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