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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
(d) Number of `alpha`-particles emitted `= (238 - 222)/(4)`
This decreases atomic number to `90 - 4 xx 2 = 82`
Since atomic number of `._83 Y^222` is `83`, this is possible if one `beta`-particle is emitted.
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