In the given nuclear reaction,how many alpha | vedclass

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(D) Let the number of $\alpha$ particles emitted be $n_{\alpha}$ and the number of $\beta$ particles emitted be $n_{\beta}$.The change in mass number

Vedclass · Accepted Answer

$7 \alpha$ particles and $4 \beta$ particles

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(D) Let the number of $\alpha$ particles emitted be $n_{\alpha}$ and the number of $\beta$ particles emitted be $n_{\beta}$.The change in mass number is given by $\Delta A = 235 - 207 = 28$.Since each $\alpha$ particle carries a mass number of $4$,the number of $\alpha$ particles is $n_{\alpha} = \frac{28}{4} = 7$.The change in atomic number is given by $\Delta Z = 92 - 82 = 10$.The emission of $n_{\alpha}$ particles decreases the atomic number by $2n_{\alpha} = 2 	imes 7 = 14$.The emission of $n_{\beta}$ particles increases the atomic number by $n_{\beta}$.Therefore,$92 - 14 + n_{\beta} = 82$.$78 + n_{\beta} = 82$,which gives $n_{\beta} = 4$.Thus,$7 \alpha$ particles and $4 \beta$ particles are emitted.

In the given nuclear reaction,how many $\alpha$ and $\beta$ particles are emitted in the decay $_{92}X^{235} \to _{82}Y^{207}$?

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