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. 82
B. 74
C. 76
D. 78

Answer

D

Explanation

Total number of `a,lpha` particles emitted = 8
`therefore` Z decreases by ` 8 xx 2 = 16`
Similarly `4beta^-` particles are emitted. `therefore` Z will increase by 4 and the emission of `2beta^(+)` (positrons) particles will decrease Z by 2.
`therefore` the Z of the resulting nucleus is 92- 16 +4-2 = 78