Question

(a) \({ }_{92} \mathrm{X}^{238} \xrightarrow{-\beta} \mathrm{X}_{1}^{-} \xrightarrow{-\alpha} \mathrm{X}_{2}^{-} \xrightarrow{-\alpha} \mathrm{X}_{3}^{-} \xrightarrow{-\beta} \mathrm{X}_{4}^{-} \xrightarrow{-\alpha} \mathrm{X}_{5}^{-} \xrightarrow{-\alpha} \mathrm{X}_{6}^{-}\)

Study the decay of \({ }_{92} \mathrm{X}^{238}\) carefully and complete the chain of mutations by writing the mass number and atomic number from X₁ to X₆.

(b) From the above transformations, identify.

1. Isotopes
2. Isobars

Answer 1

Use the laws of radioactive decay:

β decay: \[ {}_{Z}\mathrm{X}^{A} \longrightarrow {}_{Z + 1}\mathrm{Y}^{A} + {}_{-1}\mathrm{e}^{0} \]

Mass number A is unchanged; atomic number Z increases by 1.

α decay: \[ {}_{Z}\mathrm{X}^{A} \longrightarrow {}_{Z - 2}\mathrm{Y}^{A - 4} + {}_{2}\mathrm{He}^{4} \]

Mass number A decreases by 4; atomic number Z decreases by 2.

(a) Applying these step by step from \({}_{92}\mathrm{X}^{238}\):

\[ \begin{array}{|c|c|c|} \hline \textbf{Nuclide} & \textbf{Mass number (A)} & \textbf{Atomic number (Z)} \\ \hline {}_{92}\mathrm{X}^{238} & 238 & 92 \\ \hline \mathrm{X}_{1} & 238 & 93 \\ \hline \mathrm{X}_{2} & 234 & 91 \\ \hline \mathrm{X}_{3} & 230 & 89 \\ \hline \mathrm{X}_{4} & 230 & 90 \\ \hline \mathrm{X}_{5} & 226 & 88 \\ \hline \mathrm{X}_{6} & 222 & 86 \\ \hline \end{array} \]

\[ {}_{92}\mathrm{X}^{238} \xrightarrow{- \beta} {}_{93}\mathrm{X}_{1}^{238} \xrightarrow{- \alpha} {}_{91}\mathrm{X}_{2}^{234} \xrightarrow{- \alpha} {}_{89}\mathrm{X}_{3}^{230} \xrightarrow{- \beta} {}_{90}\mathrm{X}_{4}^{230} \xrightarrow{- \alpha} {}_{88}\mathrm{X}_{5}^{226} \xrightarrow{- \alpha} {}_{86}\mathrm{X}_{6}^{222} \]

(b)

1. Isotopes: None, because no two nuclides in this chain have the same atomic number but different mass numbers.

2. Isobars:

\[ {}_{92}\mathrm{X}^{238} \text{ and } {}_{93}\mathrm{X}_{1}^{238} \ (\text{both } A = 238) \]

\[ {}_{89}\mathrm{X}_{3}^{230} \text{ and } {}_{90}\mathrm{X}_{4}^{230} \ (\text{both } A = 230) \]