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A radioactive element .90 X^238 decay in...

A radioactive element `._90 X^238` decay into `._83 Y^222`. The number of `beta-`particles emitted are.

A

4

B

6

C

2

D

1

Text Solution

Verified by Experts

The correct Answer is:
D
`n_(alpha)= (238- 222)/(4)= (16)/(4)= 4`
`n_(beta)= 2n_(alpha)- Z + Z. = 2 xx 4 - 90 + 83 =1`
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