After losing a number of alpha and beta parti | vedclass

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(A) The nuclear reaction is represented as: $_{92}U^{238} 	o _{82}Pb^{206} + x(_{2}\alpha^{4}) + y(_{-1}\beta^{0})$.Number of $\alpha$ particles $(x)

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) The nuclear reaction is represented as: $_{92}U^{238} 	o _{82}Pb^{206} + x(_{2}\alpha^{4}) + y(_{-1}\beta^{0})$.Number of $\alpha$ particles $(x)$ = $\frac{238 - 206}{4} = \frac{32}{4} = 8$.Number of $\beta$ particles $(y)$ = $92 - 82 - 2x = 10 - 2(8) = 10 - 16 = -6$. Wait,let us re-calculate: $92 = 82 + 2(8) - y \implies 92 = 82 + 16 - y \implies 92 = 98 - y \implies y = 6$.Total number of particles lost = $x + y = 8 + 6 = 14$.

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

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