In the nuclear reaction _{92}U^{238} to _{82} | vedclass

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Question

(B) For the decay of $_{92}U^{238}$ to $_{82}Pb^{206}$:Number of $\alpha$-particles $(n_{\alpha})$ = $\frac{238 - 206}{4} = \frac{32}{4} = 8$.Number o

Vedclass · Accepted Answer

$8\alpha, 6\beta$

Vedclass · Answer

(B) For the decay of $_{92}U^{238}$ to $_{82}Pb^{206}$:Number of $\alpha$-particles $(n_{\alpha})$ = $\frac{238 - 206}{4} = \frac{32}{4} = 8$.Number of $\beta$-particles $(n_{\beta})$ = $2 	imes n_{\alpha} - (Z_{initial} - Z_{final}) = 2 	imes 8 - (92 - 82) = 16 - 10 = 6$.Thus,the number of $\alpha$-particles is $8$ and $\beta$-particles is $6$.

In the nuclear reaction $_{92}U^{238} \to _{82}Pb^{206}$,the number of alpha and beta particles decayed are

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