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A nucleus with atomic number \( z \) and neutron number N undergoes two decay processes. The result is a nucleus with atomic number Z-3 and neutron number N-1. Which decay processes took place?

1. Two \( \beta^{-} \) decays
2. Two \( \beta^{+} \) decays
3. \( A n \alpha-d e c a y \) and \( a \beta^{-} \) decay
4. An \( \alpha- \) decay and a \( \beta^{+} \) decay

Correct answer: An \( \alpha- \) decay and a \( \beta^{+} \) decay

Solution

An alpha decay reduces the atomic number by 2 and the neutron number by 2, while a beta-plus decay converts a proton into a neutron, so Z decreases by 1 and N increases by 1. Combined, these give Z-3 and N-1, matching the result.

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