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

Answer

1

Explanation

`._(90)X^(238) to ._(83)Y^(222)`
As mass number decreases by `238-222=16`,
therefore, number of `alpha` particle emitted `=16/4=4`
The charge number reduces to `90-4xx2=82`. To increases charge no. (83-82)=1. One beta particle must be emitted.