Question

An element `""_ZS^A` decays to `""_85R^222` after emitting 2 α particles and 1 β particle.

Find the atomic number and atomic mass of the element S.

Answer 1

\[ {}_{Z}^{A}S \rightarrow {}_{85}^{222}R + 2({}_{2}^{4}He) + {}_{-1}^{\;\;0}e \]

Given: Initial element = \({}_{Z}S^{A}\)

Final element = \({}_{85}R^{222}\)

Number of \(\alpha\)-particles emitted = 2

Number of \(\beta\)-particles emitted = 1

The atomic number (Z).

Each \(\alpha\)-particle emission decreases the atomic number by 2, while each \(\beta^{-}\)-particle emission increases it by 1.

\[ Z-2(2)+1=85 \]

\[ Z-4+1=85 \]

\[ Z=85+3\]

= 88

The atomic mass (A).

Each \(\alpha\)-particle emission decreases the mass number by 4, while \(\beta\)-particle emission does not change it.

\[ A-2(4)=222 \]

\[ A-8=222 \]

\[ A=222+8\]

= 230