Worksheet for practicing balancing nuclear equations, including beta decay, alpha emission, and other radioactive processes.

Let's solve each part of this nuclear equations worksheet step by step.

---

1. Write balanced equations for the following nuclear reactions

We need to balance both mass numbers (top number) and atomic numbers (bottom number). Remember:

- Beta decay (β⁻): A neutron turns into a proton, emitting an electron (⁰₋₁e). The atomic number increases by 1; mass number stays same.
- Alpha decay (α): Emission of a helium nucleus (⁴₂He). Mass decreases by 4, atomic number by 2.

---

#### (i) Beta decay of Sr-90

Strontium-90: $^{90}_{38}\text{Sr}$

In beta decay:
- A neutron becomes a proton → atomic number increases by 1
- Mass number remains same

So:
$$
^{90}_{38}\text{Sr} \rightarrow ^{90}_{39}\text{Y} + ^0_{-1}\beta
$$

✔ Answer: $^{90}_{38}\text{Sr} \rightarrow ^{90}_{39}\text{Y} + ^0_{-1}\beta$

---

#### (ii) Uranium-238 decays by emitting an alpha particle to form thorium-234

Uranium-238: $^{238}_{92}\text{U}$

Alpha decay emits $^4_2\text{He}$

Mass: 238 → 234 + 4 ✔
Atomic number: 92 → 90 + 2 ✔

So:
$$
^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^4_2\alpha
$$

✔ Answer: $^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^4_2\alpha$

---

#### (iii) Potassium-40 decays to calcium-40 by beta emission

Potassium-40: $^{40}_{19}\text{K}$

Decays to Calcium-40: $^{40}_{20}\text{Ca}$

This is beta decay because atomic number increases from 19 to 20.

$$
^{40}_{19}\text{K} \rightarrow ^{40}_{20}\text{Ca} + ^0_{-1}\beta
$$

✔ Answer: $^{40}_{19}\text{K} \rightarrow ^{40}_{20}\text{Ca} + ^0_{-1}\beta$

---

#### (iv) Technetium-99 decays by beta emission to form ruthenium-99

Technetium-99: $^{99}_{43}\text{Tc}$

Ruthenium-99: $^{99}_{44}\text{Ru}$

Again, beta decay: atomic number increases by 1.

$$
^{99}_{43}\text{Tc} \rightarrow ^{99}_{44}\text{Ru} + ^0_{-1}\beta
$$

✔ Answer: $^{99}_{43}\text{Tc} \rightarrow ^{99}_{44}\text{Ru} + ^0_{-1}\beta$

---

#### (v) Phosphorus-32 decays by beta emission to form sulfur-32

Phosphorus-32: $^{32}_{15}\text{P}$

Sulfur-32: $^{32}_{16}\text{S}$

Beta decay again: Z increases from 15 to 16.

$$
^{32}_{15}\text{P} \rightarrow ^{32}_{16}\text{S} + ^0_{-1}\beta
$$

✔ Answer: $^{32}_{15}\text{P} \rightarrow ^{32}_{16}\text{S} + ^0_{-1}\beta$

---

#### (vi) Francium-212 decays by alpha emission

Francium-212: $^{212}_{87}\text{Fr}$

Alpha decay → loses $^4_2\text{He}$

So new mass: 212 - 4 = 208
New atomic number: 87 - 2 = 85

Element with atomic number 85 is Astatine (At)

$$
^{212}_{87}\text{Fr} \rightarrow ^{208}_{85}\text{At} + ^4_2\alpha
$$

✔ Answer: $^{212}_{87}\text{Fr} \rightarrow ^{208}_{85}\text{At} + ^4_2\alpha$

---

2. Fill in the blanks to complete the following nuclear reactions

Use conservation of mass and atomic numbers.

---

#### (i) $^3_1\text{H} \rightarrow \_\_\_\_\_\_ + ^0_{-1}e$

Hydrogen-3 (tritium) undergoes beta decay.

Mass: 3 → ? + 0 → so missing mass = 3
Atomic number: 1 → ? + (-1) → so missing atomic number = 2

So element with atomic number 2 is Helium (He)

$$
^3_1\text{H} \rightarrow ^3_2\text{He} + ^0_{-1}e
$$

✔ Answer: $^3_2\text{He}$

---

#### (ii) $^{232}_{92}\text{U} \rightarrow \_\_\_\_\_\_ + ^{228}_{90}\text{Th}$

Mass: 232 → ? + 228 → missing mass = 4
Atomic number: 92 → ? + 90 → missing atomic number = 2

That’s an alpha particle: $^4_2\text{He}$

$$
^{232}_{92}\text{U} \rightarrow ^4_2\text{He} + ^{228}_{90}\text{Th}
$$

✔ Answer: $^4_2\text{He}$

---

#### (iii) $\_\_\_\_\_\_ \rightarrow ^0_{-1}e + ^{144}_{59}\text{Pr}$

Beta decay: daughter nucleus is Pr-144

So parent must have:
- Mass: 144 (since beta doesn't change mass)
- Atomic number: 59 - 1 = 58 (because beta emission increases Z by 1)

Element with atomic number 58 is Cerium (Ce)

So parent is $^{144}_{58}\text{Ce}$

$$
^{144}_{58}\text{Ce} \rightarrow ^0_{-1}e + ^{144}_{59}\text{Pr}
$$

✔ Answer: $^{144}_{58}\text{Ce}$

---

#### (iv) $^{65}_{30}\text{Zn} \rightarrow ^0_{-1}e + \_\_\_\_\_\_$

Beta decay: zinc-65 → ?

Mass: 65 → 65 + 0 → daughter has mass 65
Atomic number: 30 → ? + (-1) → daughter has atomic number 31

Element with atomic number 31 is Gallium (Ga)

$$
^{65}_{30}\text{Zn} \rightarrow ^0_{-1}e + ^{65}_{31}\text{Ga}
$$

✔ Answer: $^{65}_{31}\text{Ga}$

---

#### (v) $^{40}_{19}\text{K} \rightarrow \_\_\_\_\_\_ + ^{40}_{18}\text{Ar}$

Potassium-40 decays to argon-40 — this is positron emission or electron capture, but here we see that atomic number decreases from 19 to 18.

So it's positron emission (β⁺): $^0_{+1}e$ emitted

So:
$$
^{40}_{19}\text{K} \rightarrow ^0_{+1}e + ^{40}_{18}\text{Ar}
$$

✔ Answer: $^0_{+1}e$

> Note: This is not beta-minus decay — it's positron emission. But the problem says "fill in the blank", so we just match what fits.

---

#### (vi) $^7_4\text{Be} \rightarrow ^7_4\text{Be} + \_\_\_\_\_\_$

Wait — left side: $^7_4\text{Be}$
Right side: $^7_4\text{Be} + ?$

But if Be-7 is on both sides, then something must be emitted.

Actually, Beryllium-7 decays by electron capture or positron emission.

But here, the product is still $^7_4\text{Be}$? That can’t be unless it's a typo.

Wait — perhaps it's meant to be:

$^7_4\text{Be} \rightarrow ^7_3\text{Li} + ?$

But the equation shows $^7_4\text{Be}$ on both sides.

Wait — let's read carefully:

> $^7_4\text{Be} \rightarrow ^7_4\text{Be} + \_\_\_\_\_\_$

That would imply no change — which is impossible.

But maybe it's a typo in the question.

Actually, Be-7 does decay via electron capture to lithium-7:

$$
^7_4\text{Be} + ^0_{-1}e \rightarrow ^7_3\text{Li} + \nu_e
$$

But here it's written as:

$$
^7_4\text{Be} \rightarrow ^7_4\text{Be} + ?
$$

That doesn't make sense.

Alternatively, perhaps it's emission of a gamma ray?

But gamma rays don't change the nucleus — so it should be:

$$
^7_4\text{Be} \rightarrow ^7_4\text{Be} + \gamma
$$

Yes! This could represent gamma decay — where an excited nucleus emits a photon without changing identity.

So the blank is filled with γ (gamma ray).

But note: Gamma decay usually occurs after other decays. However, if Be-7 is in an excited state, it can emit gamma.

But actually, Be-7 decays primarily by electron capture to Li-7, not by gamma.

But given the format:

$$
^7_4\text{Be} \rightarrow ^7_4\text{Be} + \_\_\_\_
$$

The only way this makes sense is if the product is a gamma ray.

So:
$$
^7_4\text{Be} \rightarrow ^7_4\text{Be} + \gamma
$$

✔ Answer: $\gamma$

Or more formally: $^0_0\gamma$

---

✔ Final Answers Summary

---

Part 1: Balanced Nuclear Equations

(i) $^{90}_{38}\text{Sr} \rightarrow ^{90}_{39}\text{Y} + ^0_{-1}\beta$
(ii) $^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^4_2\alpha$
(iii) $^{40}_{19}\text{K} \rightarrow ^{40}_{20}\text{Ca} + ^0_{-1}\beta$
(iv) $^{99}_{43}\text{Tc} \rightarrow ^{99}_{44}\text{Ru} + ^0_{-1}\beta$
(v) $^{32}_{15}\text{P} \rightarrow ^{32}_{16}\text{S} + ^0_{-1}\beta$
(vi) $^{212}_{87}\text{Fr} \rightarrow ^{208}_{85}\text{At} + ^4_2\alpha$

---

Part 2: Fill in the Blanks

(i) $^3_1\text{H} \rightarrow \boxed{^3_2\text{He}} + ^0_{-1}e$
(ii) $^{232}_{92}\text{U} \rightarrow \boxed{^4_2\text{He}} + ^{228}_{90}\text{Th}$
(iii) $\boxed{^{144}_{58}\text{Ce}} \rightarrow ^0_{-1}e + ^{144}_{59}\text{Pr}$
(iv) $^{65}_{30}\text{Zn} \rightarrow ^0_{-1}e + \boxed{^{65}_{31}\text{Ga}}$
(v) $^{40}_{19}\text{K} \rightarrow \boxed{^0_{+1}e} + ^{40}_{18}\text{Ar}$
(vi) $^7_4\text{Be} \rightarrow ^7_4\text{Be} + \boxed{\gamma}$

> Note: For (vi), the reaction implies gamma emission (no change in nucleus), so γ is correct. If it were positron emission, the product would be $^7_3\text{Li}$, but since it's written as Be-7 on both sides, γ is the best fit.

---

Let me know if you'd like this formatted as a printable answer sheet!