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# 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 $m$ $=$ $35\cdot1.66\cdot10^{-27}$ $=$ $5.8\cdot10^{-26}\;kg$