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(B) The initial nucleus is ${ }_{92}^{238} U$ and the final nucleus is ${ }_{82}^{206} Pb$.Let $n_{\alpha}$ be the number of $\alpha$-particles and $n

Vedclass · Accepted Answer

$8$,$6$

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(B) The initial nucleus is ${ }_{92}^{238} U$ and the final nucleus is ${ }_{82}^{206} Pb$.Let $n_{\alpha}$ be the number of $\alpha$-particles and $n_{\beta}$ be the number of $\beta$-particles emitted.The change in mass number is $238 - 206 = 32$.Since each $\alpha$-particle emission reduces the mass number by $4$,the number of $\alpha$-particles is $n_{\alpha} = \frac{32}{4} = 8$.The change in atomic number is $92 - 82 = 10$.The emission of $8$ $\alpha$-particles would decrease the atomic number by $8 	imes 2 = 16$.Let $n_{\beta}$ be the number of $\beta$-particles. Each $\beta$-particle increases the atomic number by $1$.So,$92 - (8 	imes 2) + n_{\beta} = 82$.$92 - 16 + n_{\beta} = 82$.$76 + n_{\beta} = 82$.$n_{\beta} = 82 - 76 = 6$.Thus,the number of $\alpha$-particles is $8$ and the number of $\beta$-particles is $6$.

In the Uranium radioactive series,the initial nucleus is ${ }_{92}^{238} U$ and the final nucleus is ${ }_{82}^{206} Pb$. When the Uranium nucleus decays to lead,the number of $\alpha$-particles emitted is .......... and the number of $\beta$-particles emitted is ..........

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