After losing a number of alpha and beta parti | vedclass

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Question

(C) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.The nuclear reaction is represented as: $_{

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.The nuclear reaction is represented as: $_{92}U^{238} 	o _{82}Pb^{206} + x(_{2}\alpha^{4}) + y(_{-1}\beta^{0})$.Equating the mass number on both sides:$238 = 206 + 4x + 0y$$4x = 238 - 206$$4x = 32$$x = \frac{32}{4} = 8$.Therefore,the total number of $\alpha$-particles lost is $8$.

After losing a number of $\alpha$ and $\beta$ particles,$_{92}U^{238}$ is changed to $_{82}Pb^{206}$. The total number of $\alpha$-particles lost in this process is:

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