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Decay Help on Thorium

Submitted by SmAs on Sun, 09/13/2009 - 17:30

Well the question is...

23290Th emits 6 Alpha particles and 4 Beta particles during natural decay. State what the new atomic number and mass number of the final product will be..

I tried workin it out and got to

20874W but i ain't sure if thats right!

Help please! :)


Each alpha particle causes the mass number to decrease by 4 and the atomic number by 2
So after 6 x alpha emissions the particle will become 20878X

Each beta emission causes the mass number to stay the same but the atomic number increases by 1
So 4 beta emissions will cause 20878X to become 20882Y.

So look up the element with atomic number 82 and this will be Y


Ah i see it now, thanks!

I was taking the extra away from the atomic number