What is the respective number of alpha and be | vedclass

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(D) The number of $\alpha$ particles emitted is given by the change in mass number divided by $4$: $n_{\alpha} = \frac{A - A'}{4} = \frac{200 - 168}{4

Vedclass · Accepted Answer

$8$ and $6$

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(D) The number of $\alpha$ particles emitted is given by the change in mass number divided by $4$: $n_{\alpha} = \frac{A - A'}{4} = \frac{200 - 168}{4} = \frac{32}{4} = 8$. The number of $\beta$ particles emitted is given by the formula: $n_{\beta} = 2n_{\alpha} - (Z - Z')$. Substituting the values: $n_{\beta} = 2(8) - (90 - 80) = 16 - 10 = 6$. Therefore,$8$ $\alpha$ particles and $6$ $\beta$ particles are emitted.

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

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