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An atomic nucleus .90 Th^232 emits sever...

An atomic nucleus `._90 Th^232` emits several `alpha` and `beta` radiations and finally reduces to `._82 Pb^208`. It must have emitted.

A

`4 alpha` and `2 beta`

B

`6 alpha` and `4 beta`

C

`8 alpha` and `24 beta`

D

`4 alpha` and `16 beta`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `n_alpha = (A - A')/(4) = (232 - 208)/(4) = 6`
`n_beta = 2n_alpha - Z + Z' = 2 xx 6 - 90 + 82 = 4`.
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