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

A nucleus with Z =92 emits the following in a sequence alpha, beta, al

1.	A nucleus with Z =92 emits the following in a sequence alpha, beta, alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(+),alpha. The Z of the resulting nucleus is :
Answer» 78 82 74 76 Solution :Here, `8 ALPHA, 4 beta^(-) and 2 beta^(+)` RADIATION take PLACE, so change in atomic no. is `=2 xx 8 -4 xx 1+2 xx 1=14` New atomic no. =92-14=78

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

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