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If an element X has atomic number of 92 and it undergoes five alpha decays and two beta decays then what will be the atomic number?
The answer is 84. I was thinking that in Kaplan blue book it says alpha decay, the atomic number of parent is Z-2 which is equal to the daughter atomic number so would it not increase more than 92 since 92=Z-2 so Z daughter would be 92+2=94 for one alpha decay and 94+2=96 for next alpha decay and so it will be increasing for then next alpha decays. In beta decay it would be Z-1 for positron and Z+1 for beta negative. So for beta it would be Z-1 parent=Z daughter.
To summarize, I am not sure how the atomic number comes out to be 84 when it should be higher than 92. Can anyone explain?
Kami
The answer is 84. I was thinking that in Kaplan blue book it says alpha decay, the atomic number of parent is Z-2 which is equal to the daughter atomic number so would it not increase more than 92 since 92=Z-2 so Z daughter would be 92+2=94 for one alpha decay and 94+2=96 for next alpha decay and so it will be increasing for then next alpha decays. In beta decay it would be Z-1 for positron and Z+1 for beta negative. So for beta it would be Z-1 parent=Z daughter.
To summarize, I am not sure how the atomic number comes out to be 84 when it should be higher than 92. Can anyone explain?
Kami