Balancing Nuclear Equations worksheet with exercises on nuclear reactions and decay processes.

Let’s solve each part step by step. We’ll balance nuclear equations by making sure the total mass number (top) and total atomic number (bottom) are equal on both sides of the arrow.

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Part 1: Write balanced equations

#### (A) C-11 decays by electron capture.
Electron capture means an inner-shell electron is absorbed by the nucleus. The electron is written as \( ^0_{-1}e \).
Carbon-11 has atomic number 6 → \( ^{11}_6C \)
After capturing an electron, a proton turns into a neutron → atomic number decreases by 1 → becomes Boron (atomic number 5), mass stays same.

So:
\[
^{11}_6C + ^0_{-1}e \rightarrow ^{11}_5B
\]

✔ Balanced: Mass: 11+0=11; Atomic: 6 + (-1) = 5 → matches B-11.

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#### (B) Curium-246 reacts with carbon-12 to produce nobelium-254 and four neutrons.
Curium (Cm) atomic number = 96 → \( ^{246}_{96}Cm \)
Carbon-12 → \( ^{12}_6C \)
Nobelium (No) atomic number = 102 → \( ^{254}_{102}No \)
Neutron → \( ^1_0n \), and there are 4 of them.

Left side mass: 246 + 12 = 258
Right side: 254 + 4×1 = 258 ✔
Left side atomic: 96 + 6 = 102
Right side: 102 + 0 = 102 ✔

Equation:
\[
^{246}_{96}Cm + ^{12}_6C \rightarrow ^{254}_{102}No + 4^1_0n
\]

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#### (C) Californium-250 reacts with boron-10 to produce lawrencium-258 and two neutrons.
Californium (Cf) atomic number = 98 → \( ^{250}_{98}Cf \)
Boron-10 → \( ^{10}_5B \)
Lawrencium (Lr) atomic number = 103 → \( ^{258}_{103}Lr \)
Two neutrons → \( 2^1_0n \)

Mass left: 250 + 10 = 260
Mass right: 258 + 2 = 260 ✔
Atomic left: 98 + 5 = 103
Atomic right: 103 + 0 = 103 ✔

Equation:
\[
^{250}_{98}Cf + ^{10}_5B \rightarrow ^{258}_{103}Lr + 2^1_0n
\]

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#### (D) Fluorine-18 decays to oxygen-18 by positron emission.
Fluorine (F) atomic number = 9 → \( ^{18}_9F \)
Oxygen (O) atomic number = 8 → \( ^{18}_8O \)
Positron = \( ^0_{+1}e \) or sometimes written \( ^0_1\beta^+ \)

Decay: F-18 → O-18 + positron

Check:
Mass: 18 = 18 + 0 ✔
Atomic: 9 = 8 + 1 ✔

Equation:
\[
^{18}_9F \rightarrow ^{18}_8O + ^0_{+1}e
\]

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#### (E) Sodium-24 decays by beta emission.
Sodium (Na) atomic number = 11 → \( ^{24}_{11}Na \)
Beta particle = electron = \( ^0_{-1}e \)
When a nucleus emits a beta particle, a neutron turns into a proton → atomic number increases by 1 → becomes Magnesium (Mg, atomic number 12), mass stays same.

So:
\[
^{24}_{11}Na \rightarrow ^{24}_{12}Mg + ^0_{-1}e
\]

Check: Mass 24=24+0; Atomic 11=12 + (-1) → 11 ✔

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#### (F) Krypton-76 absorbs a beta particle to form bromine-76.
Krypton (Kr) atomic number = 36 → \( ^{76}_{36}Kr \)
Absorbs beta particle = \( ^0_{-1}e \)
Forms Bromine (Br) atomic number = 35 → \( ^{76}_{35}Br \)

Wait — if Kr (36) absorbs an electron (charge -1), then new atomic number = 36 + (-1) = 35 → which is Br. Mass unchanged.

Equation:
\[
^{76}_{36}Kr + ^0_{-1}e \rightarrow ^{76}_{35}Br
\]

Check: Mass 76+0=76; Atomic 36 + (-1)=35 ✔

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Part 2: Fill in the blanks

We’ll use conservation of mass number (top) and atomic number (bottom).

#### (A) \( ^1H_1 + ^7Li_3 \rightarrow ? \)

Mass: 1 + 7 = 8
Atomic: 1 + 3 = 4 → element with atomic number 4 is Beryllium (Be)

Answer: \( ^8Be_4 \)

But wait — actually, this reaction usually produces two alpha particles? Let me double-check common reactions.

Actually, \( ^1H + ^7Li \rightarrow 2^4He \) is a known reaction. But let’s check math:

If product is \( ^8Be \), that’s possible but unstable. However, in many textbooks, they show:

\( ^1_1H + ^7_3Li \rightarrow ^4_2He + ^4_2He \)

But the problem says “fill in the blank” — only one blank. So maybe it’s expecting \( ^8_4Be \)? But Be-8 immediately breaks into two alphas.

Looking at the format — probably expects single nucleus. But let’s see other parts.

Wait — perhaps I should just go by numbers.

Mass sum: 8, atomic sum: 4 → so \( ^8_4Be \)

But let’s hold on — maybe it's a trick. Actually, in some contexts, they write it as producing two heliums, but here only one blank. Hmm.

Wait — looking ahead, part (G) has multiple products. Here only one blank. So likely \( ^8_4Be \)

But let me verify with standard knowledge: Proton + Lithium-7 → Beryllium-8 → which decays to two alphas. But for balancing purposes, we can write \( ^8_4Be \)

I think for this worksheet, they want \( ^8_4Be \)

✔ Final answer for (A): \( ^8_4Be \)

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#### (B) \( ^7Be_4 + ^1n_0 \rightarrow ^1H_1 + ? + ^4He_2 \)

Left mass: 7 + 1 = 8
Left atomic: 4 + 0 = 4

Right so far: H-1 (mass 1, atomic 1) + He-4 (mass 4, atomic 2) → total mass 5, atomic 3

Missing: mass = 8 - 5 = 3; atomic = 4 - 3 = 1 → Hydrogen-3 (tritium) → \( ^3_1H \)

Check:
Mass: 1 + 3 + 4 = 8 ✔
Atomic: 1 + 1 + 2 = 4 ✔

Answer: \( ^3_1H \)

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#### (C) \( ^{27}Al_{13} + ^4He_2 \rightarrow ? + ^{30}P_{15} \)

Left mass: 27 + 4 = 31
Left atomic: 13 + 2 = 15

Right: P-30 has mass 30, atomic 15 → missing particle must have mass 1, atomic 0 → neutron!

Because 31 - 30 = 1; 15 - 15 = 0 → \( ^1_0n \)

Answer: \( ^1_0n \)

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#### (D) \( ^{63}Cu_{29} + ^2H_1 \rightarrow 2^1n_0 + ? \)

Left mass: 63 + 2 = 65
Left atomic: 29 + 1 = 30

Right: 2 neutrons → mass 2, atomic 0 → missing: mass 65 - 2 = 63; atomic 30 - 0 = 30 → Zinc-63? Atomic number 30 is Zn.

So \( ^{63}_{30}Zn \)

Check: Mass 63+2=65; 2*1 + 63=65 ✔ Atomic: 29+1=30; 0+30=30 ✔

Answer: \( ^{63}_{30}Zn \)

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#### (E) \( ^9Be_4 + ^4He_2 \rightarrow ? + ^{12}C_6 \)

Left mass: 9 + 4 = 13
Left atomic: 4 + 2 = 6

Right: C-12 → mass 12, atomic 6 → missing: mass 1, atomic 0 → neutron again!

\( ^1_0n \)

Check: 12 + 1 = 13; 6 + 0 = 6 ✔

Answer: \( ^1_0n \)

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#### (F) \( ^{239}Np_{93} \rightarrow ^0e_{-1} + ? \)

This is beta decay. Np-239 emits beta particle (electron).

In beta decay, atomic number increases by 1, mass same.

So daughter nucleus: mass 239, atomic number 93 + 1 = 94 → Plutonium (Pu)

So \( ^{239}_{94}Pu \)

Check: Mass 239 = 0 + 239 ✔ Atomic: 93 = -1 + 94 → 93 ✔

Answer: \( ^{239}_{94}Pu \)

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#### (G) \( ^4He_2 + ^{238}U_{92} \rightarrow 2^1n_0 + ? \)

Left mass: 4 + 238 = 242
Left atomic: 2 + 92 = 94

Right: 2 neutrons → mass 2, atomic 0 → missing: mass 242 - 2 = 240; atomic 94 - 0 = 94 → Plutonium-240

\( ^{240}_{94}Pu \)

Check: Mass 4+238=242; 2*1 + 240=242 ✔ Atomic: 2+92=94; 0+94=94 ✔

Answer: \( ^{240}_{94}Pu \)

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Now, let’s compile all answers neatly.

Final Answer:

Part 1:

(A) \( ^{11}_6C + ^0_{-1}e \rightarrow ^{11}_5B \)
(B) \( ^{246}_{96}Cm + ^{12}_6C \rightarrow ^{254}_{102}No + 4^1_0n \)
(C) \( ^{250}_{98}Cf + ^{10}_5B \rightarrow ^{258}_{103}Lr + 2^1_0n \)
(D) \( ^{18}_9F \rightarrow ^{18}_8O + ^0_{+1}e \)
(E) \( ^{24}_{11}Na \rightarrow ^{24}_{12}Mg + ^0_{-1}e \)
(F) \( ^{76}_{36}Kr + ^0_{-1}e \rightarrow ^{76}_{35}Br \)

Part 2:

(A) \( ^8_4Be \)
(B) \( ^3_1H \)
(C) \( ^1_0n \)
(D) \( ^{63}_{30}Zn \)
(E) \( ^1_0n \)
(F) \( ^{239}_{94}Pu \)
(G) \( ^{240}_{94}Pu \)