A radioactive decay chain starts from _{93}Np | vedclass

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(A) The radioactive decay process is given by: $_{93}Np^{237} \rightarrow _{90}Th^{229} + n_{\alpha}(_{2}He^{4}) + n_{\beta}(-_{1}e^{0})$.First,calcul

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Two $\alpha$-particles and one $\beta^-$-particle

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(A) The radioactive decay process is given by: $_{93}Np^{237} \rightarrow _{90}Th^{229} + n_{\alpha}(_{2}He^{4}) + n_{\beta}(-_{1}e^{0})$.First,calculate the number of $\alpha$-particles $(n_{\alpha})$: $n_{\alpha} = \frac{237 - 229}{4} = \frac{8}{4} = 2$.Next,calculate the number of $\beta$-particles $(n_{\beta})$ using the conservation of atomic number: $93 = 90 + 2(2) + n_{\beta}(-1)$.$93 = 90 + 4 - n_{\beta} \implies 93 = 94 - n_{\beta} \implies n_{\beta} = 1$.Since $n_{\beta} = 1$ is positive,it represents the emission of one $\beta^-$-particle.Therefore,the emitted particles are two $\alpha$-particles and one $\beta^-$-particle.

$A$ radioactive decay chain starts from $_{93}Np^{237}$ and produces $_{90}Th^{229}$ by successive emissions. The emitted particles can be:

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