Solved Balancing Nuclear Equations Name: Period: There are | Chegg.com - Free Printable

Let’s solve each problem one by one. We’ll use the rules from the table and examples given:

- In nuclear equations, mass number (top) and atomic number (bottom) must be conserved on both sides.
- For beta-minus decay (β⁻), a neutron turns into a proton + electron → so atomic number increases by 1, mass stays same.
- For positron emission (β⁺), a proton turns into neutron + positron → atomic number decreases by 1, mass stays same.
- Alpha decay: emits helium nucleus → mass -4, atomic number -2.
- Electron capture or other decays follow similar conservation.

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Problem 1:
³₁H → ? + ⁰₋₁e

This is beta-minus decay because it emits an electron (⁰₋₁e).
In β⁻ decay:
→ Mass number stays same → still 3
→ Atomic number increases by 1 → 1 + 1 = 2
Element with atomic number 2 is Helium (He)

So:
³₁H → ³₂He + ⁰₋₁e
Name: Beta-minus decay

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Problem 2:
²³²₉₂U → ²²⁸₉₀Th + ?

Mass: 232 → 228 + ? → missing mass = 4
Atomic number: 92 → 90 + ? → missing charge = 2
That’s an alpha particle: ⁴₂He

So:
²³²₉₂U → ²²⁸₉₀Th + ⁴₂He
Name: Alpha decay

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Problem 3:
¹⁴⁴₅₈Ce → ¹⁴⁴₅₉Pr + ?

Mass: 144 → 144 + ? → missing mass = 0
Atomic number: 58 → 59 + ? → missing charge = -1
That’s an electron: ⁰₋₁e → beta-minus decay

So:
¹⁴⁴₈Ce → ¹⁴₅₉Pr + ⁰₋₁e
Name: Beta-minus decay

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Problem 4:
⁶₃₀Zn → ? + ⁰₊₁e

Emits a positron (⁰₊₁e) → this is positron emission (β⁺ decay)
In β⁺ decay:
→ Mass stays same → 65
→ Atomic number decreases by 1 → 30 - 1 = 29
Element with atomic number 29 is Copper (Cu)

So:
⁶⁵₃₀Zn → ⁶⁵₂₉Cu + ⁰₊₁e
Name: Positron emission (or Beta-plus decay)

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Problem 5:
⁴⁰₁₉K → ⁴⁰₁₈Ar + ?

Mass: 40 → 40 + ? → missing mass = 0
Atomic number: 19 → 18 + ? → missing charge = +1
That’s a positron: ⁰₊₁e → positron emission

So:
⁴⁰₁₉K → ⁴⁰₁₈Ar + ⁰₊₁e
Name: Positron emission (or Beta-plus decay)

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Final Answer:
1. ³₂He ; Beta-minus decay
2. ⁴₂He ; Alpha decay
3. ⁰₋₁e ; Beta-minus decay
4. ⁶⁵₂₉Cu ; Positron emission
5. ⁰₊₁e ; Positron emission