When an element {}_{90}^{232}Th decays into { | vedclass

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(A) Let the number of $\alpha$-particles emitted be $n$ and the number of $\beta^{-}$-particles emitted be $m$.In an $\alpha$-decay, the mass number d

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$6, 4$

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(A) Let the number of $\alpha$-particles emitted be $n$ and the number of $\beta^{-}$-particles emitted be $m$.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 $1$.The change in mass number is: $232 - 208 = 4n + 0m = 24$.Thus, $n = 24 / 4 = 6$.The change in atomic number is: $90 - 82 = 2n - m$.Substituting $n = 6$: $8 = 2(6) - m$, which gives $8 = 12 - m$.Therefore, $m = 12 - 8 = 4$.So, $6$ $\alpha$-particles and $4$ $\beta^{-}$-particles are emitted.

When an element ${}_{90}^{232}Th$ decays into ${}_{82}^{208}Pb$, the number of $\alpha$ and $\beta^{-}$ particles emitted respectively are

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