The natural boron of atomic weight 10.81 is found to have two isotopes $\mathrm{B}^{10}$ and $\mathr

Question

The natural boron of atomic weight 10.81 is found to have two isotopes $\mathrm{B}^{10}$ and $\mathrm{B}^{11}$. The ratio of abundance of isotopes in natural boron should be\

Professor PC Thomas Classes · Answer

Let abundance of $\mathrm{B}^{10}$ bex%\ $	herefore$ Abundance of $\mathrm{B}^{11}=(100-\mathrm{x}) \%$\ $	herefore 10.81=\frac{(10 	imes \mathrm{x})+11(100-\mathrm{x})}{100}$\ or $1081=10 \mathrm{x}+1100-11 \mathrm{x}$ or $\mathrm{x}=19$\ $	herefore$ Ratio of abundance $=\frac{19}{100-19}=\frac{19}{81}$\