The end product of decay of `""_(90)Th^(232)` is `""_(82)Pb^(208)` . The number of `alpha` and `beta` particles emitted are respectively

Options

A. 6, 0
B. 3, 3
C. 4, 6
D. 6, 4

Answer

D

Explanation

(d): `""_(90)Th^(232) to ""_(82)Pb^(208)`
Change in atomic mass = 232 - 208 = 24
`alpha`-particle emitted = 24/4 = 6
Atomic number will decrease by 6 `xx` 2 = 12
Atomic number will be increase by 4 due to 4 `beta` particle
So, net decrease in atomic number = 12 - 4 = 8