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Atomic Structure - The nuclear atom

Grade 11IBChemistry

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

🔑Concepts

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The atom consists of a small, dense, positively charged nucleus containing protons and neutrons, surrounded by negatively charged electrons in shells.

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Protons and neutrons are known as nucleons. The relative mass of a proton and neutron is 11, while the relative mass of an electron is approximately 11836\frac{1}{1836} or 5×10−45 \times 10^{-4}.

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The atomic number (ZZ) represents the number of protons in the nucleus and defines the element's identity. In a neutral atom, the number of protons equals the number of electrons.

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The mass number (AA) is the total number of protons and neutrons in the nucleus.

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Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have the same chemical properties (due to the same electron configuration) but different physical properties such as density and boiling point.

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Specific isotopes (radioisotopes) have medical and industrial uses, such as 14C^{14}C for radiocarbon dating and 60Co^{60}Co for radiotherapy.

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The relative atomic mass (ArA_r) is the weighted average mass of an atom of an element compared to 112\frac{1}{12} of the mass of an atom of 12C^{12}C.

📐Formulae

A=Z+NA = Z + N

Ar=∑(isotopic mass×relative abundance)100A_r = \frac{\sum (\text{isotopic mass} \times \text{relative abundance})}{100}

Number of neutrons (N)=A−Z\text{Number of neutrons } (N) = A - Z

Charge of ion=protons−electrons\text{Charge of ion} = \text{protons} - \text{electrons}

💡Examples

Problem 1:

Calculate the number of protons, neutrons, and electrons in a gallium ion, 3171Ga3+^{71}_{31}Ga^{3+}.

Solution:

Protons = 3131, Neutrons = 4040, Electrons = 2828.

Explanation:

The atomic number Z=31Z = 31 gives the number of protons. The mass number A=71A = 71. Neutrons are calculated as A−Z=71−31=40A - Z = 71 - 31 = 40. Since the ion has a 3+3+ charge, it has lost 33 electrons from its neutral state: 31−3=2831 - 3 = 28 electrons.

Problem 2:

Chlorine has two naturally occurring isotopes: 35Cl^{35}Cl (relative abundance 75.78%75.78\%) and 37Cl^{37}Cl (relative abundance 24.22%24.22\%). Calculate the relative atomic mass (ArA_r) of chlorine to two decimal places.

Solution:

Ar=(35×75.78)+(37×24.22)100=35.48A_r = \frac{(35 \times 75.78) + (37 \times 24.22)}{100} = 35.48

Explanation:

The relative atomic mass is the weighted average of the isotopic masses. Multiply each isotopic mass by its percentage abundance, sum them up, and divide by 100100.

The nuclear atom - Revision Notes & Key Formulas | IB Grade 11 Chemistry