A nucleus with Z = 92 emits the following in a sequence: $\alpha, \beta^-, \beta^-, \alpha, \alpha, \alpha, \a

Question

A nucleus with Z = 92 emits the following in a sequence: $\alpha, \beta^-, \beta^-, \alpha, \alpha, \alpha, \alpha, \alpha, \beta^-, \beta^-, \alpha, \beta^+, \beta^+, \alpha$. The Z of the resulting nucleus is:

Accepted Answer

1. **Identify the Decay Rules:**
   - **Alpha ($\alpha$) decay:** The nucleus emits a helium nucleus (${}_{2}^{4}\mathrm{He}$). The atomic number ($Z$) decreases by 2 .
   - **Beta-minus ($\beta^-$) decay:** The nucleus emits an electron (${}_{-1}^{0}e$). The atomic number ($Z$) increases by 1 .
   - **Beta-plus ($\beta^+$) decay:** The nucleus emits a positron (${}_{+1}^{0}e$). The atomic number ($Z$) decreases by 1 .
2. **Count the Particles:**
   - Number of $\alpha$ particles ($n_\alpha$) = 8.
   - Number of $\beta^-$ particles ($n_{\beta^-}$) = 4.
   - Number of $\beta^+$ particles ($n_{\beta^+}$) = 2.
3. **Calculate Final Atomic Number:**
   - $Z_{	ext{final}} = Z_{	ext{initial}} + n_\alpha(-2) + n_{\beta^-}(+1) + n_{\beta^+}(-1)$
   - $Z_{	ext{final}} = 92 + 8(-2) + 4(1) + 2(-1)$
   - $Z_{	ext{final}} = 92 - 16 + 4 - 2$
   - $Z_{	ext{final}} = 78$.

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