Saved Bookmarks
| 1. |
A radioactive element `._90 X^238` decay into `._83 Y^222`. The number of `beta-`particles emitted are. |
|
Answer» Correct Answer - 1 `._(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. |
|