Solved BALANCING NUCLEAR REACTIONS WORKSHEET Predict the | Chegg.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 numbers.
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 number: 42 → 42 + 0 ⇒ unchanged
- Atomic number: 19 → 18 + (-1) ⇒ 19 → 18 + 1 = 19 ✔
- So, the product has mass 42, atomic number 20 → $^{42}_{20}\text{Ca}$

✔ 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 + x ⇒ x = 235
- Atomic: 94 → 2 + y ⇒ y = 92 → $^{235}_{92}\text{U}$

✔ 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 → x + 231 ⇒ x = 4
- Atomic: 92 → y + 90 ⇒ y = 2 → $^4_2\text{He}$

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

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

- Fusion of hydrogen isotopes (protium and tritium)
- Mass: 1 + 3 = 4
- Atomic: 1 + 1 = 2 → $^4_2\text{He}$

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

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5. $^6_3\text{Li} \rightarrow ^1_0\text{n} + ^3_2\text{He} + \_\_\_\_\_\_\_\_$

- Lithium-6 splits into neutron, helium-3, and another particle
- Mass: 6 → 1 + 3 + x ⇒ x = 2
- Atomic: 3 → 0 + 2 + y ⇒ y = 1 → $^2_1\text{H}$ or deuterium

✔ Answer: $^2_1\text{H}$
Type: Artificial transmutation (or fission-like, but induced)

Wait — this is actually a spontaneous reaction? No, lithium-6 doesn't decay this way naturally. But in nuclear reactions, this can be induced. This is more like artificial transmutation via neutron bombardment? Wait — no, here it's written as decay.

But actually, $^6\text{Li}$ does not undergo spontaneous fission like this. However, if we balance:

Left: $^6_3\text{Li}$
Right: $^1_0\text{n} + ^3_2\text{He} + ?$

Mass: 6 = 1 + 3 + x ⇒ x = 2
Charge: 3 = 0 + 2 + y ⇒ y = 1 ⇒ $^2_1\text{H}$

So yes, $^2_1\text{H}$ (deuterium). This is a known reaction in nuclear physics:
$^6_3\text{Li} + n \rightarrow ^3_2\text{He} + ^3_1\text{H}$ — but here it's not showing a neutron input.

Wait — this reaction as written seems unbalanced unless it’s a decay.

But $^6\text{Li}$ does not spontaneously emit a neutron and helium-3. This is likely a misrepresentation, but assuming the equation is correct, we must find what completes it.

Wait — let’s recheck:
Left: mass=6, charge=3
Right: $^1_0n$: mass=1, charge=0; $^3_2\text{He}$: mass=3, charge=2 → total mass=4, charge=2
So missing: mass=2, charge=1 → $^2_1\text{H}$

Yes, so $^2_1\text{H}$

✔ Answer: $^2_1\text{H}$
Type: Artificial transmutation (this is a common reaction when Li-6 absorbs a neutron, but here it's shown without input — possibly error, but mathematically it balances)

Alternatively, perhaps it's a typo and should include a neutron. But as written, we go with balancing.

But wait — if it's a decay, it would need to be induced. So likely artificial transmutation.

✔ Type: Artificial transmutation

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

- Alpha particle bombards aluminum → phosphorus-30
- Mass: 27 + 4 = 31 → 30 + x ⇒ x = 1
- Charge: 13 + 2 = 15 → 15 + y ⇒ y = 0 → $^1_0\text{n}$

✔ Answer: $^1_0\text{n}$
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 → x + 4 ⇒ x = 6
- Charge: 4 + 1 = 5 → y + 2 ⇒ y = 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 ($\beta^+$): proton → neutron + positron + neutrino
- Mass: 37 → 0 + x ⇒ x = 37
- Charge: 19 → (+1) + y ⇒ y = 18 → $^{37}_{18}\text{Ar}$

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

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9. $\_\_\_\_\_\_\_\_ + ^1_0\text{n} \rightarrow ^{142}_{56}\text{Ba} + ^{91}_{36}\text{Kr} + 3^1_0\text{n}$

- Fission: heavy nucleus split by neutron
- Total mass on right: 142 + 91 + 3×1 = 236
- Total charge: 56 + 36 + 0 = 92
- Left: neutron has mass 1, charge 0 → unknown nucleus has mass 235, charge 92 → $^{235}_{92}\text{U}$

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

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10. $^{238}_{92}\text{U} + ^4_2\text{He} \rightarrow \_\_\_\_\_\_\_\_ + ^1_0\text{n}$

- Uranium-238 hit by alpha → ?
- Mass: 238 + 4 = 242 → x + 1 ⇒ x = 241
- Charge: 92 + 2 = 94 → y + 0 ⇒ y = 94 → $^{241}_{94}\text{Pu}$

✔ 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 ⇒ balanced
- Charge: 6 → 7 + x ⇒ x = -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 + x = 188 + 1 ⇒ x = 2
- Charge: 75 + y = 75 + 1 ⇒ y = 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 + x = 22 ⇒ x = 0
- Charge: 11 + y = 10 ⇒ y = -1 → $^0_{-1}e$

But that would be beta decay — but sodium-22 actually undergoes positron emission.

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

But here, it’s written as reactant, so:

Left: $^{22}_{11}\text{Na} + X \rightarrow ^{22}_{10}\text{Ne}$

So mass: 22 + A = 22 ⇒ A = 0
Charge: 11 + Z = 10 ⇒ Z = -1 → $^0_{-1}e$

But that’s not physical — you don’t add electrons to make a positron decay.

Wait — maybe 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’s correct.

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

✔ Answer: $^0_{-1}e$
Type: Electron capture (but not listed — closest is artificial transmutation or β emission? Actually, electron capture is a type of decay, but not listed.)

Wait — options are:
- α emission, β emission, γ emission, positron emission, artificial transmutation, fission, fusion

Electron capture is not listed, but β emission is usually for β⁻, and positron emission is β⁺.

But here, the reaction is electron capture, which is not among the choices.

But note: $^{22}_{11}\text{Na}$ decays by positron emission:
$^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^0_{+1}e + \nu_e$

So if this is a decay, it should be positron emission.

But here it's written as: $^{22}_{11}\text{Na} + X \rightarrow ^{22}_{10}\text{Ne}$

So if it's positron emission, then X is nothing — but here it's a reactant.

So if it's electron capture, then X = $^0_{-1}e$

And electron capture is not listed. But sometimes grouped under artificial transmutation or β emission?

But electron capture is a decay process.

Wait — perhaps the problem expects positron emission, but the arrow shows reactants.

But in this case, the equation is not a decay — it's a reaction with a reactant.

So only possible explanation is electron capture, so missing reactant is $^0_{-1}e$

But since "electron capture" isn't listed, and artificial transmutation includes induced processes, but electron capture is natural.

Alternatively, maybe it's a typo.

But let’s check:
If $^{22}_{11}\text{Na} + ^0_{-1}e \rightarrow ^{22}_{10}\text{Ne}$, then yes.

But in reality, $^{22}\text{Na}$ decays by positron emission, not electron capture.

Wait — actually, both occur? No, $^{22}\text{Na}$ primarily decays by positron emission.

So why would it be written as needing an electron?

Possibly a mistake.

But if we follow the math:

Left: mass = 22 + 0 = 22
Charge = 11 + (-1) = 10 → matches right side

So yes, $^0_{-1}e$

But type — electron capture is not listed. The closest is artificial transmutation, but it's natural.

Alternatively, perhaps the problem meant to show positron emission, but wrote it backward.

But as written, it's electron capture, so we’ll go with:

✔ Answer: $^0_{-1}e$
Type: Artificial transmutation (since it's not spontaneous in this form — but technically, electron capture is natural decay)

But since electron capture is not listed, and positron emission is, but that’s different.

Wait — maybe the reaction is positron emission, but written incorrectly.

No — the arrow is from Na + X → Ne, so X must be added.

Only possibility is $^0_{-1}e$ for electron capture.

But since the option list doesn't include "electron capture", and artificial transmutation is the closest for induced reactions, but this is natural.

Alternatively, maybe it's a typo, and it's supposed to be:

$^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^0_{+1}e$ → then type: positron emission

But here it's written with a reactant.

So unless it's induced, but unlikely.

We’ll go with:

✔ Answer: $^0_{-1}e$
Type: Artificial transmutation (as best fit)

But note: electron capture is not one of the listed types. So perhaps the intended answer is positron emission, but that would require different notation.

Wait — maybe the reaction is:

$^{22}_{11}\text{Na} \rightarrow ^{22}_{10}\text{Ne} + ^0_{+1}e$ → but then no reactant.

But here, it’s written as reactant, so must be electron capture.

So I'll keep:

✔ Answer: $^0_{-1}e$
Type: Artificial transmutation (best available choice)

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

- Polonium-218 decays by alpha emission
- Mass: 218 → x + 4 ⇒ x = 214
- Charge: 84 → y + 2 ⇒ y = 82 → $^{214}_{82}\text{Pb}$

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

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15. $^{253}_{99}\text{Es} + ^4_2\text{He} \rightarrow ^1_0\text{n} + \_\_\_\_\_\_\_\_$

- Einsteinium-253 + alpha → neutron + ?
- Mass: 253 + 4 = 257 → 1 + x ⇒ x = 256
- Charge: 99 + 2 = 101 → 0 + y ⇒ y = 101 → $^{256}_{101}\text{Md}$ (Mendelevium)

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

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Final Answers:

| # | Missing Particle/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 | $^1_0\text{n}$ | 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 | $^0_{-1}e$ | $\beta$ 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, although the actual decay of $^{22}\text{Na}$ is positron emission, the given reaction implies electron capture, which requires $^0_{-1}e$. Since "electron capture" is not listed, "Artificial transmutation" is used as the closest match.

Let me know if you'd like the answers formatted differently!