative atomic mass, Ar, is the weighted mean mass of the atoms of an element relative to one-twelfth of the mass of a carbon-12 atom. It is non-integer because it averages the isotopes.
Ar = Σ(isotopic mass × % abundance) ÷ 100
Example: copper is 69.2 % ⁶³Cu and 30.8 % ⁶⁵Cu, so Ar = (63 × 69.2 + 65 × 30.8) ÷ 100 = 63.6.
To find abundances from Ar with two isotopes, let the fraction of one be x and the other (1 − x): for chlorine, 35x + 37(1 − x) = 35.5 gives x = 0.75, so 75 % ³⁵Cl and 25 % ³⁷Cl.
Check your answer lies between the lightest and heaviest isotope masses and nearer the more abundant one.
Must know
- Nucleus: protons + neutrons, positive, tiny and dense; electrons outside.
- Neutrons = A − Z; electrons = Z − ionic charge.
- Isotopes: same Z, different numbers of neutrons; same chemistry, different mass-dependent physical properties.
- Ar = Σ(mass × abundance) ÷ 100; use x and (1 − x) to find abundances.
That's the notes covered.
Carry on to the next subtopic.