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When .90 Th^228 transforms to .83 Bi^212...

When `._90 Th^228` transforms to `._83 Bi^212`, then the number of the emitted `alpha-`and `beta-`particle is, respectively.

A

`8 alpha, 7 beta`

B

`4 alpha, 7 beta`

C

`4 alpha, 4 beta`

D

`4 alpha, 1 beta`

Text Solution

Verified by Experts

The correct Answer is:
D

(d) `n_alpha = (228 - 212)/(4) = 4` and `n_(beta) = 2 xx 4 - 90 + 83 = 1`.
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