Question

An element \[{ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}}\] decays to \({ }_{88} \mathrm{Y}^{222}\) after emitting two alpha particles and one beta particle. What are the values of atomic mass A and atomic number Z in element X?

Answer 1

Each alpha particle decreases the mass number by 4 and the atomic number by 2. Each beta particle increases the atomic number by 1 without changing the mass number.

Hence, from X to Y: A − 8 = 222 and Z − 3 = 88.

Therefore, A = 230 and Z = 91

\[ {}_{91}\mathrm{Pa}^{230} \longrightarrow {}_{88}\mathrm{Y}^{222} + 2 \ {}_{2}\mathrm{He}^{4} + {}_{-1}\mathrm{e}^{0} \]