SPH3U1 Nuclear Equations Worksheet featuring 28 problems to complete nuclear equations and state the type of nuclear decay.
Nuclear equations worksheet with problems to complete and identify types of nuclear decay.
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Step-by-step solution for: Classification of Nuclear Reactions Practice Worksheet ...
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Show Answer Key & Explanations
Step-by-step solution for: Classification of Nuclear Reactions Practice Worksheet ...
Let’s solve each nuclear equation step by step. In nuclear reactions, two things must be conserved:
1. Mass number (top number) – total protons + neutrons
2. Atomic number (bottom number) – number of protons
We’ll balance both sides for each reaction and identify the missing particle or nucleus.
---
This is alpha decay (emits helium nucleus).
Left: mass = 210, atomic = 84
Right: He has mass 4, atomic 2 → so missing part must have:
Mass: 210 - 4 = 206
Atomic: 84 - 2 = 82
Element with atomic number 82 is Pb (Lead)
→ \( ^{206}_{82}Pb \)
Type: Alpha decay
---
Left: mass 8, atomic 5
Right: Be has mass 8, atomic 4 → missing particle:
Mass: 8 - 8 = 0
Atomic: 5 - 4 = +1
That’s a positron: \( ^0_{+1}e \) or \( ^0_1\beta^+ \)
Type: Positron emission
---
Beta minus decay (emits electron). Gamma doesn’t change mass/atomic.
So parent must have:
Mass: same as Pa → 234
Atomic: Pa is 91, but we emitted an electron (atomic -1), so parent was 91 + 1 = 92
Element 92 is Uranium (U)
→ \( ^{234}_{92}U \)
Type: Beta decay
---
Beta decay again.
Left: mass 14, atomic 6
Right: electron has mass 0, atomic -1 → daughter:
Mass: 14 - 0 = 14
Atomic: 6 - (-1) = 7
Element 7 is Nitrogen (N)
→ \( ^{14}_7N \)
Type: Beta decay
---
X-ray photon means no mass or charge change — this is electron capture.
In electron capture, a proton turns into neutron by capturing an inner electron.
So left side: Rb (atomic 37) captures an electron → becomes Kr (atomic 36)
But wait — the blank is on the LEFT with Rb? That doesn’t make sense unless it’s the electron being captured.
Actually, standard notation: Electron capture is written as:
\( ^A_ZX + ^0_{-1}e \rightarrow ^A_{Z-1}Y \)
So here, the blank should be the electron!
→ \( ^0_{-1}e \)
Type: Electron capture
---
Left: mass 15, atomic 8
Right: N has mass 15, atomic 7 → missing:
Mass: 0
Atomic: 8 - 7 = +1
→ Positron: \( ^0_{+1}e \)
Type: Positron emission
---
Electron capture.
Left: mass 58, atomic 28 + (-1) = 27
So product has mass 58, atomic 27 → Cobalt (Co)
→ \( ^{58}_{27}Co \)
Type: Electron capture
---
Gamma doesn’t affect balancing.
Left: mass 226, atomic 88
Right: Rn has mass 222, atomic 86 → missing:
Mass: 226 - 222 = 4
Atomic: 88 - 86 = 2
→ Alpha particle: \( ^4_2He \)
Type: Alpha decay
---
Neutron decays via beta decay.
Left: mass 1, atomic 0
Right: electron has mass 0, atomic -1 → other product:
Mass: 1 - 0 = 1
Atomic: 0 - (-1) = +1
→ Proton: \( ^1_1H \) or just \( ^1_1p \)
Type: Beta decay (of a free neutron)
---
Alpha decay.
Left: mass 238, atomic 92
Right: He has mass 4, atomic 2 → daughter:
Mass: 238 - 4 = 234
Atomic: 92 - 2 = 90
Element 90 is Thorium (Th)
→ \( ^{234}_{90}Th \)
Type: Alpha decay
---
Now let’s do the second set (no need to state decay type):
---
Left: mass = 9 + 4 = 13; atomic = 4 + 2 = 6
Right: neutron has mass 1, atomic 0 → missing:
Mass: 13 - 1 = 12
Atomic: 6 - 0 = 6
→ Carbon-12: \( ^{12}_6C \)
---
Beta decay → parent had one less atomic number.
Mass same: 239
Atomic: 94 - (-1)? Wait — in beta decay, parent atomic number is Z, daughter is Z+1.
Wait: if daughter is Pu (94), and it came from beta decay, then parent was atomic number 93.
Because: \( ^A_ZX \rightarrow ^A_{Z+1}Y + ^0_{-1}e \)
So parent: mass 239, atomic 93 → Neptunium (Np)
→ \( ^{239}_{93}Np \)
---
Left: mass 66, atomic 29
Right: Zn has mass 66, atomic 30 → missing:
Mass: 0
Atomic: 29 - 30 = -1
→ Electron: \( ^0_{-1}e \)
(Beta decay)
---
Left: Al mass 27, atomic 13 + ?
Right: Si mass 30, H mass 1 → total right mass = 31; atomic = 14 + 1 = 15
So left must also sum to mass 31, atomic 15
Al is 27,13 → missing particle: mass = 31 - 27 = 4, atomic = 15 - 13 = 2
→ Alpha particle: \( ^4_2He \)
---
Beta decay → daughter has same mass, atomic +1
Mass: 141
Atomic: 56 + 1 = 57
→ Lanthanum (La): \( ^{141}_{57}La \)
---
Right: mass = 17 + 1 = 18; atomic = 8 + 1 = 9
Left: He has mass 4, atomic 2 → missing:
Mass: 18 - 4 = 14
Atomic: 9 - 2 = 7
→ Nitrogen-14: \( ^{14}_7N \)
---
Alpha decay → parent had mass +4, atomic +2
Mass: 181 + 4 = 185
Atomic: 77 + 2 = 79
→ Gold (Au): \( ^{185}_{79}Au \)
---
Left: mass 241, atomic 95
Right: Np has mass 237, atomic 93 → missing:
Mass: 241 - 237 = 4
Atomic: 95 - 93 = 2
→ Alpha: \( ^4_2He \)
---
Right: Cf mass 246, 4 neutrons → total mass = 246 + 4×1 = 250; atomic = 98 + 0 = 98
Left: C has mass 12, atomic 6 → missing:
Mass: 250 - 12 = 238
Atomic: 98 - 6 = 92
→ Uranium-238: \( ^{238}_{92}U \)
---
Left: mass 18, atomic 9
Right: O has mass 17, atomic 8 → missing:
Mass: 18 - 17 = 1
Atomic: 9 - 8 = 1
→ Proton: \( ^1_1H \) or \( ^1_1p \)
(Though sometimes written as positron + neutrino, but based on numbers, it's proton emission — rare, but fits.)
Wait — actually, Fluorine-18 usually decays by positron emission to Oxygen-18, not Oxygen-17.
Check: If it were positron: \( ^{18}_9F \rightarrow ^{18}_8O + ^0_{+1}e \)
But here it says \( ^{17}_8O \), so mass decreased by 1, atomic by 1 → that’s proton emission.
Yes, possible in some cases.
→ \( ^1_1H \)
---
Left: H-2 mass 2, atomic 1 + ?
Right: He mass 4, n mass 1 → total mass 5; atomic 2 + 0 = 2
So missing particle: mass = 5 - 2 = 3; atomic = 2 - 1 = 1
→ Tritium: \( ^3_1H \)
(This is D-T fusion)
---
Beta decay → daughter has same mass, atomic +1
Mass: 27
Atomic: 14 + 1 = 15
→ Phosphorus: \( ^{27}_{15}P \)
---
Alpha decay → parent mass = 4 + 210 = 214; atomic = 2 + 81 = 83
→ Bismuth: \( ^{214}_{83}Bi \)
---
Beta decay → daughter mass 32, atomic 16
→ Sulfur: \( ^{32}_{16}S \)
---
Left: Pm mass 142, atomic 61 + ?
Right: Nd mass 142, atomic 60
So missing particle must reduce atomic number by 1, mass unchanged → electron capture or positron?
If it’s electron capture: add \( ^0_{-1}e \)
Then left atomic: 61 + (-1) = 60 → matches.
→ \( ^0_{-1}e \)
---
Right: mass 14 + 1 = 15; atomic 6 + 1 = 7
Left: N has mass 14, atomic 7 → missing:
Mass: 15 - 14 = 1
Atomic: 7 - 7 = 0
→ Neutron: \( ^1_0n \)
---
Left: mass 13 + 1 = 14; atomic 6 + 0 = 6
→ Carbon-14: \( ^{14}_6C \)
---
Left: mass 239 + 4 = 243; atomic 94 + 2 = 96
Right: H mass 1, 2n mass 2 → total so far 3; atomic 1 + 0 = 1
Missing: mass = 243 - 3 = 240; atomic = 96 - 1 = 95
→ Americium: \( ^{240}_{95}Am \)
---
## Final Answers:
1. \( ^{206}_{82}Pb \) — Alpha decay
2. \( ^0_{+1}e \) — Positron emission
3. \( ^{234}_{92}U \) — Beta decay
4. \( ^{14}_7N \) — Beta decay
5. \( ^0_{-1}e \) — Electron capture
6. \( ^0_{+1}e \) — Positron emission
7. \( ^{58}_{27}Co \) — Electron capture
8. \( ^4_2He \) — Alpha decay
9. \( ^1_1H \) — Beta decay
10. \( ^{234}_{90}Th \) — Alpha decay
11. \( ^{12}_6C \)
12. \( ^{239}_{93}Np \)
13. \( ^0_{-1}e \)
14. \( ^4_2He \)
15. \( ^{141}_{57}La \)
16. \( ^{14}_7N \)
17. \( ^{185}_{79}Au \)
18. \( ^4_2He \)
19. \( ^{238}_{92}U \)
20. \( ^1_1H \)
21. \( ^3_1H \)
22. \( ^{27}_{15}P \)
23. \( ^{214}_{83}Bi \)
24. \( ^{32}_{16}S \)
25. \( ^0_{-1}e \)
26. \( ^1_0n \)
27. \( ^{14}_6C \)
28. \( ^{240}_{95}Am \)
1. Mass number (top number) – total protons + neutrons
2. Atomic number (bottom number) – number of protons
We’ll balance both sides for each reaction and identify the missing particle or nucleus.
---
1. \( ^{210}_{84}Po \rightarrow \_\_\_\_ + ^4_2He \)
This is alpha decay (emits helium nucleus).
Left: mass = 210, atomic = 84
Right: He has mass 4, atomic 2 → so missing part must have:
Mass: 210 - 4 = 206
Atomic: 84 - 2 = 82
Element with atomic number 82 is Pb (Lead)
→ \( ^{206}_{82}Pb \)
Type: Alpha decay
---
2. \( ^8_5B \rightarrow ^8_4Be + \_\_\_\_ \)
Left: mass 8, atomic 5
Right: Be has mass 8, atomic 4 → missing particle:
Mass: 8 - 8 = 0
Atomic: 5 - 4 = +1
That’s a positron: \( ^0_{+1}e \) or \( ^0_1\beta^+ \)
Type: Positron emission
---
3. \( \_\_\_\_ \rightarrow ^{234}_{91}Pa + ^0_{-1}e + \gamma \)
Beta minus decay (emits electron). Gamma doesn’t change mass/atomic.
So parent must have:
Mass: same as Pa → 234
Atomic: Pa is 91, but we emitted an electron (atomic -1), so parent was 91 + 1 = 92
Element 92 is Uranium (U)
→ \( ^{234}_{92}U \)
Type: Beta decay
---
4. \( ^{14}_6C \rightarrow \_\_\_\_ + ^0_{-1}e \)
Beta decay again.
Left: mass 14, atomic 6
Right: electron has mass 0, atomic -1 → daughter:
Mass: 14 - 0 = 14
Atomic: 6 - (-1) = 7
Element 7 is Nitrogen (N)
→ \( ^{14}_7N \)
Type: Beta decay
---
5. \( \_\_\_\_ + ^{81}_{37}Rb \rightarrow ^{81}_{36}Kr + X-ray photon \)
X-ray photon means no mass or charge change — this is electron capture.
In electron capture, a proton turns into neutron by capturing an inner electron.
So left side: Rb (atomic 37) captures an electron → becomes Kr (atomic 36)
But wait — the blank is on the LEFT with Rb? That doesn’t make sense unless it’s the electron being captured.
Actually, standard notation: Electron capture is written as:
\( ^A_ZX + ^0_{-1}e \rightarrow ^A_{Z-1}Y \)
So here, the blank should be the electron!
→ \( ^0_{-1}e \)
Type: Electron capture
---
6. \( ^{15}_8O \rightarrow ^{15}_7N + \_\_\_\_ \)
Left: mass 15, atomic 8
Right: N has mass 15, atomic 7 → missing:
Mass: 0
Atomic: 8 - 7 = +1
→ Positron: \( ^0_{+1}e \)
Type: Positron emission
---
7. \( ^{58}_{28}Ni + ^0_{-1}e \rightarrow \_\_\_\_ \)
Electron capture.
Left: mass 58, atomic 28 + (-1) = 27
So product has mass 58, atomic 27 → Cobalt (Co)
→ \( ^{58}_{27}Co \)
Type: Electron capture
---
8. \( ^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + \_\_\_\_ + \gamma \)
Gamma doesn’t affect balancing.
Left: mass 226, atomic 88
Right: Rn has mass 222, atomic 86 → missing:
Mass: 226 - 222 = 4
Atomic: 88 - 86 = 2
→ Alpha particle: \( ^4_2He \)
Type: Alpha decay
---
9*. \( ^1_0n \rightarrow \_\_\_\_ + ^0_{-1}e \)
Neutron decays via beta decay.
Left: mass 1, atomic 0
Right: electron has mass 0, atomic -1 → other product:
Mass: 1 - 0 = 1
Atomic: 0 - (-1) = +1
→ Proton: \( ^1_1H \) or just \( ^1_1p \)
Type: Beta decay (of a free neutron)
---
10. \( ^{238}_{92}U \rightarrow \_\_\_\_ + ^4_2He \)
Alpha decay.
Left: mass 238, atomic 92
Right: He has mass 4, atomic 2 → daughter:
Mass: 238 - 4 = 234
Atomic: 92 - 2 = 90
Element 90 is Thorium (Th)
→ \( ^{234}_{90}Th \)
Type: Alpha decay
---
Now let’s do the second set (no need to state decay type):
---
11. \( ^9_4Be + ^4_2He \rightarrow \_\_\_\_ + ^1_0n \)
Left: mass = 9 + 4 = 13; atomic = 4 + 2 = 6
Right: neutron has mass 1, atomic 0 → missing:
Mass: 13 - 1 = 12
Atomic: 6 - 0 = 6
→ Carbon-12: \( ^{12}_6C \)
---
12. \( \_\_\_\_ \rightarrow ^{239}_{94}Pu + ^0_{-1}e \)
Beta decay → parent had one less atomic number.
Mass same: 239
Atomic: 94 - (-1)? Wait — in beta decay, parent atomic number is Z, daughter is Z+1.
Wait: if daughter is Pu (94), and it came from beta decay, then parent was atomic number 93.
Because: \( ^A_ZX \rightarrow ^A_{Z+1}Y + ^0_{-1}e \)
So parent: mass 239, atomic 93 → Neptunium (Np)
→ \( ^{239}_{93}Np \)
---
13. \( ^{66}_{29}Cu \rightarrow ^{66}_{30}Zn + \_\_\_\_ \)
Left: mass 66, atomic 29
Right: Zn has mass 66, atomic 30 → missing:
Mass: 0
Atomic: 29 - 30 = -1
→ Electron: \( ^0_{-1}e \)
(Beta decay)
---
14. \( ^{27}_{13}Al + \_\_\_\_ \rightarrow ^{30}_{14}Si + ^1_1H \)
Left: Al mass 27, atomic 13 + ?
Right: Si mass 30, H mass 1 → total right mass = 31; atomic = 14 + 1 = 15
So left must also sum to mass 31, atomic 15
Al is 27,13 → missing particle: mass = 31 - 27 = 4, atomic = 15 - 13 = 2
→ Alpha particle: \( ^4_2He \)
---
15. \( ^{141}_{56}Ba \rightarrow \_\_\_\_ + ^0_{-1}e \)
Beta decay → daughter has same mass, atomic +1
Mass: 141
Atomic: 56 + 1 = 57
→ Lanthanum (La): \( ^{141}_{57}La \)
---
16. \( \_\_\_\_ + ^4_2He \rightarrow ^{17}_8O + ^1_1p \)
Right: mass = 17 + 1 = 18; atomic = 8 + 1 = 9
Left: He has mass 4, atomic 2 → missing:
Mass: 18 - 4 = 14
Atomic: 9 - 2 = 7
→ Nitrogen-14: \( ^{14}_7N \)
---
17. \( \_\_\_\_ \rightarrow ^{181}_{77}Ir + ^4_2He \)
Alpha decay → parent had mass +4, atomic +2
Mass: 181 + 4 = 185
Atomic: 77 + 2 = 79
→ Gold (Au): \( ^{185}_{79}Au \)
---
18. \( ^{241}_{95}Am \rightarrow \_\_\_\_ + ^{237}_{93}Np \)
Left: mass 241, atomic 95
Right: Np has mass 237, atomic 93 → missing:
Mass: 241 - 237 = 4
Atomic: 95 - 93 = 2
→ Alpha: \( ^4_2He \)
---
19. \( \_\_\_\_ + ^{12}_6C \rightarrow ^{246}_{98}Cf + 4^1_0n \)
Right: Cf mass 246, 4 neutrons → total mass = 246 + 4×1 = 250; atomic = 98 + 0 = 98
Left: C has mass 12, atomic 6 → missing:
Mass: 250 - 12 = 238
Atomic: 98 - 6 = 92
→ Uranium-238: \( ^{238}_{92}U \)
---
20. \( ^{18}_9F \rightarrow ^{17}_8O + \_\_\_\_ \)
Left: mass 18, atomic 9
Right: O has mass 17, atomic 8 → missing:
Mass: 18 - 17 = 1
Atomic: 9 - 8 = 1
→ Proton: \( ^1_1H \) or \( ^1_1p \)
(Though sometimes written as positron + neutrino, but based on numbers, it's proton emission — rare, but fits.)
Wait — actually, Fluorine-18 usually decays by positron emission to Oxygen-18, not Oxygen-17.
Check: If it were positron: \( ^{18}_9F \rightarrow ^{18}_8O + ^0_{+1}e \)
But here it says \( ^{17}_8O \), so mass decreased by 1, atomic by 1 → that’s proton emission.
Yes, possible in some cases.
→ \( ^1_1H \)
---
21. \( ^2_1H + \_\_\_\_ \rightarrow ^4_2He + ^1_0n + energy \)
Left: H-2 mass 2, atomic 1 + ?
Right: He mass 4, n mass 1 → total mass 5; atomic 2 + 0 = 2
So missing particle: mass = 5 - 2 = 3; atomic = 2 - 1 = 1
→ Tritium: \( ^3_1H \)
(This is D-T fusion)
---
22. \( ^{27}_{14}Si \rightarrow ^0_{-1}e + \_\_\_\_ \)
Beta decay → daughter has same mass, atomic +1
Mass: 27
Atomic: 14 + 1 = 15
→ Phosphorus: \( ^{27}_{15}P \)
---
23. \( \_\_\_\_ \rightarrow ^4_2He + ^{210}_{81}Tl \)
Alpha decay → parent mass = 4 + 210 = 214; atomic = 2 + 81 = 83
→ Bismuth: \( ^{214}_{83}Bi \)
---
24. \( ^{32}_{15}P \rightarrow \_\_\_\_ + ^0_{-1}e \)
Beta decay → daughter mass 32, atomic 16
→ Sulfur: \( ^{32}_{16}S \)
---
25. \( ^{142}_{61}Pm + \_\_\_\_ \rightarrow ^{142}_{60}Nd \)
Left: Pm mass 142, atomic 61 + ?
Right: Nd mass 142, atomic 60
So missing particle must reduce atomic number by 1, mass unchanged → electron capture or positron?
If it’s electron capture: add \( ^0_{-1}e \)
Then left atomic: 61 + (-1) = 60 → matches.
→ \( ^0_{-1}e \)
---
26. \( ^{14}_7N + \_\_\_\_ \rightarrow ^{14}_6C + ^1_1p \)
Right: mass 14 + 1 = 15; atomic 6 + 1 = 7
Left: N has mass 14, atomic 7 → missing:
Mass: 15 - 14 = 1
Atomic: 7 - 7 = 0
→ Neutron: \( ^1_0n \)
---
27. \( ^{13}_6C + ^1_0n \rightarrow \_\_\_\_ \)
Left: mass 13 + 1 = 14; atomic 6 + 0 = 6
→ Carbon-14: \( ^{14}_6C \)
---
28. \( ^{239}_{94}Pu + ^4_2He \rightarrow ^1_1H + 2^1_0n + \_\_\_\_ \)
Left: mass 239 + 4 = 243; atomic 94 + 2 = 96
Right: H mass 1, 2n mass 2 → total so far 3; atomic 1 + 0 = 1
Missing: mass = 243 - 3 = 240; atomic = 96 - 1 = 95
→ Americium: \( ^{240}_{95}Am \)
---
## Final Answers:
1. \( ^{206}_{82}Pb \) — Alpha decay
2. \( ^0_{+1}e \) — Positron emission
3. \( ^{234}_{92}U \) — Beta decay
4. \( ^{14}_7N \) — Beta decay
5. \( ^0_{-1}e \) — Electron capture
6. \( ^0_{+1}e \) — Positron emission
7. \( ^{58}_{27}Co \) — Electron capture
8. \( ^4_2He \) — Alpha decay
9. \( ^1_1H \) — Beta decay
10. \( ^{234}_{90}Th \) — Alpha decay
11. \( ^{12}_6C \)
12. \( ^{239}_{93}Np \)
13. \( ^0_{-1}e \)
14. \( ^4_2He \)
15. \( ^{141}_{57}La \)
16. \( ^{14}_7N \)
17. \( ^{185}_{79}Au \)
18. \( ^4_2He \)
19. \( ^{238}_{92}U \)
20. \( ^1_1H \)
21. \( ^3_1H \)
22. \( ^{27}_{15}P \)
23. \( ^{214}_{83}Bi \)
24. \( ^{32}_{16}S \)
25. \( ^0_{-1}e \)
26. \( ^1_0n \)
27. \( ^{14}_6C \)
28. \( ^{240}_{95}Am \)
Parent Tip: Review the logic above to help your child master the concept of nuclear decay equations worksheet.
