# Nucleus and electronic cloud of atoms and ions

## Atoms (neutral!)

Number of particles:
- Number of protons = Z
- Number of protons + number of neutrons = A
- Number of electrons = number of protons = Z

Charges:
The proton charge is:
Charge($p^+$) = 1 elementary positive charge =
$1\;e$ $=$ $1.6\cdot 10^{-19}\;C$
The charge of the electron is:
Charge($e^-$) = 1 elementary negative charge =
$-1\;e$ $=$ $-1.6\cdot 10^{-19}\;C$
The charge of the neutron is zero.
In a neutral atom, the number of protons equal to the number of electrons. So its overall charge is zero.

Masses:
Proton and neutron have approximately the same mass:
$m(p^+)$ $=$ $m(n^o)$ =
1 atomic mass unit = 1 u =
1 Dalton = 1 Da =
$\frac{1}{6\cdot 10^{23}}$ $ =$ $1.66\cdot 10^{-27}\;kg$
The electron mass is much smaller:
$m(e^-)$ $=$
$\frac{1}{1830}\;u$ $=$ $9.11\cdot10^{-31}\;kg$
As an atom contains protons and neutrons and as the mass of electrons is negligible, we have:
- Mass of the atom (in u) = A

Example 1:

The chlorine atom $_{17}^{35}Cl $ comprises 17 $p^+$, 17 $e^-$ and $35$ $-$ $17$ $=$ $18\;n^o$:
It has a mass of $m$ $=$ $35\cdot1.66\cdot10^{-27}$ $=$
$5.8\cdot10^{-26}\;kg$

Example 2:

The hydrogen atom $_{1}^{1}H $ comprises 1 $p^+$, 1 $e^-$ and $1$ $-$ $2$ $=$ $0$ $n^o$:
By definition of the Avogadro number $N$, it has a mass of $\frac{1}{N}\; g$, so
$m(p^+)$ $ =$ $ \frac{1}{N}\; g$
Comparing with the value given above, we find:
$m(p^+)$ $=$ $\frac{1}{N}\;g$ $=$
$1\cdot1.66\cdot10^{-24 }\;g$ ,
and the value of Avogadro's number is found again:
$N$ $=$ $6\cdot10^{23}$

## Ions

Particle numbers and charge:
- Number of protons = Z
- Number of protons + number of neutrons = A
The number of protons inside an ion differs from the number of electrons.
If Q is the ion charge, then:
- Number of electrons = Z - Q

Masses:
As ions comprise a sum A of protons and neutrons and as the mass of electrons is negligible, we have:
- Mass of ion (in u) = A

Example 1:

The chloride ion $_{17}^{35}Cl^-$ comprises
$17\;p^+$,
$17$ $-$ $(-1)$ $=$ $18\;e^-$
and
$35$ $-$ $17$ $=$ $18\;n^o$:
It has a mass of $m$ $=$ $35\cdot1.66\cdot10^{-27}$ $=$
$5.8\cdot10^{-26}\;kg$