The element with atomic number 84 and mass nu | vedclass

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(C) Let the reaction be $_{84}X^{218} 	o _{84}Y^{214} + x_{2}\alpha^{4} + y_{-1}\beta^{0}$.Number of $\alpha$-particles emitted is given by the chang

Vedclass · Accepted Answer

$1, 2$

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(C) Let the reaction be $_{84}X^{218} 	o _{84}Y^{214} + x_{2}\alpha^{4} + y_{-1}\beta^{0}$.Number of $\alpha$-particles emitted is given by the change in mass number: $x = \frac{218 - 214}{4} = \frac{4}{4} = 1$.For atomic number conservation: $84 = 84 + 2x - y$.Substituting $x = 1$: $84 = 84 + 2(1) - y$,which gives $y = 2$.Therefore,the number of $\alpha$-particles is $1$ and $\beta$-particles is $2$.

The element with atomic number $84$ and mass number $218$ changes to another element with atomic number $84$ and mass number $214$. The number of $\alpha$ and $\beta$-particles emitted are respectively:

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