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The end product of decay of ""(90)Th^(23...

The end product of decay of `""_(90)Th^(232)` is `""_(82)Pb^(208)` . The number of `alpha` and `beta` particles emitted are respectively

A

6,0

B

3,3

C

4,6

D

6,4

Text Solution

Verified by Experts

The correct Answer is:
D
`""_(90)Th^(232) to ""+_(82) Pb^(208)+A He_4 +B_(-1) e^(n)`
Balancing the mass and atomic number A=6,B=4
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