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.
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
THANK YOU SO MUCH
To calculate the number of alpha and beta decays in the given series, we need to understand the decay modes of Ra-227 and Pb-203.
Ra-227 undergoes a series of alpha decays until it reaches Pb-203. Each alpha decay involves the emission of an alpha particle, which consists of two protons and two neutrons (equivalent to a helium nucleus). Therefore, each alpha decay reduces the atomic number of the nuclide by 2 and the mass number by 4.
Since Ra-227 becomes Pb-203, there are a total of (227 - 203) / 4 = 6 alpha decays.
Now, let's determine the number of beta decays. Beta decay occurs when a nucleus undergoes the transformation of a neutron into a proton, or a proton into a neutron, along with the emission of an electron (beta particle) or a positron (positron emission).
However, from the given information, we only know that Ra-227 becomes Pb-203 without any specific details about the number of beta decays occurring in between.
To find the number of beta decays, we would need additional information about the decay modes of the nuclides involved in the series. Different nuclides undergo different types and numbers of beta decays during their decay series.