To determine which of the given options represents a beta decay, we need to understand what happens during a beta decay process. In beta decay:
- A neutron in the unstable nucleus is transformed into a proton, resulting in the emission of a beta particle (an electron, [tex]\( e^- \)[/tex], or a positron, [tex]\( e^+ \)[/tex]), and an antineutrino or neutrino.
### Option Analysis:
Option A:
[tex]\[
{}_{83}^{212} Bi \rightarrow {}_{84}^{212} Po + {}_{-1}^{0} e
\][/tex]
- This reaction shows an increase in the atomic number from 83 (Bismuth) to 84 (Polonium) while the atomic mass number stays at 212.
- This meets the criteria for beta decay because the atomic number increases by 1, signifying a neutron has been converted to a proton, and a beta particle ([tex]\( e^- \)[/tex]) has been emitted.
Option B:
[tex]\[
{}_{93}^{235} Np \rightarrow {}_{91}^{231} Pa + \alpha
\][/tex]
- This reaction involves the emission of an alpha particle ([tex]\( {}_{2}^{4} He \)[/tex]).
- Alpha decay involves the loss of 2 protons and 2 neutrons, corresponding to a decrease in atomic number by 2 and mass number by 4.
- This is not beta decay.
Option C:
[tex]\[
{}_{88}^{220} Rn \rightarrow {}_{84}^{216} Po + {}_{2}^{4} He
\][/tex]
- This also represents the emission of an alpha particle ([tex]\( {}_{2}^{4} He \)[/tex]).
- As explained above, this is not beta decay, but rather alpha decay.
Option D:
[tex]\[
{}_{28}^{80} Ni \rightarrow {}_{28}^{80} Ni + \gamma
\][/tex]
- This reaction shows no change in the atomic number or mass number of the nucleus and involves the emission of a gamma photon ([tex]\( \gamma \)[/tex]).
- Gamma decay involves the release of energy from an excited nucleus but no change in its proton or neutron count.
- This is not beta decay.
### Conclusion:
Among the provided options, Option A represents a beta decay. Thus, the correct answer is:
1. A. [tex]\[ {}_{83}^{212} Bi \rightarrow {}_{84}^{212} Po + {}_{-1}^{0} e \][/tex]
