What is the respective number of alpha and be | vedclass

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(D) In an $\alpha$-decay,the mass number decreases by $4$ and the atomic number decreases by $2$. In a $\beta$-decay,the mass number remains unchanged

Vedclass · Accepted Answer

$8$ and $6$

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(D) In an $\alpha$-decay,the mass number decreases by $4$ and the atomic number decreases by $2$. In a $\beta$-decay,the mass number remains unchanged and the atomic number increases by $1$.Let $n_{\alpha}$ be the number of $\alpha$-particles and $n_{\beta}$ be the number of $\beta$-particles emitted.The change in mass number is given by: $200 - 168 = 4 	imes n_{\alpha}$.Therefore,$n_{\alpha} = \frac{32}{4} = 8$.The change in atomic number is given by: $90 - 80 = 2 	imes n_{\alpha} - 1 	imes n_{\beta}$.Substituting $n_{\alpha} = 8$: $10 = 2(8) - n_{\beta}$.$10 = 16 - n_{\beta}$,which gives $n_{\beta} = 6$.Thus,the number of $\alpha$ and $\beta$ particles emitted are $8$ and $6$ respectively.

What is the respective number of $\alpha$ and $\beta$ particles emitted in the following radioactive decay: $_{90}X^{200} \to _{80}Y^{168}$?

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