Solved BALANCING NUCLEAR REACTIONS WORKSHEET Predict the | Chegg.com - Free Printable

Let’s solve each nuclear reaction step by step. We’ll balance the mass numbers (top) and atomic numbers (bottom), then identify the type of reaction.

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1. ⁴²₁₉K → ⁰₋₁e + ______

- Mass: 42 = 0 + ? → ? = 42
- Atomic number: 19 = -1 + ? → ? = 20
- Element with atomic number 20 is Calcium (Ca)
→ Product: ⁴²₂₀Ca
- Type: Beta decay (β⁻ emission) — electron emitted, neutron turns into proton.

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2. ²³⁹₉Pu → ⁴₂He + ______

- Mass: 239 = 4 + ? → ? = 235
- Atomic number: 94 = 2 + ? → ? = 92
- Element 92 is Uranium (U)
→ Product: ²³⁵₉₂U
- Type: Alpha decay (α emission)

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3. ²³⁵₉₂U → ______ + ²³¹₉₀Th

- Mass: 235 = ? + 231 → ? = 4
- Atomic number: 92 = ? + 90 → ? = 2
→ Particle: ⁴₂He (alpha particle)
- Type: Alpha decay

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4. ¹₁H + ³₁H → ______

- Mass: 1 + 3 = 4
- Atomic number: 1 + 1 = 2
→ Product: ⁴₂He
- Type: Fusion (two light nuclei combine)

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5. ⁶₃Li → ¹₀n → ⁴₂He + ______

Wait — this notation is odd. It says “⁶₃Li → ¹₀n → ...” — that doesn’t make sense as written. Probably it means:

⁶₃Li + ¹₀n → ⁴₂He + ______

That makes more sense for a nuclear reaction.

So:
- Mass: 6 + 1 = 4 + ? → ? = 3
- Atomic number: 3 + 0 = 2 + ? → ? = 1
→ Product: ³₁H (tritium)
- Type: Artificial transmutation (neutron bombardment)

*(Note: If the arrow was meant to be “+”, we assume that. Otherwise, if it's truly “→ n →”, it’s invalid. But in context, likely typo.)*

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6. ²⁷₁₃Al + ⁴₂He → ³⁰₁₅P + ______

- Mass: 27 + 4 = 30 + ? → ? = 1
- Atomic number: 13 + 2 = 15 + ? → ? = 0
→ Product: ¹₀n (neutron)
- Type: Artificial transmutation

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7. ⁹₄Be + ¹₁H → ______ + ⁴₂He

- Mass: 9 + 1 = ? + 4 → ? = 6
- Atomic number: 4 + 1 = ? + 2 → ? = 3
→ Product: ⁶₃Li
- Type: Artificial transmutation

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8. ³⁷₁₉K → ⁰₊₁e + ______

- Mass: 37 = 0 + ? → ? = 37
- Atomic number: 19 = +1 + ? → ? = 18
→ Element 18 is Argon (Ar)
→ Product: ³⁷₁₈Ar
- Type: Positron emission (β⁺)

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9. ______ + ¹₀n → ¹⁴²₅₆Ba + ⁹¹₃₆Kr + 3¹₀n

Left side: ? + 1n
Right side: Ba (142) + Kr (91) + 3n → total mass = 142 + 91 + 3 = 236
Atomic numbers: 56 + 36 + 0 = 92

So left side must have mass 236 - 1 = 235? Wait — no:

Actually, reactant + neutron → products including 3 neutrons.

So:

Mass: A + 1 = 142 + 91 + 3 → A + 1 = 236 → A = 235
Atomic number: Z + 0 = 56 + 36 → Z = 92

→ Reactant: ²³⁵₉₂U
- Type: Fission (heavy nucleus splits after neutron capture)

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10. ²³⁸₉₂U + ⁴₂He → ______ + ¹₀n

- Mass: 238 + 4 = ? + 1 → ? = 241
- Atomic number: 92 + 2 = ? + 0 → ? = 94
→ Element 94 is Plutonium (Pu)
→ Product: ²⁴¹₉₄Pu
- Type: Artificial transmutation

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11. ¹⁴₆C → ¹⁴₇N + ______

- Mass: 14 = 14 + ? → ? = 0
- Atomic number: 6 = 7 + ? → ? = -1
→ Particle: ⁰₋₁e (beta particle)
- Type: Beta decay (β⁻)

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12. ¹⁸⁷₇Re + ______ → ¹⁸₇₅Re + ¹₁H

Left: Re-187 + ?
Right: Re-188 + H-1 → mass 188 + 1 = 189; atomic 75 + 1 = 76

So left must sum to same.

Mass: 187 + A = 189 → A = 2
Atomic: 75 + Z = 76 → Z = 1

→ Particle: ²₁H (deuterium)

But wait — product has ¹₁H, which is proton. So maybe it’s:

Re-187 + ? → Re-188 + p

Then:

Mass: 187 + A = 188 + 1 → A = 2
Atomic: 75 + Z = 75 + 1 → Z = 1

Yes → ²₁H

But let’s check: Is this realistic? Actually, sometimes it’s written as adding a deuteron and emitting a proton — yes, possible.

Alternatively, could it be neutron? Let’s see:

If ? = ¹₀n → mass: 187+1=188, but right is 188+1=189 → no.

So must be ²₁H.

→ Reactant: ²₁H
- Type: Artificial transmutation

*(Note: Some might interpret as Re-187 absorbing something to become Re-188 and emit proton — so incoming particle must supply extra mass and charge. Deuteron fits.)*

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13. ²²₁₁Na + ______ → ²²₁₀Ne

Mass: 22 + A = 22 → A = 0
Atomic: 11 + Z = 10 → Z = -1

→ Particle: ⁰₋₁e (electron) — but that would be electron capture or beta plus?

Wait — if Na-22 becomes Ne-22, it loses one proton → so it must emit a positron or capture an electron.

But here it’s written as “+ ______ →”, so likely electron capture: Na-22 captures an electron to become Ne-22.

In electron capture:
²²₁₁Na + ⁰₋₁e → ²²₁₀Ne

Yes! That balances.

→ Reactant: ⁰₋₁e
- Type: Electron capture (sometimes grouped under positron emission category, but technically different — however, worksheet may list as "positron emission" loosely? But strictly, electron capture is separate. Looking at options: they list “positron emission” — but electron capture is not listed. Hmm.)

Wait — the problem says: “Determine the type... (α emission, β emission, γ emission, positron emission, artificial transmutation, fission, or fusion)”

Electron capture isn't listed. But in many curricula, electron capture is considered similar to positron emission (both reduce atomic number by 1). However, since it’s not explicitly listed, and the reaction requires an electron on the left, perhaps they expect “positron emission” — but that’s incorrect because positron emission emits a positron, doesn’t consume an electron.

Actually, looking back at #8: they had positron emission as ⁰₊₁e emitted.

Here, we’re consuming ⁰₋₁e — so it’s electron capture.

But since “electron capture” isn’t in the list, and the only way to get from Na to Ne is losing a proton, which can happen via positron emission OR electron capture.

In positron emission: ²²₁₁Na → ²²₁₀Ne + ⁰₊₁e

But here the equation is written as: ²²₁₁Na + ______ → ²²₁₀Ne

So it must be electron capture.

Perhaps the worksheet expects us to write the missing particle and call it “electron capture” even if not listed? Or maybe they consider it under “positron emission”? Unlikely.

Looking at standard classification: electron capture is a type of radioactive decay, often taught alongside positron emission.

Since the instruction says “determine the type” from the given list, and “electron capture” isn’t there, but “positron emission” is — and both result in same daughter — perhaps they want “positron emission” for any process that converts proton to neutron? But that’s inaccurate.

Wait — let’s check actual decay of Na-22: it decays by β⁺ (positron emission) to Ne-22, NOT by electron capture primarily. Na-22 half-life 2.6 years, decays 90% by β⁺, 10% by EC? Actually, I think it’s mostly β⁺.

But in this equation, it’s written with a blank on the left, implying a reactant — so it must be electron capture.

To resolve: perhaps the problem has a typo, and it should be → instead of +.

But as written, we must fill the blank.

So: ²²₁₁Na + ⁰₋₁e → ²²₁₀Ne

Type: Since “electron capture” isn’t listed, and the closest is “positron emission”, but that’s wrong. Alternatively, maybe they consider it “artificial transmutation”? No, it’s natural decay.

I think we should go with “electron capture” even if not listed, or note it.

But looking at the worksheet instructions: it says “(α emission, β emission, γ emission, positron emission, artificial transmutation, fission, or fusion)”

None fit perfectly. But “positron emission” is for emitting positron, not capturing electron.

However, in some contexts, they group them. To be safe, let’s see what the expected answer is.

Alternative interpretation: maybe the blank is for the emitted particle, but the arrow is wrong.

No — the format is consistent: for decay, it’s parent → daughter + particle.

For #11: ¹⁴C → ¹⁴N + ____ — that’s decay.

For #13: ²²Na + ____ → ²²Ne — this suggests a reaction where Na absorbs something to become Ne.

That only makes sense for electron capture.

Perhaps in this context, they accept “positron emission” as the type, even though the equation is written differently.

I recall that in some worksheets, they might write electron capture as requiring an electron on left.

And for type, since it’s not listed, but “positron emission” is the only one that reduces atomic number, perhaps they mean that.

To avoid confusion, let’s calculate correctly and assign based on common practice.

Upon second thought: in many high school texts, when they say “positron emission”, they include both emission and capture under that umbrella, but technically not.

Let me check online or standard: Na-22 decays by β⁺ emission: ²²Na → ²²Ne + e⁺ + ν

So the correct equation should be: ²²₁₁Na → ²²₁₀Ne + ⁰₊₁e

But here it’s written as + _____ → , which is backwards.

Probably a typo in the worksheet. Likely, it should be:

²²₁₁Na → ²²₁₀Ne + ______

Then: mass 22=22+0, atomic 11=10+1 → so ⁰₁e, positron emission.

That makes sense, and matches the pattern of other decays.

In #1,3,8,11,14 — all are decays with → and products on right.

#13 is written with + on left, which is unusual for decay.

Compare to #12: ¹⁸⁷Re + _____ → ¹⁸⁸Re + ¹H — that’s a reaction, not decay.

Similarly, #13 might be intended as a reaction, but Na to Ne by adding what? Only electron capture fits.

But to match the worksheet's likely intent, and since Na-22 commonly decays by positron emission, I think it's a formatting error, and it should be:

²²₁₁Na → ²²₁₀Ne + ______

Then product is ⁰₊₁e, type: positron emission.

I'll go with that, as it's standard.

So for #13: assume it's decay: ²²₁₁Na → ²²₁₀Ne + ⁰₊₁e

Type: positron emission

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14. ²¹⁸₈₄Po → ______ + ⁴₂He

- Mass: 218 = ? + 4 → ? = 214
- Atomic number: 84 = ? + 2 → ? = 82
→ Element 82 is Lead (Pb)
→ Product: ²¹⁴₈₂Pb
- Type: Alpha decay

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15. ²⁵³₉₉Es + ⁴₂He → ¹₀n + ______

- Mass: 253 + 4 = 1 + ? → ? = 256
- Atomic number: 99 + 2 = 0 + ? → ? = 101
→ Element 101 is Mendelevium (Md)
→ Product: ²⁵⁶₁₀₁Md
- Type: Artificial transmutation

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Now, compiling all answers:

1. ⁴²₂₀Ca ; beta emission
2. ²³⁵₉₂U ; alpha emission
3. ⁴₂He ; alpha emission
4. ⁴₂He ; fusion
5. ³₁H ; artificial transmutation *(assuming Li + n → He + H)*
6. ¹₀n ; artificial transmutation
7. ⁶₃Li ; artificial transmutation
8. ³⁷₁₈Ar ; positron emission
9. ²³⁵₉₂U ; fission
10. ²⁴¹₉₄Pu ; artificial transmutation
11. ⁰₋₁e ; beta emission
12. ²₁H ; artificial transmutation
13. ⁰₊₁e ; positron emission *(assuming decay format)*
14. ²¹⁴₈₂Pb ; alpha emission
15. ²⁵⁶₁₀₁Md ; artificial transmutation

For #5, if the original is "⁶Li → ¹n → ⁴He + ___", it's malformed. But in context, likely "⁶Li + ¹n → ⁴He + ___", so we'll go with that.

Final Answer:

1. \( ^{42}_{20}\text{Ca} \); beta emission
2. \( ^{235}_{92}\text{U} \); alpha emission
3. \( ^{4}_{2}\text{He} \); alpha emission
4. \( ^{4}_{2}\text{He} \); fusion
5. \( ^{3}_{1}\text{H} \); artificial transmutation
6. \( ^{1}_{0}\text{n} \); artificial transmutation
7. \( ^{6}_{3}\text{Li} \); artificial transmutation
8. \( ^{37}_{18}\text{Ar} \); positron emission
9. \( ^{235}_{92}\text{U} \); fission
10. \( ^{241}_{94}\text{Pu} \); artificial transmutation
11. \( ^{0}_{-1}\text{e} \); beta emission
12. \( ^{2}_{1}\text{H} \); artificial transmutation
13. \( ^{0}_{+1}\text{e} \); positron emission
14. \( ^{214}_{82}\text{Pb} \); alpha emission
15. \( ^{256}_{101}\text{Md} \); artificial transmutation