A radioactive element _{90}{X^{238}} decay in | vedclass

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Question

(d) Number of $\alpha - $particles emitted $ = \frac{{238 - 222}}{4} = 4$

This decreases atomic number to $90 - 4 	imes 2 = 82$

Since atomic nu

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(d) Number of $\alpha - $particles emitted $ = \frac{{238 - 222}}{4} = 4$

This decreases atomic number to $90 - 4 	imes 2 = 82$

Since atomic number of $_{83}{Y^{222}}$ is $83,$

this is possible if one $\beta - $ particle is emitted.

A radioactive element $_{90}{X^{238}}$ decay into $_{83}{Y^{222}}$. The number of $\beta - $ particles emitted are

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