Nuclear Decay Worksheet Answers Chemistry Best Sale | pacomaeurope.com - Free Printable

Let's solve each nuclear reaction step by step. We will:

1. Balance the mass numbers (top numbers) and atomic numbers (bottom numbers) on both sides.
2. Identify the missing particle based on conservation of mass and atomic number.
3. Determine the type of nuclear reaction.

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1. $^{42}_{19}\text{K} \rightarrow ^0_{-1}e + \_\_\_\_\_\_\_\_$

- Beta emission ($\beta^-$): a neutron turns into a proton, emitting an electron.
- Mass: 42 → 42 + ? → ? = 42
- Atomic number: 19 → -1 + ? → ? = 20
- So, product is $^{42}_{20}\text{Ca}$

✔ Answer: $^{42}_{20}\text{Ca}$
✔ Type: $\beta$ emission

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2. $^{239}_{94}\text{Pu} \rightarrow ^4_2\text{He} + \_\_\_\_\_\_\_\_$

- Alpha decay: emits $^4_2\text{He}$
- Mass: 239 → 4 + ? → ? = 235
- Atomic number: 94 → 2 + ? → ? = 92
- So, product is $^{235}_{92}\text{U}$

✔ Answer: $^{235}_{92}\text{U}$
✔ Type: $\alpha$ emission

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3. $^{235}_{92}\text{U} \rightarrow \_\_\_\_\_\_\_\_ + ^{231}_{90}\text{Th}$

- Uranium-235 decays to Thorium-231
- Mass: 235 → ? + 231 → ? = 4
- Atomic number: 92 → ? + 90 → ? = 2
- So, emitted particle is $^4_2\text{He}$ (alpha)

✔ Answer: $^4_2\text{He}$
✔ Type: $\alpha$ emission

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4. $^1_1\text{H} + ^3_1\text{H} \rightarrow \_\_\_\_\_\_\_\_$

- Fusion of deuterium and tritium
- Mass: 1 + 3 = 4 → product has mass 4
- Atomic number: 1 + 1 = 2 → product has atomic number 2
- So, $^4_2\text{He}$ (helium nucleus)

✔ Answer: $^4_2\text{He}$
✔ Type: Fusion

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5. $^6_3\text{Li} \rightarrow ^1_0n + ^3_2\text{He} + \_\_\_\_\_\_\_\_$

- Lithium-6 splits into neutron and helium-3
- Mass: 6 → 1 + 3 + ? → ? = 2
- Atomic number: 3 → 0 + 2 + ? → ? = 1
- So, missing particle has mass 2, atomic number 1 → $^2_1\text{H}$ (deuterium)

✔ Answer: $^2_1\text{H}$
✔ Type: Artificial transmutation (or fission-like, but this is induced decay) — actually, this is a fission-type reaction, but not spontaneous. More precisely, it’s a nuclear reaction induced by bombardment, so likely artificial transmutation

Wait — this looks like a spontaneous decay? But lithium-6 doesn't decay this way naturally. This reaction is usually induced by neutron capture or similar. But since no reactant is shown on left, perhaps it's written as a decay? Actually, $^6_3\text{Li}$ can undergo neutron-induced fission.

But here, it's written as: $^6_3\text{Li} \rightarrow ^1_0n + ^3_2\text{He} + ?$

Let’s balance:
- Left: mass=6, atomic=3
- Right: n (mass=1), He-3 (mass=3), so total mass = 1+3 = 4 → need 2 more mass
- Atomic: 0 + 2 = 2 → need 1 more

So missing: mass=2, atomic=1 → $^2_1\text{H}$ (deuterium)

Yes: $^6_3\text{Li} \rightarrow ^1_0n + ^3_2\text{He} + ^2_1\text{H}$

This is a known reaction: lithium-6 absorbs a neutron and splits into two alpha particles? No — wait, this is not correct.

Wait! Actually, lithium-6 captures a neutron to form lithium-7, which then decays. But in some cases, $^6\text{Li}$ can be used in fusion reactions.

But here, no neutron on the left. The reaction is written as a decay, but $^6\text{Li}$ does not spontaneously emit a neutron and helium.

Wait — maybe it's a typo? Or perhaps we are to assume it's a reaction?

Alternatively, could this be: $^6_3\text{Li} \rightarrow ^3_2\text{He} + ^3_1\text{H}$? That would be a splitting, but not common.

But according to mass/charge:

We have:
- Left: mass 6, charge 3
- Right: $^1_0n$ (mass 1, charge 0), $^3_2\text{He}$ (mass 3, charge 2), so total right: mass 4, charge 2
- Need: mass 2, charge 1 → $^2_1\text{H}$

So yes: $^6_3\text{Li} \rightarrow ^1_0n + ^3_2\text{He} + ^2_1\text{H}$

But this is not a natural decay path. However, in nuclear physics, $^6\text{Li}$ can be used in fusion or transmutation.

But since it's written as a single species decaying, and it's balanced, we accept it.

Actually, this is not correct — $^6\text{Li}$ does not decay this way.

Wait — let’s recheck: Is there a known reaction?

Yes! In neutron capture, $^6\text{Li} + n \rightarrow ^3\text{He} + ^3\text{H}$

But here, no neutron on left.

So perhaps the reaction is miswritten.

But if we go purely by balancing:

Left: $^6_3\text{Li}$ → mass 6, atomic 3
Right: $^1_0n$ (1,0), $^3_2\text{He}$ (3,2), and unknown

So unknown must have mass = 6 - 1 - 3 = 2
Atomic = 3 - 0 - 2 = 1 → $^2_1\text{H}$

So answer is $^2_1\text{H}$

Even though not a natural decay, for the purpose of this worksheet, it's acceptable.

✔ Answer: $^2_1\text{H}$
✔ Type: Artificial transmutation (since it's not natural decay; likely induced)

But wait — if it's just one nucleus breaking apart, it might be fission, but fission is for heavy nuclei.

Lithium is light. This is more like artificial transmutation.

But without a projectile, it's odd.

Perhaps the reaction is meant to be: $^6_3\text{Li} + ^1_0n \rightarrow ^3_2\text{He} + ^3_1\text{H}$ — that's standard.

But here, no neutron on left.

So unless it's a typo, we'll proceed with the given.

So we’ll say: $^2_1\text{H}$

✔ Answer: $^2_1\text{H}$
✔ Type: Artificial transmutation (forced splitting)

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6. $^{27}_{13}\text{Al} + ^4_2\text{He} \rightarrow ^{30}_{15}\text{P} + \_\_\_\_\_\_\_\_$

- Aluminum bombarded with alpha particle → phosphorus
- Mass: 27 + 4 = 31 → 30 + ? → ? = 1
- Atomic: 13 + 2 = 15 → 15 + ? → ? = 0
- So, missing particle: mass 1, atomic 0 → $^1_0n$ (neutron)

✔ Answer: $^1_0n$
✔ Type: Artificial transmutation

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7. $^9_4\text{Be} + ^1_1\text{H} \rightarrow \_\_\_\_\_\_\_\_ + ^4_2\text{He}$

- Beryllium-9 + proton → ?
- Mass: 9 + 1 = 10 → ? + 4 → ? = 6
- Atomic: 4 + 1 = 5 → ? + 2 → ? = 3
- So, product: mass 6, atomic 3 → $^6_3\text{Li}$

✔ Answer: $^6_3\text{Li}$
✔ Type: Artificial transmutation

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

- Positron emission: proton → neutron + positron + neutrino
- Mass: 37 → 0 + ? → ? = 37
- Atomic: 19 → +1 + ? → ? = 18
- So, product: $^{37}_{18}\text{Ar}$

✔ Answer: $^{37}_{18}\text{Ar}$
✔ Type: Positron emission

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9. $\_\_\_\_\_\_\_\_ + ^1_0n \rightarrow ^{142}_{56}\text{Ba} + ^{91}_{36}\text{Kr} + 3^1_0n$

- Fission: heavy nucleus splits after absorbing neutron
- Total mass on right: 142 + 91 + 3(1) = 236
- Total atomic: 56 + 36 + 0 = 92
- Left: neutron (mass 1, atomic 0), so missing nucleus: mass = 236 - 1 = 235, atomic = 92 - 0 = 92 → $^{235}_{92}\text{U}$

✔ Answer: $^{235}_{92}\text{U}$
✔ Type: Fission

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10. $^{238}_{92}\text{U} + ^4_2\text{He} \rightarrow \_\_\_\_\_\_\_\_ + ^1_0n$

- Uranium-238 bombarded with alpha → ?
- Mass: 238 + 4 = 242 → ? + 1 → ? = 241
- Atomic: 92 + 2 = 94 → ? + 0 → ? = 94
- So, product: $^{241}_{94}\text{Pu}$

✔ Answer: $^{241}_{94}\text{Pu}$
✔ Type: Artificial transmutation

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11. $^{14}_6\text{C} \rightarrow ^{14}_7\text{N} + \_\_\_\_\_\_\_\_$

- Carbon-14 decays to nitrogen-14
- Mass: 14 → 14 + ? → ? = 0
- Atomic: 6 → 7 + ? → ? = -1
- So, emitted particle: mass 0, atomic -1 → $^0_{-1}e$ (beta minus)

✔ Answer: $^0_{-1}e$
✔ Type: $\beta$ emission

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12. $^{187}_{75}\text{Re} + \_\_\_\_\_\_\_\_ \rightarrow ^{188}_{75}\text{Re} + ^1_1\text{H}$

- Rhenium-187 + ? → Rhenium-188 + proton
- Mass: 187 + ? = 188 + 1 → ? = 2
- Atomic: 75 + ? = 75 + 1 → ? = 1
- So, missing: mass 2, atomic 1 → $^2_1\text{H}$ (deuterium)

✔ Answer: $^2_1\text{H}$
✔ Type: Artificial transmutation

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13. $^{22}_{11}\text{Na} + \_\_\_\_\_\_\_\_ \rightarrow ^{22}_{10}\text{Ne}$

- Sodium-22 → Neon-22
- Mass: 22 + ? = 22 → ? = 0
- Atomic: 11 + ? = 10 → ? = -1
- So, missing: mass 0, atomic -1 → $^0_{-1}e$ (beta emission)

But wait — sodium-22 decays by positron emission to neon-22.

Check: $^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^0_{+1}e + \nu_e$

So here, it's written as: Na + X → Ne

So X must be $^0_{+1}e$? But that doesn't make sense — you don’t add a positron to sodium.

Wait — the reaction is: $^{22}_{11}\text{Na} + \_\_\_\_\_\_\_\_ \rightarrow ^{22}_{10}\text{Ne}$

But this implies a bombardment.

But Na-22 decays to Ne-22 via positron emission.

So if it's a reaction, then:

Mass: 22 + ? = 22 → ? = 0
Atomic: 11 + ? = 10 → ? = -1 → $^0_{-1}e$

But that would be beta minus, which is wrong.

Alternatively, maybe it's: $^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^0_{+1}e$ — but no reactant on left.

But here, the format is: Na + X → Ne

So X must have mass 0, atomic -1 → $^0_{-1}e$ — but that's impossible.

Wait — perhaps it's a typo, and it should be:

$^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^0_{+1}e$ — but the blank is on the left.

Alternatively, maybe it's: $^{22}_{11}\text{Na} + ^0_{-1}e \rightarrow ^{22}_{10}\text{Ne} + \text{something}$ — but not matching.

Wait — perhaps it's electron capture?

Electron capture: nucleus captures orbital electron → $^{22}_{11}\text{Na} + ^0_{-1}e \rightarrow ^{22}_{10}\text{Ne} + \nu_e$

Yes! That makes sense.

So: $^{22}_{11}\text{Na} + ^0_{-1}e \rightarrow ^{22}_{10}\text{Ne} + \nu_e$

But the problem says: $^{22}_{11}\text{Na} + \_\_\_\_\_\_\_\_ \rightarrow ^{22}_{10}\text{Ne}$

It doesn't show the neutrino, so we assume it's implied.

So missing reactant is $^0_{-1}e$

✔ Answer: $^0_{-1}e$
✔ Type: Electron capture (which is a type of artificial transmutation or decay — but technically, electron capture is a decay process, but often classified under β⁺ decay or internal conversion)

But in many contexts, electron capture is considered a type of β decay.

But strictly, it's electron capture.

However, in many textbooks, it's grouped under positron emission or beta decay.

But here, since it's not emitting a positron, but capturing an electron, it's electron capture.

But the options given are: α, β, γ, positron emission, artificial transmutation, fission, fusion.

"Positron emission" is listed, but electron capture is not.

But sometimes electron capture is called K-capture, and may be grouped under artificial transmutation.

But Na-22 decays via positron emission, not electron capture.

Wait — actually, sodium-22 decays primarily by positron emission, not electron capture.

So why is this written as Na + e⁻ → Ne?

That would be electron capture, but Na-22 does not do that.

So perhaps it's a mistake.

But let's check: If we want: Na + X → Ne

Then:

Mass: 22 + A = 22 → A = 0
Atomic: 11 + Z = 10 → Z = -1 → X = $^0_{-1}e$

So mathematically, it's correct.

But physically, sodium-22 does not decay via electron capture — it decays via positron emission.

So this reaction is not how it happens.

But perhaps it's a hypothetical or different isotope.

Alternatively, maybe it's a reaction where a neutron is captured or something.

Wait — another possibility: Neon-22 is formed from sodium-22 via positron emission, so the reverse is not physical.

So unless it's electron capture, but that’s not typical.

But in any case, mathematically, the only way to balance is:

X = $^0_{-1}e$

And the reaction is electron capture, which is a type of decay, but not listed directly.

But "positron emission" is listed, not "electron capture".

So perhaps the intended answer is electron capture, but since it's not in the list, maybe they expect artificial transmutation.

But electron capture is a natural decay process, not artificial.

So perhaps the reaction is incorrect.

Wait — maybe it's: $^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^0_{+1}e$ — but then the blank is on the left, so it's not.

Given the format, the only possible answer is $^0_{-1}e$, even if it's not the actual decay mode.

But in reality, electron capture is not the primary decay mode for Na-22.

Alternatively, perhaps the reaction is: $^{22}_{11}\text{Na} + ^0_{-1}e \rightarrow ^{22}_{10}\text{Ne} + \nu_e$

So the missing reactant is $^0_{-1}e$

And the type is electron capture — but since it's not in the list, perhaps it's artificial transmutation or β emission?

But electron capture is not β emission.

Alternatively, maybe the problem intends positron emission, but the arrow is reversed.

I think there's a mistake in the question.

But for now, we'll go with the balance.

✔ Answer: $^0_{-1}e$
✔ Type: Artificial transmutation (since it involves capture of an external electron, though rare)

But better: electron capture — but since not listed, perhaps β decay is expected? No.

Alternatively, maybe it's a typo, and it should be $^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^0_{+1}e$ — then no reactant needed.

But the blank is on the left.

So we'll go with $^0_{-1}e$ and label it electron capture, but since not in options, perhaps artificial transmutation.

But electron capture is a decay process, not artificial.

So perhaps the intended answer is positron emission, but the reaction is written backward.

Wait — look at the arrow: Na + X → Ne

So it's forming Ne from Na, which is what happens in decay.

But in decay, it's Na → Ne + e⁺

So to write it as a reaction, it's not Na + X → Ne

Unless X is a neutron or something.

No.

Another possibility: Neon-22 is produced when sodium-22 emits a positron, so the missing product is $^0_{+1}e$, but here the blank is on the left.

So I think the reaction is incorrectly written.

But given the format, the only way to balance is:

X = $^0_{-1}e$

So we'll accept it.

✔ Answer: $^0_{-1}e$
✔ Type: Artificial transmutation (as it requires external electron)

But note: this is not accurate for Na-22, but for the sake of the problem, we go with balance.

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14. $^{218}_{84}\text{Po} \rightarrow \_\_\_\_\_\_\_\_ + ^4_2\text{He}$

- Polonium-218 decays by alpha emission
- Mass: 218 → ? + 4 → ? = 214
- Atomic: 84 → ? + 2 → ? = 82
- So, $^{214}_{82}\text{Pb}$

✔ Answer: $^{214}_{82}\text{Pb}$
✔ Type: $\alpha$ emission

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15. $^{253}_{99}\text{Es} + ^4_2\text{He} \rightarrow ^1_0n + \_\_\_\_\_\_\_\_$

- Einsteinium-253 + alpha → neutron + ?
- Mass: 253 + 4 = 257 → 1 + ? → ? = 256
- Atomic: 99 + 2 = 101 → 0 + ? → ? = 101
- So, product: $^{256}_{101}\text{Md}$ (Mendelevium)

✔ Answer: $^{256}_{101}\text{Md}$
✔ Type: Artificial transmutation

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## ✔ Final Answers:

| # | Missing Product/Reactant | Type of Nuclear Reaction |
|---|--------------------------|--------------------------|
| 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 | $^2_1\text{H}$ | Artificial transmutation |
| 6 | $^1_0n$ | 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 | $^0_{-1}e$ | $\beta$ emission |
|12 | $^2_1\text{H}$ | Artificial transmutation |
|13 | $^0_{-1}e$ | Artificial transmutation (electron capture) |
|14 | $^{214}_{82}\text{Pb}$ | $\alpha$ emission |
|15 | $^{256}_{101}\text{Md}$ | Artificial transmutation |

> Note: For #13, the reaction is unusual because sodium-22 typically decays by positron emission, not electron capture. However, mathematically, the only way to balance is with $^0_{-1}e$. It may represent electron capture, which is a valid decay mode for some isotopes, but not for Na-22. Still, for the purpose of balancing, we accept it. The type is best labeled as artificial transmutation since it involves capture of an external electron.