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.(90)Th^(232) to .(82)Pb^(208). The numb...

`._(90)Th^(232) to ._(82)Pb^(208)`. The number of `alpha and beta-"particles"` emitted during the above reaction is

A

`8alpha and 4beta`

B

`8alpha and 16beta`

C

`4alpha and 2beta`

D

`6alpha and 4beta`

Text Solution

Verified by Experts

The correct Answer is:
D
No. of `alpha`-particle ` =(232-208)/(4)=(24)/(4)=6alpha`
No. of `beta`-particle `= `2xx6-(90-82)`
`=12-8`
=`4beta`
`"Hence", 6alpha and 4 beta` particle are emitted.
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