If `""_(a)X^(b)` emits firstly a positron, then two `alpha` and two `beta` and in the last `alpha` is emitted and finally it converts to `""_(d)^(c)Y`. The correct relation is

Options

A. `a=C+12, d=b-5`
B. `a=c+8, d=b-1`
C. `a=c+6 , d=b-2`
D. `a=c+4, d=b-2`

Answer

A

Explanation

`""_(b)^(a)X to ""_(d)^(c)Y+underset("positron")(""_(+1)^(0)e)+underset(alpha-"particle")(3""_(2)^(4)He) + underset(beta-"particle")(2""_(-1)^(0)e)`
a=c+12
b=d+1+6-2
b=d+5
d=b-5