Certainly! Let's complete and balance each of the given nuclear equations step-by-step.
### Part 1 of 3
The equation:
[tex]\[ {}_{80}^{178} \text{Hg} \rightarrow {}_{+1}^{0} \text{e} + \square \][/tex]
This reaction is a beta-plus decay (positron emission), where a proton in the nucleus of mercury (Hg) is converted into a neutron, emitting a positron. In such reactions:
1. The mass number (A) remains the same.
2. The atomic number (Z) decreases by 1.
So, starting with:
- Mass number: 178
- Atomic number: 80
The resulting element will have:
- Mass number: 178
- Atomic number: 80 - 1 = 79
The element with atomic number 79 is Gold (Au).
Thus, the completed and balanced equation is:
[tex]\[ {}_{80}^{178} \text{Hg} \rightarrow {}_{+1}^{0} \text{e} + {}_{79}^{178} \text{Au} \][/tex]
### Part 2 of 3
The equation:
[tex]\[ {}_{83}^{210} \text{Bi} \rightarrow {}_{2}^{4} \text{He} + \square \][/tex]
This reaction is an alpha decay, where a bismuth (Bi) nucleus emits an alpha particle (a helium nucleus). In such reactions:
1. The mass number decreases by 4 (A - 4).
2. The atomic number decreases by 2 (Z - 2).
So, starting with:
- Mass number: 210
- Atomic number: 83
The resulting element will have:
- Mass number: 210 - 4 = 206
- Atomic number: 83 - 2 = 81
The element with atomic number 81 is Thallium (Tl).
Thus, the completed and balanced equation is:
[tex]\[ {}_{83}^{210} \text{Bi} \rightarrow {}_{2}^{4} \text{He} + {}_{81}^{206} \text{Tl} \][/tex]
### Part 3 of 3
The equation:
[tex]\[ {}_{83}^{214} \text{Bi} \rightarrow {}_{-1}^{0} \text{e} + \square \][/tex]
This reaction is a beta-minus decay (electron emission), where a neutron in the nucleus of bismuth (Bi) is converted into a proton, emitting an electron. In such reactions:
1. The mass number (A) remains the same.
2. The atomic number (Z) increases by 1.
So, starting with:
- Mass number: 214
- Atomic number: 83
The resulting element will have:
- Mass number: 214
- Atomic number: 83 + 1 = 84
The element with atomic number 84 is Polonium (Po).
Thus, the completed and balanced equation is:
[tex]\[ {}_{83}^{214} \text{Bi} \rightarrow {}_{-1}^{0} \text{e} + {}_{84}^{214} \text{Po} \][/tex]
That's the step-by-step solution for each part of the given nuclear equations.
