The isotope _{92}U^{235} decays in a number o | vedclass

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(C) The nuclear decay reaction is represented as: $_{92}U^{235} 	o _{82}Pb^{207} + x(_{2}He^{4}) + y(_{-1}\beta^{0})$.Number of $\alpha$-particles $(

Vedclass · Accepted Answer

$7\alpha, 4\beta$

Vedclass · Answer

(C) The nuclear decay reaction is represented as: $_{92}U^{235} 	o _{82}Pb^{207} + x(_{2}He^{4}) + y(_{-1}\beta^{0})$.Number of $\alpha$-particles $(x)$ is calculated by the change in mass number: $x = \frac{235 - 207}{4} = \frac{28}{4} = 7$.Number of $\beta$-particles $(y)$ is calculated by the change in atomic number: $92 = 82 + 2(x) - y \implies 92 = 82 + 2(7) - y \implies 92 = 82 + 14 - y \implies 92 = 96 - y \implies y = 4$.Thus,the emitted particles are $7\alpha$ and $4\beta$.

The isotope $_{92}U^{235}$ decays in a number of steps to an isotope of lead $_{82}Pb^{207}$. The groups of particles emitted in this process will be

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