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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
`alpha`-particle = `""_(2)He^(4)`
`beta`-particle = `""_(-1)e^(0)` and nucleus `= ""_(Z)X^(A)`
Change in A occurs only due to `alpha`-emission.
Change in A = 228 - 212 = 16
This change is due to `4 alpha`
Again change in Z = 90 - 83 = 7
Change in Z due to `4 alpha` = 8
`therefore` Change in Z due to `beta = 8 - 7 = 1`
This is due to one `beta`
Hence particle emitted `= 4 alpha, 1 beta`
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