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The _{90}Th^{232} atom has successive \alpha and \beta decays to the end product _{82}\mathrm{Pb}^{208}. The number of \alpha and \beta particles emitted in the process respectively are:

Option: 1

4,6


Option: 2

4,4


Option: 3

6,2


Option: 4

6,4


Answers (1)

best_answer

If n number of \alpha particle is emitted then mass number decreases by 4 n.

\Rightarrow 4 n=232-208=24

\Rightarrow n=6

\Rightarrow 6 \alpha \text { particle are emitted. }

each \alpha particle reduced b atomic number by 2 and \beta particle increases atomic number by 1 .

If number of \beta particle emitted is m then new atomic no.

=90-6 \times 2+m

\Rightarrow 78+m=82 \Rightarrow m=4 .

\text { number of } \beta \text { particle }=4 \text {. }

Posted by

Gautam harsolia

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