Atomic weight of ` Ne` is ` 20.2 Ne` is a mixuture of ` Ne^(20)` and ` Ne^(22)`. Relative abunduance fo heavior isotope is :

Options

A. `1:9`
B. `9:1`
C. `70%`
D. `80%`

Answer

B

Explanation

Let `W%` of `Ne^(20) = x`
`W%` of `Ne^(22) = 100 - x`
`20.2 = (x xx 20 +(100 -x)22)/(100)`
`2020 = 20x +2200 - 22x`
`2x = 2200 - 2020`
`2x = 180`
`2x = 180`
`x = 90%`
`w%` of `Ne^(20) = 90%`
`w%` of `Ne^(22) = 100 - 90 = 10%`
`Ne^(20): Ne^(20) = 90: 10 = 9:1`