A nucleus with Z =92 emits the following in a sequence:
`alpha,beta^(-),beta^(-),alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(+),alpha`. The Z of the resulting nucleus is

Options

A. `76`
B. `78`
C. `82`
D. `74`

Answer

B

Explanation

`8 alpha` have been emitted
`4 beta^(-)` have been emitted
`2 beta^(+)` have emitted
`alpha` reduces atomic number by 2
`beta^(-)` increases atomic number by 1
`beta^(+)` increases by 1
So, `Z_(eff) = 92 - (8 xx 2) + (4 xx 1) - (2 xx 1) = 96 - 18 = 78`