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A radioactive disintegration of ""(90)Th...

A radioactive disintegration of `""_(90)Th^(232)` yields `""_(82)Pb^(208)` in the end . The number of `alpha` and `beta` -particle emitted will be

A

`6 alpha , 6 beta`

B

`6alpha , 4 beta`

C

`6alpha , 5beta`

D

`5alpha , 6 beta`

Text Solution

Verified by Experts

The correct Answer is:
B
No. of `alpha` particle `=(232-208)/4 = 6`
No. of `beta`-particle = 4
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