Radioactive decay problem Hey could someone please explain why the answer is C. I thought if the answer is C, the atomic number would be 44, not 50. I'm not sure I'm approaching this correctly. Thanks for any advice. 
 A: The decay processes in question are

*

*Alpha decay (emission of a $^4_2\text{He}$) decreases the atomic mass number $A$ by 4 and decreases the atomic number $Z$ by 2.

*Beta+ decay (emission of a positron or absorption of an electron) leaves the atomic mass number unchanged but decreases the atomic number by 1. A proton in the nucleus changes to a neutron.

*Beta- decay (emission of an electron) similarly leaves the atomic mass number unchanged but increases the atomic number by 1. A neutron in the nucleus changes to a proton.

In the question, the atomic mass number decreases by four over a series of decays while the atomic number increases by one. Since the only choices involve alpha decay and beta decay, there must be exactly one alpha decay so as to decrease the atomic mass number by four. This eliminates options (B) and (D). Option (A) would leave the atomic number unchanged as the two protons lost by alpha decay are offset by the two protons gained by two $\beta+$ decays. Since the atomic number increases by one, one can rule out option (A) as well.
That leaves (C) as the only viable solution amongst the four choices. Does it fit? There is one alpha decay in that option, so the atomic mass has decreased by 4 (check). There are three $\beta-$ decays, so the atomic number has decreased by 2 due to the alpha decay but has increased by 3 due to the three $\beta-$ decays, a net increase of one. This is indeed a match.
