In a radioactive decay chain,{ }_{90}^{232} T | vedclass

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Question

(B) The decay process is represented as: ${ }_{90}^{232} Th \rightarrow { }_{82}^{212} Pb + N_{\alpha} { }_{2}^{4} He + N_{\beta} { }_{-1}^{0} e$.Firs

Vedclass · Accepted Answer

$A, D$

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(B) The decay process is represented as: ${ }_{90}^{232} Th \rightarrow { }_{82}^{212} Pb + N_{\alpha} { }_{2}^{4} He + N_{\beta} { }_{-1}^{0} e$.First,consider the change in mass number $(\Delta A)$:$\Delta A = 232 - 212 = 20$.Since each $\alpha$-particle has a mass number of $4$,the number of $\alpha$-particles emitted is $N_{\alpha} = \frac{20}{4} = 5$.Next,consider the change in atomic number $(\Delta Z)$:$\Delta Z = 90 - 82 = 8$.In the decay,the change in atomic number is given by $2N_{\alpha} - N_{\beta} = \Delta Z$.Substituting the values: $2(5) - N_{\beta} = 8$.$10 - N_{\beta} = 8$,which gives $N_{\beta} = 2$.Thus,$N_{\alpha} = 5$ and $N_{\beta} = 2$. The correct statements are $(A)$ and $(C)$.

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