Question
What are isotopes? Explain with an example how the average mass can be obtained from the relative proportions of different isotopes of the same element.
✓
Answer
→The atoms for which atomic number Z is same but atomic mass number A is different then such type of atoms are called the isotopes of each other.
→Almost every element is a mixture of many isotopes. The relative abundance of different isotopes differs from element to element.
→For example :
(i) Chlorine has two isotopes having masses $34.98 u$ and $36.98 u$. The relative abundances of these isotopes are $75.4 \%$ and $24.6 \%$.
→Average mass of Chlorine
$\begin{array}{l}
=\frac{75.4 \times 34.98+24.6 \times 36.98}{100} \\
=35.47 u
\end{array}$
→This mass is almost equal to the atomic mass of chlorine.
(ii) Hydrogen also has three isotopes. Their masses are $1.0078 u, 2.0141 u$ and $3.0160 u$.
→The nucleus of the lightest atom of hydrogen has a relative abundance of $99.985 \%$, is called proton but there is no neutron.
→The other two isotopes of hydrogen are called deuterium (mass $=2.0141 u$ ) and tritium (mass $=3.0160 u$ ).
→Tritium nuclei being unstable, do not occur naturally and are produced artificially in laboratories.
→The relative abundance of deuterium is so small (relative abundance of hydrogen is $99.985 \%$ ) that the masses of deuterium and tritium are neglected when calculating the average mass of hydrogen.
→Hydrogen $\left({ }_1 H ^1\right)$ nucleus has only one proton and do not have neutrons.
→Mass of proton (mass of ${ }_1 H ^{ 1}$ )
$\begin{array}{l}
=\frac{1.0078 \times 99.985}{100} \\
=1.00727 u \\
=1.00727 \times 1.660539 \times 10^{-27} kg \\
=1.67262 \times 10^{-27} kg
\end{array}$
→This value is equal to the value obtained by subtracting the mass of an electron from the mass of a hydrogen atom.
→Almost every element is a mixture of many isotopes. The relative abundance of different isotopes differs from element to element.
→For example :
(i) Chlorine has two isotopes having masses $34.98 u$ and $36.98 u$. The relative abundances of these isotopes are $75.4 \%$ and $24.6 \%$.
→Average mass of Chlorine
$\begin{array}{l}
=\frac{75.4 \times 34.98+24.6 \times 36.98}{100} \\
=35.47 u
\end{array}$
→This mass is almost equal to the atomic mass of chlorine.
(ii) Hydrogen also has three isotopes. Their masses are $1.0078 u, 2.0141 u$ and $3.0160 u$.
→The nucleus of the lightest atom of hydrogen has a relative abundance of $99.985 \%$, is called proton but there is no neutron.
→The other two isotopes of hydrogen are called deuterium (mass $=2.0141 u$ ) and tritium (mass $=3.0160 u$ ).
→Tritium nuclei being unstable, do not occur naturally and are produced artificially in laboratories.
→The relative abundance of deuterium is so small (relative abundance of hydrogen is $99.985 \%$ ) that the masses of deuterium and tritium are neglected when calculating the average mass of hydrogen.
→Hydrogen $\left({ }_1 H ^1\right)$ nucleus has only one proton and do not have neutrons.
→Mass of proton (mass of ${ }_1 H ^{ 1}$ )
$\begin{array}{l}
=\frac{1.0078 \times 99.985}{100} \\
=1.00727 u \\
=1.00727 \times 1.660539 \times 10^{-27} kg \\
=1.67262 \times 10^{-27} kg
\end{array}$
→This value is equal to the value obtained by subtracting the mass of an electron from the mass of a hydrogen atom.
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