Mass number of a radioactive element is 232 and its atomic number is 90. When this element undergoes certain nuclear reactions, it transforms into an isotope of lead with a mass number 208 and an atomic number 82. Determine the number of alpha and beta dacay that can occur.

Explanation

Mass number A = 232
Atomic number Z = 90
`""_90X^(232) to ""_82Y^(208) +n_(alpha) ""_2He^(4) + n""_(beta-1)e^(0)`
`n_(alpha) = (A-A')/(4) = (232-208)/(4) = (24)/(4) = 6`
`n_(beta) = Z'-Z +2n_(alpha)`
=82-90+2(6)
82-90+12
`n_(beta) = 94-90`
`n_(beta) = 4`
No. of alpha decay = 6
No. of beta decay =4