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.
I know that there are 6 alpha decays happening. I just do not get the beta decay part.
To determine the number of beta decays in this series, we need to understand the process of beta decay and the changes that occur during each decay step.
In beta decay, a neutron in the nucleus is transformed into a proton, and an electron (referred to as a beta particle) and an electron antineutrino are emitted. This process can be represented as follows:
n → p + e- + ν̄e
Now, let's analyze the decays in the given series:
Ra-227 → Rn-223 + α
Rn-223 → Po-219 + α
Po-219 → Pb-215 + α
Pb-215 → Bi-211 + β
Bi-211 → Tl-207 + α
Tl-207 → Pb-203 + α
Based on these decay steps, we observe that β decay occurs only in the fourth decay, where Pb-215 transforms into Bi-211. This means that there is one beta decay in this series.
Therefore, in the given series, the number of alpha decays is 6, and the number of beta decays is 1.