Nb of protons = Z
Nb of electrons (neutral atom) = Z
Nb of neutron = A-Z
H | He | C | N | O | |
---|---|---|---|---|---|
Z | 1 | 2 | 6 | 7 | 8 |
A | 1 | 4 | 12 | 14 | 16 |
$$ \underbrace{\text{π€’ Repulsive } \qquad {\color{cyan}\text{or}} \qquad \text{π Attractive }}{\normalsize\text{𧲠Atoms bind together}} \\ \overbrace{\underbrace{{\text{π¨βπ©βπ§βπ¦ In large numbers}}}{\text{π§ Form solids}} \quad {\color{cyan}\text{or}} \quad \underbrace{\text{π§πΌβπ€βπ§πΏ In small numbers}}_{\text{π Form molecules}}} $$
Steep dependence
<aside> <img src="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/97271424-4a68-4e9f-b2eb-e0c2af3a2769/Ions.png" alt="https://s3-us-west-2.amazonaws.com/secure.notion-static.com/97271424-4a68-4e9f-b2eb-e0c2af3a2769/Ions.png" width="40px" /> Ions: Atoms are ionised if they donβt have the same amount of electrons as protons ( $Nb(e^-)\ne Z$ )
</aside>
It is a long range force, so to calculate the total binding energy we need to sum for all nearby atoms
$$ V(r)=\frac{e^2}{4\pi\epsilon_0}\sum_i \frac{1}{|r-r_i|} $$
ποΈ Example: Na$^+$Cl$^-$, Melting temp ~ 1100 K which is very high since ionic bonds are strong
ποΈ Example: hydrogen
$$ \text{if we have: }\quad {e^-}\overset{\vec d}{\rightarrow}e^+ \quad\text{ inst. dipole moment: } \vec p = e\vec d $$