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# The sum of number of neutrons and protons in all the isotopes of hydrogen is:A. $2$ B. $4$ C. $5$ D. $6$

Last updated date: 12th Sep 2024
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Hint:
We know that the three natural isotopes of hydrogen are $_1^1{\rm{H}}$, $_1^2{\rm{H}}$ and$_1^3{\rm{H}}$ that differ in mass number.

Complete step by step solution
We can define isotopes of a given element as variants that have equal number of protons $\left( {{{\rm{n}}_{\rm{p}}}} \right)$ but different number of neutrons $\left( {{{\rm{n}}_{\rm{n}}}} \right)$. This gives rise to the same atomic number $\left( {\rm{Z}} \right)$ for all the isotopes as we can define atomic number being equal to the number of protons present in a given atom. The expression can be shown as below:
${\rm{Z}} = {{\rm{n}}_{\rm{p}}}$

On the other hand, the mass number $\left( {\rm{A}} \right)$ is different for all the isotopes as we can define the mass number being equal to the sum of the number of protons and neutrons present in a given atom. The expression can be shown as below:
${\rm{A}} = {{\rm{n}}_{\rm{p}}} + {{\rm{n}}_{\rm{n}}}$

We can take carbon for example, it has three natural isotopes that can be represented as: $_6^{12}{\rm{C, }} \, _6^{13}{\rm{C and }} \, _6^{14}{\rm{C}}$
For all of the three isotopes, the atomic number is equal $\left( {{\rm{Z}} = {\rm{6}}} \right)$ as the number of protons is equal for all three $\left( {{{\rm{n}}_{\rm{p}}} = 6} \right)$ but $_6^{12}{\rm{C}}$ isotope has $\left( {{\rm{A}} = {\rm{12}}} \right)$, $_6^{13}{\rm{C}}$ has $\left( {{\rm{A}} = {\rm{13}}} \right)$ and $_6^{14}{\rm{C}}$ isotope has $\left( {{\rm{A}} = {\rm{14}}} \right)$ as the number of neutrons are different; $\left( {{{\rm{n}}_{\rm{n}}} = 6} \right)$ for $_6^{12}{\rm{C}}$, $\left( {{{\rm{n}}_{\rm{n}}} = 7} \right)$ for $_6^{13}{\rm{C}}$ and $\left( {{{\rm{n}}_{\rm{n}}} = 8} \right)$ for $_6^{14}{\rm{C}}$.

Now, we will have a look at the three natural isotopes of ${\rm{H}}$ that can be represented as $_1^1{\rm{H}}$, $_1^2{\rm{H}}$ and $_1^3{\rm{H}}$. As we can see that that all three of them have $\left( {{\rm{Z}} = 1} \right)$ with $\left( {{{\rm{n}}_{\rm{p}}} = 1} \right)$ but the number of neutron(s) is different for each being $0,1 \,{\rm{ and 2}}$ giving $\left( {{\rm{A}} = {\rm{1,2 \, and 3}}} \right)$ respectively.
So, if we add the mass number of each isotope we will get the sum of $\left( {{{\rm{n}}_{\rm{n}}}} \right)$and $\left( {{{\rm{n}}_{\rm{p}}}} \right)$ in all the isotopes as follows:
${\rm{sum = }}{\left( {{{\rm{n}}_{\rm{n}}} + {{\rm{n}}_{\rm{p}}}} \right)_{_1^1{\rm{H}}}} + {\left( {{{\rm{n}}_{\rm{n}}} + {{\rm{n}}_{\rm{p}}}} \right)_{_1^2{\rm{H}}}} + {\left( {{{\rm{n}}_{\rm{n}}} + {{\rm{n}}_{\rm{p}}}} \right)_{_1^3{\rm{H}}}}\\ = {{\rm{A}}_{_1^1{\rm{H}}}} + {{\rm{A}}_{_1^2{\rm{H}}}} + {{\rm{A}}_{_1^3{\rm{H}}}}\\ = 1 + 2 + 3\\ = 6$

Hence, the correct option is D.

Note:

We have to be careful while taking a number of neutrons for each as $_1^1{\rm{H}}$ has no neutron at all.