In a series of decays (hypothetical), the nuclide Ra-227 beomes Pb-203. How many alphas and how many betas are emitted in this series? Give two answers: first, give the number of alpha decays; then, the number of beta decays.

Question

I know that there are 6 alpha decays happening. I just do not get the beta decay part.

Answer 1

88Ra227 ==> 2He4 + ?X?
227-203 = 24; therefore, the number of 2He4 must be 6 since 6x4 = 24. That lets us assign ?values to X. The number of protons must be 88-12 = 76 which is Os so the full equation with alpha particles is
88Ra227 --> 6*2He4 + 76Os203. Now to assign the beta particles.
We want to go from element 76 to element 82; therefore, we need 82-76 = 6. Then
76Os203 ==> 6*-1e0 + 82Pb203

Answer 2

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