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.
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Answer
$\rightarrow$ 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.
$\rightarrow$ Almost every element is a mixture of many isotopes.
The relative abundance of different isotopes differs from element to element.
$\rightarrow$ 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 $24.6 \%$.
$\rightarrow$ Average mass of Chlorine
$=\frac{75.4 \times 34.98+24.6 \times 36.98}{100}$
$=35.47 u$
$\rightarrow$ 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 $3.0160 u$.
$\rightarrow$ The nucleus of the lightest atom of hydrogen has a relative abundance of $99.985 \%$, is called proton but there is no neutron.
$\rightarrow$ The other two isotopes of hydrogen are called deuterium $($mass $=2.0141 u )$ and tritium $($mass $=3.0160 u )$.
$\rightarrow$ Tritium nuclei being unstable, do not occur naturally and are produced artificially in laboratories.
$\rightarrow$ 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.
$\rightarrow$ Hydrogen $\left({ }_1 H ^1\right)$ nucleus has only one proton and do not have neutrons.
$\rightarrow$ Mass of proton $($mass of ${ }_1 H ^{ 1})$
$=\frac{1.0078 \times 99.985}{100}$
$=1.00727 u$
$=1.00727 \times 1.660539 \times 10^{-27} \ kg$
$=1.67262 \times 10^{-27} \ kg$
$\rightarrow$ This value is equal to the value obtained by subtracting the mass of an electron from the mass of a hydrogen atom.
$\rightarrow$ Almost every element is a mixture of many isotopes.
The relative abundance of different isotopes differs from element to element.
$\rightarrow$ 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 $24.6 \%$.
$\rightarrow$ Average mass of Chlorine
$=\frac{75.4 \times 34.98+24.6 \times 36.98}{100}$
$=35.47 u$
$\rightarrow$ 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 $3.0160 u$.
$\rightarrow$ The nucleus of the lightest atom of hydrogen has a relative abundance of $99.985 \%$, is called proton but there is no neutron.
$\rightarrow$ The other two isotopes of hydrogen are called deuterium $($mass $=2.0141 u )$ and tritium $($mass $=3.0160 u )$.
$\rightarrow$ Tritium nuclei being unstable, do not occur naturally and are produced artificially in laboratories.
$\rightarrow$ 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.
$\rightarrow$ Hydrogen $\left({ }_1 H ^1\right)$ nucleus has only one proton and do not have neutrons.
$\rightarrow$ Mass of proton $($mass of ${ }_1 H ^{ 1})$
$=\frac{1.0078 \times 99.985}{100}$
$=1.00727 u$
$=1.00727 \times 1.660539 \times 10^{-27} \ kg$
$=1.67262 \times 10^{-27} \ kg$
$\rightarrow$ 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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