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
Answer & explanation
Correct answer: option 1
Let the percentage of B10 atoms be x, then Average atomic weight = $\frac{10 x+11(100-x)}{100}$ = 10.81 ⇒ x = 19
∴ $\frac{N_{B^{10}}}{N_{B^{11}}}=\frac{19}{81}$