Question

A radioactive nucleus \(^A_ZX\) undergoes successive disintegrations, emitting x α–particles followed by (positive) β –particles. The mass number (A₁), and atomic number (z₁), of the final daughter nucleus, are

(1) A₁ = (A–2x); z₁ = (z–2x+y)

(2) A₁ = (A–4x); z₁ = (z–2x–y)

(3) A₁ = (A–4x); z₁ = (z–2x+y)

(4) A₁ = (A–2x); z₁ = (z–x+y)

Answer 1

(2) A₁ = (A–4x); z₁ = (z–2x–y)

The emission of an α–particle decreases mass number by 4 and atomic number by 2. Therefore; after emission of x, α–particles

The emission of a positive β–particle does not change the mass number but the atomic number decreases by one. Therefore, after emission of y positive β–particles, we have