An atomic nucleus _{90}Th^{232} emits several | vedclass

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(B) Let the number of $\alpha$-particles emitted be $n$ and the number of $\beta$-particles emitted be $m$.The change in mass number is given by $232

Vedclass · Accepted Answer

$6 \,\alpha$ and $4 \,\beta$

Vedclass · Answer

(B) Let the number of $\alpha$-particles emitted be $n$ and the number of $\beta$-particles emitted be $m$.The change in mass number is given by $232 - 208 = 24$.Since each $\alpha$-particle reduces the mass number by $4$,the number of $\alpha$-particles emitted is $n = \frac{24}{4} = 6$.The change in atomic number is given by $90 - 82 = 8$.Each $\alpha$-particle reduces the atomic number by $2$,and each $\beta$-particle increases it by $1$.Therefore,$90 - 2n + m = 82$.Substituting $n = 6$ into the equation: $90 - 2(6) + m = 82$.$90 - 12 + m = 82 \Rightarrow 78 + m = 82$.$m = 82 - 78 = 4$.Thus,the nucleus emitted $6 \,\alpha$ and $4 \,\beta$ particles.

An atomic nucleus $_{90}Th^{232}$ emits several $\alpha$ and $\beta$-radiations and finally reduces to $_{82}Pb^{208}$. It must have emitted:

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