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Atomic Structure - Subatomic Particles

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An atom consists of a dense central nucleus containing protons (p+p^+) and neutrons (n0n^0), surrounded by electrons (e−e^-) in energy levels or shells.

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The Atomic Number (ZZ) represents the number of protons in the nucleus. This number defines the identity of the element.

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The Mass Number (AA) represents the total number of protons and neutrons in the nucleus, often referred to as nucleons.

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Subatomic particle properties: Protons (Relative Mass: 11, Relative Charge: +1+1), Neutrons (Relative Mass: 11, Relative Charge: 00), Electrons (Relative Mass: 11836\frac{1}{1836} or ≈0\approx 0, Relative Charge: −1-1).

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Isotopes are atoms of the same element that have the same number of protons but a different number of neutrons (e.g., 612C{}^{12}_{6}C and 614C{}^{14}_{6}C).

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Ions are charged particles formed when an atom loses or gains electrons. A cation (positive ion) forms when an atom loses e−e^-, and an anion (negative ion) forms when an atom gains e−e^-.

📐Formulae

A=Z+nA = Z + n

n=A−Zn = A - Z

Charge=Number of Protons−Number of Electrons\text{Charge} = \text{Number of Protons} - \text{Number of Electrons}

Relative Atomic Mass(Ar)=∑(Isotope Mass×Abundance)100\text{Relative Atomic Mass} (A_r) = \frac{\sum (\text{Isotope Mass} \times \text{Abundance})}{100}

💡Examples

Problem 1:

Calculate the number of protons, neutrons, and electrons in a magnesium ion: 1224Mg2+{}^{24}_{12}Mg^{2+}.

Solution:

p+=12p^+ = 12, n0=12n^0 = 12, e−=10e^- = 10.

Explanation:

The atomic number Z=12Z = 12 indicates 1212 protons. The mass number A=24A = 24, so neutrons are A−Z=24−12=12A - Z = 24 - 12 = 12. The 2+2+ charge indicates the loss of 22 electrons from the neutral atom (12−2=1012 - 2 = 10).

Problem 2:

An unknown isotope of Chlorine has 1717 protons and 2020 neutrons. Represent this using nuclide notation ZAX{}^{A}_{Z}X.

Solution:

1737Cl{}^{37}_{17}Cl

Explanation:

The number of protons gives the atomic number Z=17Z = 17, which corresponds to the element Chlorine (ClCl). The mass number is the sum of protons and neutrons: A=17+20=37A = 17 + 20 = 37.

Problem 3:

Copper has two isotopes: 63Cu{}^{63}Cu with an abundance of 69%69\% and 65Cu{}^{65}Cu with an abundance of 31%31\%. Calculate the average relative atomic mass (ArA_r).

Solution:

Ar=63.62A_r = 63.62

Explanation:

Using the formula: Ar=(63×69)+(65×31)100=4347+2015100=63.62A_r = \frac{(63 \times 69) + (65 \times 31)}{100} = \frac{4347 + 2015}{100} = 63.62.