Atomic mass number of an element is `232` and its atomic number is `90`. The end product of this radiaoctive element is an isotope of lead (atomic mass `208` and atomic number `82`.) The number of `alpha`-and `beta` -particles emitted are.

Options

A. `alpha = 3, beta = 3`
B. `alpha = 6, beta = 4`
C. `alpha = 6, beta = 0`
D. `alpha = 4, beta = 6`.

Answer

B

Explanation

(b) `n_alpha = (A - A')/(4) = (232 - 208)/(4) = 6`
and `n_beta = (2 n_alpha - Z + Z') = (2 xx 6 - 90 + 82) = 4`.