Question
Atomic weight of boron is 10.81 and it has two isotopes $$_5^{10}B$$ and $$_5^{11}B.$$ Then, the ratio of atoms of $$_5^{10}B$$ and $$_5^{11}B$$ in nature would be
A.
$$19:81$$
B.
$$10:11$$
C.
$$15:16$$
D.
$$81:19$$
Answer :
$$19:81$$
Solution :
Let $${n_1}$$ and $${n_2}$$ be the number of atoms in $$_5^{10}B$$ and $$_5^{11}B$$ isotopes.
Atomic weight $$ = \frac{{{n_1} \times \left( {{\text{At}}{\text{.}}\,{\text{wt}}{\text{.}}\,{\text{of }}_5^{10}\;B} \right) + {n_2} \times \left( {{\text{At}}{\text{.}}\,{\text{wt}}{\text{.}}\,{\text{of }}_5^{11}\;B} \right)}}{{{n_1} + {n_2}}}$$
$$\eqalign{
& {\text{or}}\,\,10.81 = \frac{{{n_1} \times 10 + {n_2} \times 11}}{{{n_1} + {n_2}}} \cr
& {\text{or}}\,\,10.81\,{n_1} + 10.81\,{n_2} = 10\,{n_1} + 11\,{n_2} \cr
& {\text{or}}\,\,0.81\,{n_1} = 0.19\,{n_2} \cr
& {\text{or}}\,\,\frac{{{n_1}}}{{{n_2}}} = \frac{{0.19}}{{0.81}} = \frac{{19}}{{81}} \cr} $$
NOTE
Atomic weight of an atom having two or more isotopes is the average of the total weight of two of more isotopes