Balancing Nuclear Equations Lesson Plans & Worksheets - Free Printable

Let’s solve each part step by step.

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Problem 1: Balance the nuclear equations

We need to make sure that both the mass numbers (top) and atomic numbers (bottom) are equal on both sides of the equation.

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a.
₁₂²⁶Mg + ₂⁴He → _Z^A X + ₀¹n

Left side:
- Mass: 26 + 4 = 30
- Atomic number: 12 + 2 = 14

Right side:
- We have a neutron: mass 1, atomic number 0
- So unknown particle must have:
- Mass: 30 - 1 = 29
- Atomic number: 14 - 0 = 14 → element with Z=14 is Silicon (Si)

✔ Answer: ₁₄²⁹Si

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b.
_? + ₅₉²⁷Co → ₅₆²⁵Mn + ₂⁴He

Wait — this looks backwards. Usually we write reactants → products. But here it's written as “something + Co → Mn + He”

So let’s treat it as:

X + ₂₇⁵⁹Co → ₂₅⁵⁶Mn + ₂⁴He

Left: X + Co → right: Mn + He

Mass balance:
A_X + 59 = 56 + 4 → A_X = 60 - 59 = 1

Atomic number balance:
Z_X + 27 = 25 + 2 → Z_X = 27 - 27 = 0

So particle has mass 1, charge 0 → that’s a neutron!

✔ Answer: ₀¹n

But wait — the blank is before the arrow? Let me check original:

Original says:
“b. ___ + ₅₉²⁷Co → ₅₆²⁵Mn + ₂⁴He”

Yes — so yes, it’s a neutron hitting cobalt to produce manganese and alpha.

✔ Final for b: ₀¹n

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c.
₉₂²³⁵U + ___ → ₃₈⁹⁰Sr + ₅¹⁴³Xe + 3 ₀¹n

Left: U-235 + ? → Sr-90 + Xe-143 + 3 neutrons

Mass left: 235 + A_?
Mass right: 90 + 143 + 3×1 = 236

So: 235 + A_? = 236 → A_? = 1

Atomic number left: 92 + Z_?
Right: 38 + 54 + 0 = 92

So: 92 + Z_? = 92 → Z_? = 0

Again, neutron!

✔ Answer: ₀¹n

This is typical fission reaction triggered by neutron.

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d.
₉₂²³⁸U + ₂⁴He → ₂Jₙ + ___

Wait — probably typo. Likely meant:
₉₂²³⁸U + ₂⁴He → _Z^A X + ₀¹n ? Or maybe two particles?

Looking at notation: “→ ₂Jₙ + ___” — likely “→ _Z^A X + ₀¹n” or similar.

Actually, looking again:
"d. ₉₂²³⁸U + ₂⁴He → ₂Jₙ + ___"

Probably “₂Jₙ” is a typo for “₀¹n” (neutron). Maybe J is miswritten n?

Assume:
₉₂²³⁸U + ₂⁴He → ₀¹n + _Z^A X

Then:

Mass: 238 + 4 = 242 → right: 1 + A → A = 241
Atomic: 92 + 2 = 94 → right: 0 + Z → Z = 94 → Plutonium (Pu)

✔ Answer: ₉₄²⁴¹Pu

But if “₂Jₙ” means something else... Hmm.

Alternative interpretation: Maybe “₂Jₙ” is meant to be “₂⁴He”? No, already used.

Wait — perhaps it’s “→ _Z^A X + ₂⁴He”? But then why write “₂Jₙ”?

Another possibility: In some fonts, “n” looks like “J”. Probably “₀¹n”.

I’ll go with that.

✔ d: ₉₄²⁴¹Pu

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e.
₂₀⁴⁶Ca + ___ → ₁₈⁴⁰Ar + ₂⁴He

Left: Ca-46 + ? → Ar-40 + He-4

Mass: 46 + A = 40 + 4 = 44 → A = -2? That can’t be.

Wait — that doesn’t work.

Unless it’s emission? But arrow is from left to right.

Perhaps it’s:
₂₀⁴⁶Ca → ₁₈⁴⁰Ar + ₂⁴He + ?

That would make sense — decay.

Check:

Mass: 46 → 40 + 4 = 44 → missing 2
Atomic: 20 → 18 + 2 = 20 → okay

So missing particle: mass 2, charge 0 → that’s not standard.

Wait — maybe it’s two neutrons?

Try:
₂₀⁴⁶Ca → ₁₈⁴⁰Ar + ₂⁴He + 2 ₀¹n

Mass: 46 = 40 + 4 + 2 → 46 = 46 ✔
Atomic: 20 = 18 + 2 + 0 → 20 = 20 ✔

Perfect.

But the equation is written as:
“e. ₂₀⁴⁶Ca + ___ → ₁₈⁴⁰Ar + ₂⁴He”

So if we add something on left, but we need to remove mass? Doesn't make sense.

Unless it’s a reaction where calcium absorbs something and emits argon and helium? Unlikely.

More likely: typo in direction. Should be decay:
₂₀⁴⁶Ca → ₁₈⁴⁰Ar + ₂⁴He + 2 ₀¹n

But since the problem writes “+ ___ →”, perhaps they mean what is emitted besides Ar and He?

In that case, the blank is on the product side? But it’s written on reactant side.

Look back at original image description — user said “filling in the blanks”, and for e it’s:

“e. ₂₀⁴⁶Ca + ___ → ₁₈⁴⁰Ar + ₂⁴He”

This implies adding something to Ca to get Ar + He.

But mass: 46 + A = 40 + 4 = 44 → A = -2 impossible.

Unless it’s not addition — maybe it’s a mistake, and it should be:

₂₀⁴⁶Ca → ₁₈⁴⁰Ar + ₂⁴He + ___

Then blank is 2 ₀¹n

Given context, I think it’s a formatting error. Most likely, the blank is for the additional product.

In many textbooks, such problems sometimes put the blank on wrong side.

Alternatively, perhaps it’s electron capture or something, but unlikely.

Another idea: maybe “+ ___” is for energy or gamma, but mass wouldn’t balance.

Best guess: intended to be decay, so blank is 2 neutrons on product side.

But since the blank is on reactant side, perhaps no solution? That can’t be.

Wait — let’s calculate difference:

From Ca-46 to Ar-40 + He-4: loss of 2 protons and 2 neutrons? But He takes 2p+2n, so total change: Ca loses 2p and 2n to become Ar? But Ar has 18p, Ca has 20p — yes, losing 2p and 2n gives Ar-40? 46 - 4 = 42, but Ar is 40 — inconsistency.

Ca-46: 20p, 26n
Ar-40: 18p, 22n
He-4: 2p, 2n
Total products: 20p, 24n — but Ca had 26n, so 2 extra neutrons.

So yes: ₂₀⁴⁶Ca → ₁₈⁴⁰Ar + ₂⁴He + 2 ₀¹n

Therefore, if the equation is written as “Ca + ___ → Ar + He”, it must be that ___ is nothing, and there are additional products. But the blank is only one.

Perhaps the blank is for the two neutrons, but placed incorrectly.

I think safest is to assume the blank is for the missing product, even though written on left.

Maybe in some notations, but I doubt.

Another possibility: it’s a reaction with a photon or something, but mass won’t balance.

Let’s look at f: similar issue.

f.
₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n

Left: Cm-246 + ? → Cf-254 + 4n

Mass: 246 + A = 254 + 4 = 258 → A = 12
Atomic: 96 + Z = 98 + 0 = 98 → Z = 2 → Helium? But He is 4, not 12.

Z=2, A=12? Not possible.

Cf is 98, Cm is 96, so increase of 2 in atomic number.

To go from Cm to Cf, need to add 2 protons.

Common way: bombard with alpha particle (He-4), which adds 2p and 2n.

Here: Cm-246 + He-4 → Cf-250? But we have Cf-254 and 4 neutrons.

Calculate:

If Cm-246 + X → Cf-254 + 4n

Mass: 246 + A_X = 254 + 4 = 258 → A_X = 12
Atomic: 96 + Z_X = 98 → Z_X = 2

So particle with Z=2, A=12 — that’s not real. Carbon-12 has Z=6.

Mistake.

Perhaps it’s Cm-246 + C-12 → Cf-254 + 4n? Check:

Mass: 246 + 12 = 258; 254 + 4 = 258 ✔
Atomic: 96 + 6 = 102; Cf is 98, plus 0 from neutrons = 98 — not match.

102 vs 98 — off by 4.

Not good.

Another common reaction: use carbon to make heavier elements.

For example, Cm-246 + C-12 → No-254 + 4n? Nobelium is 102.

Cf is 98.

Perhaps it’s Cm-246 + O-18 or something.

Let’s solve properly.

Let the unknown be _Z^A X

Then:

Mass: 246 + A = 254 + 4*1 = 258 → A = 12

Atomic: 96 + Z = 98 + 0 = 98 → Z = 2

So _2^12 X — but no such nucleus. Closest is He-4, but A=4.

Perhaps the 4 neutrons are not all from the reaction? Or typo.

Maybe it’s Cm-246 + He-4 → Cf-249 + n or something.

Standard reaction: to make Cf, often use Cm + He.

For example: ₉₆²⁴⁴Cm + ₂⁴He → ₉₈²⁴⁷Cf + ₁¹H or something.

But here masses don't match.

Perhaps "4 ₀¹n" is incorrect.

Another idea: maybe it’s Cm-246 + C-12 → Cf-254 + 4n, but atomic numbers: 96+6=102, Cf is 98, so need to emit 4 protons or something, not neutrons.

Not making sense.

Let’s calculate the difference in atomic number: from 96 to 98, so +2, so likely alpha particle.

Mass: Cm-246 + He-4 = 250, but Cf-254 is heavier, so impossible.

Unless it’s multiple steps, but for single equation.

Perhaps the Cf is 250, not 254.

Check online or standard knowledge.

Upon second thought, a common reaction is:

₉₆²⁴⁶Cm + ₆¹²C → ₁₀₂²⁵⁴No + 4 ₀¹n (Nobelium)

But here it’s written as Cf, which is 98.

Typo in problem? Probably meant No, not Cf.

Because 96 + 6 = 102, and 246 + 12 = 258, 254 + 4 = 258, perfect.

And Nobelium is element 102.

Whereas Cf is 98.

So likely, "₉₈²⁵⁴Cf" is a typo, should be "₁₀₂²⁵⁴No".

Otherwise, no solution.

Given that, I'll assume it's No.

So for f: ₆¹²C

But the problem says Cf, so perhaps not.

Another possibility: Cm-246 + Be-9 or something.

Let's try Z=4, A=8: Be-8 unstable.

Or B-11: Z=5, A=11.

96+5=101, not 98.

To get to 98, need Z=2, as before.

Perhaps it's Cm-246 + He-4 -> Cf-249 + n, but here it's Cf-254 and 4n.

Difference in mass: 254 - 246 = 8, plus 4 from neutrons, so incoming particle must provide 12 mass units, and 2 protons.

So only possibility is carbon-12, but then atomic number 96+6=102, not 98.

Unless the product is not Cf, but something else.

Perhaps "₉₈²⁵⁴Cf" is wrong, and it's ₁₀₂²⁵⁴No.

I think that's the most reasonable assumption.

So I'll go with ₆¹²C for f.

But to match the problem, let's see the answer format.

Perhaps for e and f, the blank is for the bombarding particle, and for e, it's impossible, so maybe e is decay.

Let's list what we have:

a. ₁₄²⁹Si

b. ₀¹n

c. ₀¹n

d. ₉₄²⁴¹Pu (assuming ₂Jₙ is ₀¹n)

e. ?

f. ?

For e: ₂₀⁴⁶Ca + ___ → ₁₈⁴⁰Ar + ₂⁴He

As calculated, mass deficit, so perhaps it's ₂₀⁴⁶Ca → ₁₈⁴⁰Ar + ₂⁴He + 2 ₀¹n, so if the blank is for the 2 neutrons, but it's on left.

Maybe the "+" is a mistake, and it's just Ca -> Ar + He + ___, so blank is 2 ₀¹n.

Similarly for f, if we assume Cf is typo for No, then ₆¹²C.

Perhaps for f, it's Cm-246 + He-4 -> Cf-249 + n, but not matching.

Another idea: in f, "4 ₀¹n" might be "1 ₀¹n" or something.

Let's calculate what would work for Cf.

Suppose Cm-246 + X -> Cf-254 + 4n

A_X = 254 +4 -246 = 12

Z_X = 98 -96 = 2

So _2^12 X — not real.

If it's Cm-246 + X -> Cf-250 + 0n, then A_X=4, Z_X=2, so He-4.

But it says Cf-254 and 4n.

Perhaps the 4n is for a different reaction.

I recall that californium is made by bombarding curium with carbon, but produces nobelium.

For example: ²⁴⁶Cm + ¹²C → ²⁵⁴No + 4n

Yes, that's standard.

So likely, "Cf" is a typo, and it should be "No".

I'll proceed with that.

So for f: ₆¹²C

For e, similarly, perhaps it's a decay, so the blank is for the emitted particles, but since it's on left, maybe no answer, but that can't be.

Another thought for e: perhaps " + ___ " is for energy, but usually not.

Or perhaps it's electron capture, but then atomic number decreases by 1, not 2.

Ca to Ar is decrease of 2 in Z, so alpha decay or something.

Alpha decay of Ca-46 would give Ar-42, not Ar-40.

Ca-46 -> Ar-42 + He-4, but here it's Ar-40, so not.

So must be with additional neutrons.

I think for e, the intended answer is that the blank is for 2 neutrons on the product side, but since the blank is on reactant side, perhaps the equation is miswritten, and we should put the missing product.

To resolve, let's look at the last part of problem 1: "g. ₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n" — same as f, but g is separate.

In the user's text, it's:

"d. ₉₂²³⁸U + ₂⁴He → ₂Jₙ + ___ "

"e. ₂₀⁴⁶Ca + ___ → ₁₈⁴⁰Ar + ₂⁴He"

"f. ₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n"

"g. ₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n" — wait, f and g are the same? No, in user's message, it's:

After d, e, f, then g is "₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n" — but f is also that? Let's read:

User wrote:

"d. ₉₂²³⁸U + ₂⁴He → ₂Jₙ + ___ "

"e. ₂₀⁴⁶Ca + ___ → ₁₈⁴⁰Ar + ₂⁴He"

"f. ₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n"

"g. ₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n" — oh, f and g are identical? That can't be.

In the initial post, it's:

"d. ₉₂²³⁸U + ₂⁴He → ₂Jₙ + ___ "

"e. ₂₀⁴⁶Ca + ___ → ₁₈⁴⁰Ar + ₂⁴He"

"f. ₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n"

"g. ₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n" — yes, f and g are the same. Probably a copy-paste error.

Perhaps g is different.

In the user's message: "g. ₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n" — same as f.

But earlier in d, there is "2J_n", which might be "2 0^1 n" i.e., two neutrons.

For d: ₉₂²³⁸U + ₂⁴He → 2 ₀¹n + ___

Then mass: 238 + 4 = 242; right: 2*1 + A = 2 + A, so A=240

Atomic: 92 + 2 = 94; right: 0 + Z, so Z=94 → Pu-240

So d: ₉₄²⁴⁰Pu

And "2J_n" is "2 0^1 n" — two neutrons.

That makes sense.

For e: ₂₀⁴⁶Ca + ___ → ₁₈⁴Ar + ₂⁴He

Still problem.

Perhaps for e, it's ₂₀⁴⁶Ca → ₁₈⁴⁰Ar + ₂⁴He + 2 ₀¹n, so if the blank is for the 2 neutrons, but on left, maybe the "+" is for the products.

I think for e, the intended blank is for the 2 neutrons on the product side, so we can write the answer as 2 ₀¹n, even though position is wrong.

Similarly for f and g, if they are the same, perhaps one is different.

In user's message, after f, it's "g. ₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n" — same as f.

But perhaps g is for a different thing.

Let's assume that for e, the blank is for the missing product, so 2 ₀¹n.

For f and g, since they are identical, perhaps it's a mistake, and g is for something else, but in the text, it's the same.

Perhaps "g" is "₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n" and f is the same, so same answer.

But for f, as discussed, with Cf, it doesn't work, so likely Cf is typo for No.

I think for the sake of completing, I'll use:

For e: the missing part is 2 ₀¹n (emitted)

For f and g: ₆¹²C (assuming Cf is No)

But to match, let's do calculations as per given.

Another idea for e: perhaps " + ___ " is for a positron or something, but mass won't balance.

I found a better way: in some cases, for light nuclei, but Ca-46 to Ar-40 + He-4 requires emitting 2 neutrons, so the equation should be:

₂₀⁴⁶Ca → ₁₈⁴⁰Ar + ₂⁴He + 2 ₀¹n

So if the blank is on the left, perhaps it's a mistake, and we should ignore the "+" and put the answer as 2 ₀¹n for the products.

Similarly for others.

Perhaps for e, the blank is for nothing, and there are additional products, but the blank is for the coefficient or something.

I think the best is to state the missing particle(s).

So for e: 2 ₀¹n (on product side)

For f: if we force it, with Cf, no solution, so assume No, so ₆¹²C

But let's look at the answer choices or standard.

Perhaps for f, it's Cm-246 + He-4 -> Cf-249 + n, but not.

I recall that ²⁴⁶Cm + ¹²C -> ²⁵⁴No + 4n is correct, and Cf is 98, No is 102, so likely typo.

So I'll go with that.

Now for problem 2.

Problem 2: Write balanced nuclear equations

a. Lead-210 decays by beta emission.

Beta emission: neutron turns into proton, so atomic number increases by 1, mass same.

Lead-210: ₈₂²¹⁰Pb

After beta decay: Z=83, A=210 → Bismuth-210

Equation: ₈₂²¹⁰Pb → ₈₃²¹⁰Bi + ₋₁⁰e + antineutrino (but usually omitted in basic problems)

So: ₈₂²¹⁰Pb → ₈₃²¹⁰Bi + ₋₁⁰β

b. An atom is bombarded with an alpha particle to give two protons and another nucleus.

So: X + ₂⁴He → 2 ₁¹H + Y

But "two protons" means 2 ₁¹H

So general: _Z^A X + ₂⁴He → 2 ₁¹H + _Z'^A' Y

But we need specific? The problem doesn't specify which atom, so perhaps it's general, but usually they give the target.

Read: "An atom is bombarded..." — no specification, so perhaps it's for any, but that doesn't make sense.

Perhaps it's implying a general form, but likely, it's incomplete.

In many problems, they say "when aluminum is bombarded" etc.

Here, no target specified, so perhaps we need to choose or it's general.

But for balanced equation, we can write with variables, but probably not.

Perhaps "an atom" means we can use a placeholder, but better to assume a common one.

Perhaps it's for the reaction that produces two protons, like in some experiments.

For example, ₇¹⁴N + ₂⁴He → ₈¹⁷O + ₁¹H, but that's one proton.

To get two protons, perhaps ₅¹⁰B + ₂⁴He → ₇¹³N + ₁¹H, still one.

Or ₃⁶Li + ₂⁴He → ₅B + ₁¹H, not two.

Actually, a common reaction is: ₄Be + ₂⁴He → ₆¹²C + ₁¹H, still one.

To get two protons, perhaps: ₅¹⁰B + ₂⁴He → ₇¹³N + ₁¹H, then N decays, but not direct.

Perhaps: ₆¹²C + ₂⁴He → ₈¹⁵O + ₁¹H, not two.

I recall that in some cases, like with nitrogen: ₇¹⁴N + ₂⁴He → ₈¹⁷O + ₁¹H

For two protons, perhaps it's ₈¹⁶O + ₂⁴He → ₁₀¹⁹Ne + ₁¹H, not.

Another idea: perhaps "give two protons" means the products include two protons, so for example, ₅¹⁰B + ₂⁴He → ₇¹³N + ₁¹H, but only one.

Unless it's ₃⁶Li + ₂⁴He → ₅⁹B + ₁¹H, same.

Perhaps it's a reaction like: ₄Be + ₂⁴He → ₆¹²C + ₁¹H, and then C decays, but not.

I think a standard reaction that emits two protons is rare, but for example, in spallation, but for homework, perhaps they mean something like:

Let's assume the target is given, but it's not.

Perhaps "an atom" means we can use a generic, but better to pick a common one.

Upon searching my memory, a reaction like: ₇¹⁴N + ₂⁴He → ₉¹⁷F + ₁¹H, not two.

Perhaps it's ₈¹⁶O + ₂⁴He → ₁₀¹⁹Ne + ₁¹H.

To get two protons, perhaps: ₅¹⁰B + ₂⁴He → ₇¹³N + ₁¹H, and then ₇¹³N -> ₆¹³C + e+ + ν, but not direct.

I think for this level, they might mean a reaction like: when boron-10 is bombarded with alpha, it gives nitrogen-13 and a proton, but not two.

Another possibility: "two protons" means the nucleus emits two protons, but in bombardment, it's the product.

Perhaps it's: X + α → Y + 2p

So for example, ₆¹²C + ₂⁴He → ₈¹⁵O + ₁¹H, not 2p.

Let's calculate: suppose X + He -> Y + 2 ₁¹H

Then mass: A_X + 4 = A_Y + 2

Atomic: Z_X + 2 = Z_Y + 2, so Z_X = Z_Y

So the nucleus Y has the same atomic number as X, but mass less by 2.

So for example, if X is carbon-12, Y is carbon-10, but carbon-10 is unstable.

Or oxygen-16 + He -> oxygen-18 + 2H? Mass 16+4=20, 18+2=20, atomic 8+2=10, 8+2=10, yes.

So ₈¹⁶O + ₂⁴He → ₈¹⁸O + 2 ₁¹H

But oxygen-18 is stable, and this reaction may occur, but typically not common.

Perhaps with lighter elements.

For homework, perhaps they expect a specific one, but since not specified, I'll use this.

Or perhaps it's for aluminum or something.

Another common one: ₁₃²⁷Al + ₂⁴He → ₁₅³⁰P + ₁¹H, not two.

I think for the sake of time, I'll use ₈¹⁶O + ₂⁴He → ₈¹⁸O + 2 ₁¹H

But let's see part c and d.

c. An atom is bombarded with carbon nucleus to produce Es-253 and a proton.

Es is einsteinium, Z=99.

So: X + ₆¹²C → ₉₉²⁵³Es + ₁¹H

Then mass: A_X + 12 = 253 + 1 = 254 → A_X = 242

Atomic: Z_X + 6 = 99 + 1 = 100 → Z_X = 94 → Plutonium

So ₉₄²⁴²Pu + ₆¹²C → ₉₉²⁵³Es + ₁¹H

d. Iron nucleus undergoes alpha decay producing Fe-222.

Iron is Fe, Z=26.

Alpha decay: emits He-4, so daughter has Z=24, A=222-4=218? But it says producing Fe-222, which is iron, so same element? That can't be.

"producing Fe-222" — but if it's alpha decay, the daughter should have lower atomic number.

Unless it's not decay, but the problem says "undergoes alpha decay".

Perhaps "Fe-222" is a typo, and it's the daughter nucleus.

Read: "Iron nucleus undergoes alpha decay producing Fe-222"

But if it's alpha decay, the product should not be iron; it should be chromium or something.

For example, if iron-226 decays by alpha, it becomes chromium-222.

But here it says "producing Fe-222", which is confusing.

Perhaps "Fe-222" is the parent, and it decays to something else.

The sentence: "Iron nucleus undergoes alpha decay producing Fe-222" — that doesn't make sense because if it's undergoing decay, Fe-222 is the result, but then it's not decaying further.

Perhaps it's "an iron nucleus decays by alpha emission to produce a nucleus with mass 222", but it says "Fe-222", implying the product is iron-222, which would require the parent to have Z=28, nickel.

Let's parse: "Iron nucleus undergoes alpha decay producing Fe-222"

This is ambiguous.

In English, "producing Fe-222" likely means the product is Fe-222, but for alpha decay, the product has atomic number reduced by 2, so if product is Fe (Z=26), then parent must be Ni (Z=28).

So probably, "iron nucleus" is a mistake, and it's nickel or something.

Perhaps "Fe-222" is the parent.

Let's read carefully: "d. Iron nucleus undergoes alpha decay producing Fe-222."

This is poorly worded. Likely, it means an iron nucleus (say Fe-A) undergoes alpha decay to produce a nucleus, and that nucleus is not specified, but it says "producing Fe-222", which is confusing.

Perhaps "Fe-222" is the daughter, but then it should be Cr-222 if parent is Fe.

Assume that "Fe-222" is the daughter nucleus, but then it should be labeled as the element with Z=24.

But it says "Fe-222", so perhaps it's a typo, and it's the parent.

In many problems, they say "X decays to Y", so here "iron nucleus decays by alpha to produce [daughter]", and "Fe-222" might be the daughter, but then it's not iron.

I think the most reasonable interpretation is that the parent is an iron isotope, and it decays by alpha emission to a daughter nucleus, and the daughter is not Fe, but the problem says "producing Fe-222", which is likely a mistake.

Perhaps "Fe-222" is the parent, and it decays to something else.

Let's look at the mass: if it's alpha decay, and it produces a nucleus, and they call it Fe-222, but for alpha decay, the mass decreases by 4, so if daughter is Fe-222, parent is Fe-226.

But then the daughter is still iron, which is impossible for alpha decay.

Unless it's not alpha decay, but the problem says "alpha decay".

Another possibility: "producing Fe-222" means the product is iron-222, but that would require the parent to have Z=28, so perhaps "iron nucleus" is wrong, and it's nickel.

I think for the sake of progress, I'll assume that the parent is an iron isotope, and it decays by alpha to a daughter with mass number A-4, and atomic number 24, and they meant to say the daughter is Cr-222 or something, but it says Fe-222.

Perhaps "Fe-222" is the parent, and it decays to Cr-218 or something.

Let's calculate: if parent is Fe-226, alpha decay to Cr-222.

But the problem says "producing Fe-222", so perhaps it's a typo, and it's "from Fe-226" or "to Cr-222".

Given that, and since it's "Fe-222", perhaps the parent is Fe-226, and it produces Cr-222, but they wrote Fe by mistake.

I think the best is to assume that the daughter is not Fe, but the mass is 222, and atomic number 24.

So for d: ₂₆^A Fe → ₂₄^{A-4} Cr + ₂⁴He

And they say "producing Fe-222", but likely "producing a nucleus with mass 222", so A-4 = 222, so A=226.

So ₂₆²²⁶Fe → ₂₄²²²Cr + ₂⁴He

And "Fe-222" is a misnomer; it should be Cr-222.

So I'll go with that.

Now back to b.

For b: "An atom is bombarded with an alpha particle to give two protons and another nucleus."

So as above, X + ₂⁴He → 2 ₁¹H + Y

With Z_X = Z_Y, A_X +4 = A_Y +2, so A_Y = A_X +2

So for example, if X is oxygen-16, Y is oxygen-18, as I had.

Or if X is carbon-12, Y is carbon-14, but carbon-14 is radioactive.

Or neon-20 + He -> neon-22 + 2H, etc.

For homework, perhaps they expect a specific one, but since not specified, I'll use ₈¹⁶O + ₂⁴He → ₈¹⁸O + 2 ₁¹H

Or to make it nice, use a stable one.

Another common one is with beryllium, but usually gives one proton.

I think it's fine.

So now let's compile answers.

First, problem 1:

a. ₁₄²⁹Si

b. ₀¹n

c. ₀¹n

d. ₉₄²⁴⁰Pu (since 2 ₀¹n on right)

e. 2 ₀¹n (assuming on product side)

f. ₆¹²C (assuming Cf is No)

g. same as f, so ₆¹²C

But f and g are identical in the problem, so perhaps g is different, but in text, same.

In user's message, g is "₉₆²⁴⁶Cm + ___ → ₉₈²⁵⁴Cf + 4 ₀¹n" same as f.

Perhaps it's a duplicate, or perhaps for g, it's different.

Maybe "g" is for the decay or something, but written the same.

I'll assume f and g are the same, so same answer.

For e, to be precise, since the blank is on left, and we need to add something, but mass doesn't allow, perhaps it's a different interpretation.

Another idea for e: perhaps " + ___ " is for a gamma ray or energy, but usually not included in balancing.

Or perhaps it's for the Q-value, but not.

I think for e, the intended answer is that the blank is for the 2 neutrons, so we'll put 2 ₀¹n.

Similarly for others.

Now for problem 2:

a. ₈₂²¹⁰Pb → ₈₃²¹⁰Bi + ₋₁⁰β

b. ₈¹⁶O + ₂⁴He → ₈¹⁸O + 2 ₁¹H (or other, but this works)

c. ₉₄²⁴²Pu + ₆¹²C → ₉₉²⁵³Es + ₁¹H

d. ₂₆²²⁶Fe → ₂₄²²²Cr + ₂⁴He (assuming "Fe-222" is a mistake for the daughter)

Now to write final answers.

But for problem 1, let's list:

a. ₁₄²⁹Si

b. ₀¹n

c. ₀¹n

d. ₉₄²⁴⁰Pu

e. 2 ₀¹n (but on product side)

f. ₆¹²C

g. ₆¹²C

For e, since the blank is on left, and we have to fill, perhaps it's not possible, but I think in context, it's accepted.

Perhaps for e, it's a different reaction.

Another thought: in some cases, for light nuclei, but I think it's fine.

So now for the final answer, since the user wants the answer, and no explanation needed, but according to rules, I need to provide reasoning first, then final answer.

But in the response, I have to follow the structure.

Since ExplainAnswer is false, TeachTopic false, GeneratePractice false, so only Final Answer section.

But the instruction says: "Complete ALL reasoning and explanations BEFORE writing the Final Answer." and "If an option is false, completely omit that section."

So for this case, since all options are false, I should only have Final Answer, but I need to do reasoning first in my mind, then output only Final Answer.

The rule: "CRITICAL: First work through the problem step by step, check your calculations carefully, and ONLY THEN provide the final answer."

And "Response format rules: → Always follow the structure below. → If an option is false, completely omit that section."

So for this, since ExplainAnswer, TeachTopic, GeneratePractice are all false, I should only output the Final Answer section.

But I have done the reasoning above.

So now, let's box the final answers.

For problem 1:

a. \boxed{_{14}^{29}\text{Si}}

b. \boxed{_0^1\text{n}}

c. \boxed{_0^1\text{n}}

d. \boxed{_{94}^{240}\