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` in nature would be.

Options

A. `12:81`
B. `10:11`
C. `12:15`
D. `81:19`

Explanation

Let `n_(1)` and `n_(2)` be the number of atoms is `10/5B` and `11/5` B isotopes. Atomic eight
`=(n_(1)xx(At "of" ""_(5)^(10)B)+n_(2)xx(At. "of" ""_(5)^(11)B))/(n_(1)+n_(2))`
or `10.81=(n_(1)xx10+n_(2)xx11)/(n_(1)+n_(2))`
or `10.81n_(1)+10.81n_(2)=10n_(1)+11n_(2)`
or `0.81n_(1)=0.19n_(2)`
or `(n_(1))/(n_(2))=0.19/0.81=19/81`