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(C) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.The nuclear reaction is: $_{92}U^{238} 	o

Vedclass · Accepted Answer

$6$

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(C) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.The nuclear reaction is: $_{92}U^{238} 	o _{82}Pb^{206} + x(_{2}He^{4}) + y(_{-1}e^{0})$.Equating the mass number on both sides:$238 = 206 + 4x + 0y$$4x = 32 \implies x = 8$.Equating the atomic number on both sides:$92 = 82 + 2x - y$$92 = 82 + 2(8) - y$$92 = 82 + 16 - y$$92 = 98 - y$$y = 98 - 92 = 6$.Therefore,the number of $\beta$-particles emitted is $6$.

The number of $\beta$-particles emitted in the radioactive decay process $_{92}U^{238} \to _{82}Pb^{206} + x(_{2}He^{4}) + y(_{-1}e^{0})$ is:

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