Atomic weight of boron is 10.81 and it has two isotopes ${ }_5 B^{10}$ and ${ }_5 B^{11}$. Then ratio of ${ }_5 B^{10}: {}_5 B^{11}$ in nature would be

19 : 81 10 : 11 15 : 16 81 : 19

19 : 81

Let the percentage of B10 atoms be x, then Average atomic weight = $\frac{10 x+11(100-x)}{100}$ = 10.81 ⇒ x = 19  ∴ $\frac{N_{B^{10}}}{N_{B^{11}}}=\frac{19}{81}$