Fill in the missing components in nuclear reactions and decay processes.

Let's solve each nuclear reaction step by step. We'll use the conservation of mass number (A) and atomic number (Z) to determine the missing particles or nuclei.

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1. $^{42}_{15}K \rightarrow ^{0}_{-1}e + \_\_\_\_\_\_\_\_\_\_$

This is beta decay ($\beta^-$), where a neutron turns into a proton, emitting an electron and an antineutrino.

- Mass number: 42 → 42
- Atomic number: 15 → 16 (since $-1$ from electron, so daughter nucleus must be $+1$)

So, the product has:
- A = 42
- Z = 16 → Element with atomic number 16 is Sulfur (S)

✔ Answer: $^{42}_{16}S$

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2. $^{239}_{94}Pu \rightarrow ^{4}_{2}He + \_\_\_\_\_\_\_\_\_\_$

Alpha decay: emits $^4_2He$, so subtract 4 from mass and 2 from atomic number.

- A: 239 - 4 = 235
- Z: 94 - 2 = 92 → Uranium (U)

✔ Answer: $^{235}_{92}U$

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3. $^{238}_{92}U \rightarrow \_\_\_\_\_\_\_\_\_\_ + ^{234}_{90}Th$

We know $^{238}_{92}U$ decays to $^{234}_{90}Th$, so we need the emitted particle.

- Mass difference: 238 - 234 = 4
- Charge difference: 92 - 90 = 2

So, it's an alpha particle: $^4_2He$

✔ Answer: $^{4}_{2}He$

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4. $^{1}_{1}H + ^{3}_{1}H \rightarrow \_\_\_\_\_\_\_\_\_\_$

This is a fusion reaction: deuterium and tritium fuse.

Total:
- A: 1 + 3 = 4
- Z: 1 + 1 = 2 → So, $^4_2He$ (helium-4)

✔ Answer: $^{4}_{2}He$

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5. $^{6}_{3}Li + ^{1}_{0}n \rightarrow ^{4}_{2}He + \_\_\_\_\_\_\_\_\_\_$

Neutron capture by lithium-6.

- Total A: 6 + 1 = 7 → 4 + ? → ? = 3
- Total Z: 3 + 0 = 3 → 2 + ? → ? = 1

So, unknown has A=3, Z=1 → Hydrogen-3 → Tritium

✔ Answer: $^{3}_{1}H$

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6. $^{27}_{13}Al + ^{4}_{2}He \rightarrow ^{30}_{15}P + \_\_\_\_\_\_\_\_\_\_$

Alpha particle hits aluminum.

- Total A: 27 + 4 = 31 → 30 + ? → ? = 1
- Total Z: 13 + 2 = 15 → 15 + ? → ? = 0

So, particle has A=1, Z=0 → Neutron

✔ Answer: $^{1}_{0}n$

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7. $^{9}_{4}Be + ^{1}_{1}H \rightarrow \_\_\_\_\_\_\_\_\_\_ + ^{4}_{2}He$

Beryllium-9 reacts with proton.

- A: 9 + 1 = 10 → ? + 4 → ? = 6
- Z: 4 + 1 = 5 → ? + 2 → ? = 3

So, unknown has A=6, Z=3 → Lithium-6

✔ Answer: $^{6}_{3}Li$

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8. $^{37}_{19}K \rightarrow ^{0}_{-1}e + \_\_\_\_\_\_\_\_\_\_$

Beta decay again.

- A: 37 → 37
- Z: 19 → 20 (since electron has -1 charge, daughter nucleus gains +1)

Z = 20 → Calcium (Ca)

✔ Answer: $^{37}_{20}Ca$

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9. $\_\_\_\_\_\_\_\_\_\_ + ^{1}_{0}n \rightarrow ^{142}_{56}Ba + ^{91}_{36}Kr + 3^{1}_{0}n$

This is nuclear fission — likely uranium splitting.

Total products:
- A: 142 + 91 + 3×1 = 236
- Z: 56 + 36 + 0 = 92

Reactants: one neutron (A=1, Z=0) and unknown nucleus.

So, unknown nucleus must have:
- A = 236 - 1 = 235
- Z = 92 - 0 = 92

→ $^{235}_{92}U$

✔ Answer: $^{235}_{92}U$

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10. $^{238}_{92}U + ^{4}_{2}He \rightarrow \_\_\_\_\_\_\_\_\_\_ + ^{1}_{0}n$

Alpha particle hits uranium-238.

- A: 238 + 4 = 242 → ? + 1 → ? = 241
- Z: 92 + 2 = 94 → ? + 0 → ? = 94

So, unknown has A=241, Z=94 → Plutonium-241

✔ Answer: $^{241}_{94}Pu$

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✔ Final Answers:

1. $^{42}_{16}S$
2. $^{235}_{92}U$
3. $^{4}_{2}He$
4. $^{4}_{2}He$
5. $^{3}_{1}H$
6. $^{1}_{0}n$
7. $^{6}_{3}Li$
8. $^{37}_{20}Ca$
9. $^{235}_{92}U$
10. $^{241}_{94}Pu$

Let me know if you'd like explanations in terms of decay types or real-world applications!