1.

A radioactive element `._90 X^238` decay into `._83 Y^222`. The number of `beta-`particles emitted are.A. 4B. 6C. 2D. 1

Answer» Correct Answer - 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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