# Alpha decay

Discussion in 'DAT Discussions' started by Kami, Mar 7, 2007.

1. ### Kami Member 10+ Year Member

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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

2. ### fancymylotus A Whole New World Dentist 10+ Year Member

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Huh. Each alpha decay knocks 2 off the atomic number and four off the mass number. Each beta decay(I'm assuming this is NOT the positron one), adds 1 to the Z. So 92-10+2=84.

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3. ### ddhm 2+ Year Member

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In alpha decay, the parent nucleus ejects a Helium atom (Z=2, A=4). For every alpha decay, the parental Z is reduced by 2, so for five consecutive alpha decays, the daughter Z is 92-5(2)=82.

In beta decay, a parental neutron decays into a proton while ejecting an electron. The newly evolved proton will increase the parental Z by 1, so two consecutive beta decays will result in a daughter Z of 82+2(1)=84.

4. OP

### Kami Member 10+ Year Member

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Its not a positron emission. I think your explanation is right. Thanks.

Kami

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