The radioactive nuclide $$^{232}_{90}Th$$ decays by one $ ext{α}$ emission followed by two $ ext{β}^-$ emissions - AQA - A-Level Physics - Question 10 - 2020 - Paper 1

When the nuclide decays by one $ext{α}$ emission, it emits an alpha particle (consisting of 2 protons and 2 neutrons). This decreases the atomic number by 2 and the mass number by 4.

Thus, $^{232}_{90}Th \rightarrow ^{228}_{88}Rn + 2$

In the first $ext{β}^-$ emission, a neutron is converted into a proton, increasing the atomic number by 1 while the mass number remains the same.

Thus, $^{228}_{88}Rn \rightarrow ^{228}_{89}Th + e^-$

In the second $ext{β}^-$ emission, another neutron is converted into a proton, again increasing the atomic number by 1.

Thus, $$^{228}{89}Th \rightarrow ^{228}{90}U + e^-$.

The resulting nuclide after one α emission followed by two β emissions is $^{228}_{90}Th$, which is option C.