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Structure of the Atom - Isotopes and Isobars

Grade 9CBSE

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

🔑Concepts

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Isotopes are defined as the atoms of the same element, having the same atomic number (ZZ) but different mass numbers (AA).

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Chemical properties of isotopes are identical because they have the same number of protons and electrons, which determines their chemical behavior.

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Physical properties of isotopes (like boiling point and density) differ due to the difference in their masses.

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Fractional atomic masses of elements, such as Chlorine (35.5 u35.5 \text{ u}), are due to the existence of isotopes in specific natural abundance.

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Isobars are atoms of different elements which have different atomic numbers (ZZ) but the same mass number (AA).

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Isobars have different chemical properties because they belong to different elements and have different numbers of protons and electrons.

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Application of Isotopes: An isotope of Uranium (92235U^{235}_{92}U) is used as fuel in nuclear reactors, an isotope of Cobalt (2760Co^{60}_{27}Co) is used in the treatment of cancer, and an isotope of Iodine (53131I^{131}_{53}I) is used in the treatment of goitre.

📐Formulae

Mass Number (A)=Number of Protons (p)+Number of Neutrons (n)\text{Mass Number } (A) = \text{Number of Protons } (p) + \text{Number of Neutrons } (n)

Atomic Number (Z)=Number of Protons (p)\text{Atomic Number } (Z) = \text{Number of Protons } (p)

Average Atomic Mass=∑i=1n(Mass of Isotopei×Percentage Abundancei100)\text{Average Atomic Mass} = \sum_{i=1}^{n} \left( \text{Mass of Isotope}_i \times \frac{\text{Percentage Abundance}_i}{100} \right)

💡Examples

Problem 1:

Chlorine occurs in nature in two isotopic forms with masses 35 u35 \text{ u} and 37 u37 \text{ u} in the ratio of 3:13:1. Calculate the average atomic mass of the Chlorine atom.

Solution:

Average Atomic Mass=(35×75100)+(37×25100)\text{Average Atomic Mass} = \left( 35 \times \frac{75}{100} \right) + \left( 37 \times \frac{25}{100} \right) Average Atomic Mass=(35×34)+(37×14)\text{Average Atomic Mass} = \left( 35 \times \frac{3}{4} \right) + \left( 37 \times \frac{1}{4} \right) Average Atomic Mass=105+374=1424=35.5 u\text{Average Atomic Mass} = \frac{105 + 37}{4} = \frac{142}{4} = 35.5 \text{ u}

Explanation:

To find the average atomic mass, we take the weighted average of the masses of all naturally occurring isotopes based on their percentage abundance.

Problem 2:

Identify the relationship between Calcium (2040Ca^{40}_{20}Ca) and Argon (1840Ar^{40}_{18}Ar).

Solution:

The atoms 2040Ca^{40}_{20}Ca and 1840Ar^{40}_{18}Ar have the same mass number (A=40A = 40) but different atomic numbers (ZCa=20,ZAr=18Z_{Ca} = 20, Z_{Ar} = 18). Therefore, they are Isobars.

Explanation:

Since the total number of nucleons (protons + neutrons) is the same (4040) despite the different number of protons, these different elements are classified as isobars.