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. `15 : 16`
B. `19 : 81`
C. `81 : 19`
D. `20 : 53`

Answer

B

Explanation

`N_(1)/N_(2)=ratio`
Average weight`=(N_(1)W_(1)+N_(2)W_(2))/(N_(1)+N_(2))`
`10.81=(10N_(1)+11N_(2))/(N_(1)+N_(2))`
`10.81N_(1)=10.81N_(2)=10N_(1)+11N_(2)`
`0.81N_(1)=0.19N_(2)implies(N_(1))/(N_(2))=(19)/(81)`