To determine which particle completes the given nuclear equation:
[tex]\[
{ }_{56}^{124} \text{Ba} \longrightarrow{ }_{55}^{124} \text{Cs} +\text{ ? }
\][/tex]
we need to analyze the changes in atomic and mass numbers.
1. Initial Analysis:
- The barium (Ba) nucleus has an atomic number of 56 and a mass number of 124.
- The cesium (Cs) nucleus has an atomic number of 55 and a mass number of 124.
2. Change in Atomic Number:
- The atomic number decreases by 1 (from 56 to 55). This implies that one proton is lost in the process.
3. Change in Mass Number:
- The mass number remains unchanged at 124. This means the particle emitted should not alter the overall mass number.
4. Identifying the Emitted Particle:
- Let's go through the options and see which one fits the criteria.
- Option 1: [tex]${ }_{+1}^0 e$[/tex] (positron)
- This particle has no mass number and a positive charge.
- Emitting a positron would change the atomic number in the opposite direction (increase by 1), so it does not suit our situation.
- Option 2: [tex]${ }_{-1}^0 e$[/tex] (beta particle)
- This particle has no mass number and a negative charge.
- Emitting a beta particle corresponds to converting a neutron into a proton, thereby reducing the atomic number by 1, which matches our situation (Ba to Cs).
- Option 3: [tex]${ }_0^1 n$[/tex] (neutron)
- This particle has a mass number of 1 and no charge.
- Emitting a neutron would lower the mass number, which is not observed in our nuclear reaction (mass number remains 124).
- Option 4: [tex]${ }_1^1 H$[/tex] (hydrogen nucleus or proton)
- This particle has a mass number of 1 and a positive charge.
- Emitting a proton would also lower the mass number and decrease the atomic number by 1 more than needed, which doesn’t fit our scenario.
5. Conclusion:
- The only particle that fits the criteria of keeping the mass number the same and adjusting the atomic number from 56 to 55 is the beta particle ([tex]${ }_{-1}^0 e$[/tex]).
Therefore, the correct option is:
[tex]\[
{ }_{-1}^0 e
\][/tex]
