Complete the nuclear equations on this worksheet by balancing mass and atomic numbers.

Let's solve each nuclear equation step by step. In nuclear reactions, both mass number (A) and atomic number (Z) must be conserved.

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1. $ ^{210}_{84}\text{Po} \rightarrow \_\_\_\_\_\_ + ^4_2\text{He} $

- Alpha decay: Po emits an alpha particle ($^4_2\text{He}$)
- Mass: $210 - 4 = 206$
- Atomic number: $84 - 2 = 82$
- Element with Z=82 is Pb (Lead)

✔ Answer: $^{206}_{82}\text{Pb}$

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2. $ ^8_5\text{B} \rightarrow ^8_4\text{Be} + \_\_\_\_\_\_$

- Boron-8 decays to Beryllium-8
- Mass: $8 = 8 + x$ → $x = 0$ → so the product has mass 0
- Charge: $5 = 4 + y$ → $y = 1$
- Particle with charge +1 and mass 0 is a positron: $^0_{+1}e$

✔ Answer: $^0_{+1}e$

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3. $ \_\_\_\_\_\_ \rightarrow ^{234}_{91}\text{Pa} + ^0_{-1}e + \gamma $

- Beta decay: emission of electron ($^0_{-1}e$) and gamma ray
- The parent nucleus must have:
- Mass: $234 + 0 + 0 = 234$
- Atomic number: $91 + 1 = 92$
- Element with Z=92 is Uranium (U)

✔ Answer: $^{234}_{92}\text{U}$

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4. $ ^{14}_6\text{C} \rightarrow \_\_\_\_\_\_ + ^0_{-1}e $

- Carbon-14 undergoes beta decay
- Mass: $14 = x + 0$ → $x = 14$
- Atomic number: $6 = y + (-1)$ → $y = 7$
- Element with Z=7 is Nitrogen (N)

✔ Answer: $^{14}_7\text{N}$

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5. $ \_\_\_\_\_\_ + ^{81}_{37}\text{Rb} \rightarrow ^{81}_{36}\text{Kr} + \text{X-ray photon} $

- Rb-81 becomes Kr-81
- X-ray photon suggests electron capture
- Electron capture: nucleus captures an inner-shell electron
- So, the missing particle is $^0_{-1}e$, but it’s captured from outside
- But in nuclear equations, we write: $^0_{-1}e$ on left side
- However, here the reaction shows a particle being added to Rb to form Kr

Wait — Rb (Z=37) → Kr (Z=36): atomic number decreases by 1 → this is electron capture

So:

$$
^0_{-1}e + ^{81}_{37}\text{Rb} \rightarrow ^{81}_{36}\text{Kr} + \text{X-ray}
$$

✔ Answer: $^0_{-1}e$

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6. $ ^{15}_8\text{O} \rightarrow ^{15}_7\text{N} + \_\_\_\_\_\_$

- Oxygen-15 decays to Nitrogen-15
- Mass: $15 = 15 + x$ → $x = 0$
- Charge: $8 = 7 + y$ → $y = 1$
- So emitted particle has charge +1, mass 0 → positron: $^0_{+1}e$

✔ Answer: $^0_{+1}e$

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7. $ ^{58}_{28}\text{Ni} + ^0_{-1}e \rightarrow \_\_\_\_\_\_$

- This is electron capture by Ni-58
- Capture of electron reduces atomic number by 1
- Mass remains same
- New Z: $28 - 1 = 27$
- Element with Z=27 is Cobalt (Co)

✔ Answer: $^{58}_{27}\text{Co}$

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8. $ ^{226}_{88}\text{Ra} \rightarrow ^{222}_{86}\text{Rn} + \_\_\_\_\_\_ + \gamma $

- Ra-226 decays to Rn-222 via alpha decay
- Mass: $226 = 222 + x$ → $x = 4$
- Charge: $88 = 86 + y$ → $y = 2$
- So: $^4_2\text{He}$ (alpha particle)

Gamma ray is also emitted, which is common in alpha decay

✔ Answer: $^4_2\text{He}$

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9. $ ^1_0n \rightarrow \_\_\_\_\_\_ + ^0_{-1}e $

- Free neutron decays into proton and electron (beta decay)
- Also emits antineutrino, but not shown here
- So: $^1_0n \rightarrow ^1_1p + ^0_{-1}e + \bar{\nu}_e$
- But only two products shown → missing is proton

Mass: $1 = x + 0$ → $x = 1$
Charge: $0 = y + (-1)$ → $y = 1$

✔ Answer: $^1_1\text{H}$ or $^1_1p$

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10. $ ^{238}_{92}\text{U} \rightarrow \_\_\_\_\_\_ + ^4_2\text{He} $

- Uranium-238 undergoes alpha decay
- Mass: $238 - 4 = 234$
- Charge: $92 - 2 = 90$
- Element with Z=90 is Thorium (Th)

✔ Answer: $^{234}_{90}\text{Th}$

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11. $ ^9_4\text{Be} + ^4_2\text{He} \rightarrow \_\_\_\_\_\_ + ^1_0n $

- This is a nuclear fusion reaction
- Total mass: $9 + 4 = 13$, minus 1 → product has mass 12
- Total charge: $4 + 2 = 6$, minus 0 → product has charge 6
- Element with Z=6 is Carbon (C)

✔ Answer: $^{12}_6\text{C}$

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12. $ \_\_\_\_\_\_ \rightarrow ^{239}_{94}\text{Pu} + ^0_{-1}e $

- Beta decay: Pu-239 formed from parent nucleus
- Mass: $x = 239 + 0 = 239$
- Charge: $y = 94 + 1 = 95$
- Element with Z=95 is Americium (Am)

✔ Answer: $^{239}_{95}\text{Am}$

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13. $ ^{66}_{29}\text{Cu} \rightarrow ^{66}_{30}\text{Zn} + \_\_\_\_\_\_$

- Copper-66 decays to Zinc-66
- Atomic number increases by 1 → beta minus decay
- Emitted: $^0_{-1}e$

Mass: $66 = 66 + 0$ → OK
Charge: $29 = 30 + (-1)$ → OK

✔ Answer: $^0_{-1}e$

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14. $ ^{27}_{13}\text{Al} + \_\_\_\_\_\_ \rightarrow ^{30}_{14}\text{Si} + ^1_1\text{H} $

- Aluminum-27 reacts to form Silicon-30 and proton
- Let’s find the missing particle: $^A_ZX$

Mass balance: $27 + A = 30 + 1 = 31$ → $A = 4$
Charge balance: $13 + Z = 14 + 1 = 15$ → $Z = 2$

So particle is $^4_2\text{He}$ → alpha particle

✔ Answer: $^4_2\text{He}$

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✔ Final Answers:

1. $^{206}_{82}\text{Pb}$
2. $^0_{+1}e$
3. $^{234}_{92}\text{U}$
4. $^{14}_7\text{N}$
5. $^0_{-1}e$
6. $^0_{+1}e$
7. $^{58}_{27}\text{Co}$
8. $^4_2\text{He}$
9. $^1_1\text{H}$ (or $^1_1p$)
10. $^{234}_{90}\text{Th}$
11. $^{12}_6\text{C}$
12. $^{239}_{95}\text{Am}$
13. $^0_{-1}e$
14. $^4_2\text{He}$

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