Free Printable Balancing Nuclear Equations Worksheets - Free Printable

Problem: Balancing Nuclear Equations

#### Part 1: Write balanced equations for the following nuclear reactions.

To balance nuclear equations, we need to ensure that the total number of protons (atomic number) and the total number of nucleons (mass number) are conserved on both sides of the equation.

---

(i) Beta decay of Sr-90

Beta decay involves the emission of a beta particle ($\beta^-$), which is essentially an electron ($e^-$). During this process, a neutron in the nucleus is converted into a proton, increasing the atomic number by 1 while keeping the mass number constant.

The general form of beta decay is:
$$
{}_{Z}^{A}X \rightarrow {}_{Z+1}^{A}Y + {}_{-1}^{0}e + \bar{\nu}_e
$$
where $\bar{\nu}_e$ is an antineutrino (not typically shown in basic equations).

For Sr-90:
- Strontium (Sr) has an atomic number $Z = 38$.
- The daughter nucleus will have an atomic number $Z + 1 = 39$, which corresponds to Yttrium (Y).
- The mass number remains $A = 90$.

The balanced equation is:
$$
{}_{38}^{90}\text{Sr} \rightarrow {}_{39}^{90}\text{Y} + {}_{-1}^{0}e
$$

---

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

Alpha decay involves the emission of an alpha particle, which is a helium nucleus (${}_{2}^{4}\text{He}$). During this process, the atomic number decreases by 2, and the mass number decreases by 4.

The general form of alpha decay is:
$$
{}_{Z}^{A}X \rightarrow {}_{Z-2}^{A-4}Y + {}_{2}^{4}\text{He}
$$

For Uranium-238:
- Uranium (U) has an atomic number $Z = 92$ and mass number $A = 238$.
- The daughter nucleus will have an atomic number $Z - 2 = 90$, which corresponds to Thorium (Th), and a mass number $A - 4 = 234$.

The balanced equation is:
$$
{}_{92}^{238}\text{U} \rightarrow {}_{90}^{234}\text{Th} + {}_{2}^{4}\text{He}
$$

---

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

This is similar to the beta decay described in part (i). A neutron in the nucleus is converted into a proton, increasing the atomic number by 1 while keeping the mass number constant.

For Potassium-40:
- Potassium (K) has an atomic number $Z = 19$.
- The daughter nucleus will have an atomic number $Z + 1 = 20$, which corresponds to Calcium (Ca).
- The mass number remains $A = 40$.

The balanced equation is:
$$
{}_{19}^{40}\text{K} \rightarrow {}_{20}^{40}\text{Ca} + {}_{-1}^{0}e
$$

---

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

Again, this is a beta decay process. A neutron in the nucleus is converted into a proton, increasing the atomic number by 1 while keeping the mass number constant.

For Technetium-99:
- Technetium (Tc) has an atomic number $Z = 43$.
- The daughter nucleus will have an atomic number $Z + 1 = 44$, which corresponds to Ruthenium (Ru).
- The mass number remains $A = 99$.

The balanced equation is:
$$
{}_{43}^{99}\text{Tc} \rightarrow {}_{44}^{99}\text{Ru} + {}_{-1}^{0}e
$$

---

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

This is another beta decay process. A neutron in the nucleus is converted into a proton, increasing the atomic number by 1 while keeping the mass number constant.

For Phosphorus-32:
- Phosphorus (P) has an atomic number $Z = 15$.
- The daughter nucleus will have an atomic number $Z + 1 = 16$, which corresponds to Sulfur (S).
- The mass number remains $A = 32$.

The balanced equation is:
$$
{}_{15}^{32}\text{P} \rightarrow {}_{16}^{32}\text{S} + {}_{-1}^{0}e
$$

---

(vi) Francium-212 decays by alpha emission

This is an alpha decay process. An alpha particle (${}_{2}^{4}\text{He}$) is emitted, decreasing the atomic number by 2 and the mass number by 4.

For Francium-212:
- Francium (Fr) has an atomic number $Z = 87$ and mass number $A = 212$.
- The daughter nucleus will have an atomic number $Z - 2 = 85$, which corresponds to Astatine (At), and a mass number $A - 4 = 208$.

The balanced equation is:
$$
{}_{87}^{212}\text{Fr} \rightarrow {}_{85}^{208}\text{At} + {}_{2}^{4}\text{He}
$$

---

#### Part 2: Fill in the blanks to complete the following nuclear reactions.

We will use the principles of conservation of mass number and atomic number to fill in the blanks.

---

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

This is a beta decay process. The mass number remains the same ($A = 3$), and the atomic number increases by 1 ($Z = 1 + 1 = 2$). The element with $Z = 2$ and $A = 3$ is Helium-3 (${}_{2}^{3}\text{He}$).

The completed equation is:
$$
{}_{1}^{3}\text{H} \rightarrow {}_{2}^{3}\text{He} + {}_{-1}^{0}e
$$

---

(ii) ${}_{92}^{232}\text{U} \rightarrow \_\_\_\_ + {}_{90}^{238}\text{Th}$

This is an alpha decay process. The mass number decreases by 4 ($A = 232 - 4 = 228$), and the atomic number decreases by 2 ($Z = 92 - 2 = 90$). The missing particle is an alpha particle (${}_{2}^{4}\text{He}$).

The completed equation is:
$$
{}_{92}^{232}\text{U} \rightarrow {}_{2}^{4}\text{He} + {}_{90}^{238}\text{Th}
$$

---

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

This is a beta decay process in reverse (positron emission or electron capture). The mass number remains the same ($A = 144$), and the atomic number decreases by 1 ($Z = 59 + 1 = 60$). The element with $Z = 60$ and $A = 144$ is Neodymium-144 (${}_{60}^{144}\text{Nd}$).

The completed equation is:
$$
{}_{60}^{144}\text{Nd} \rightarrow {}_{-1}^{0}e + {}_{59}^{144}\text{Pr}
$$

---

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

This is a beta decay process. The mass number remains the same ($A = 65$), and the atomic number increases by 1 ($Z = 30 + 1 = 31$). The element with $Z = 31$ and $A = 65$ is Gallium-65 (${}_{31}^{65}\text{Ga}$).

The completed equation is:
$$
{}_{30}^{65}\text{Zn} \rightarrow {}_{-1}^{0}e + {}_{31}^{65}\text{Ga}
$$

---

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

This is a beta decay process. The mass number remains the same ($A = 40$), and the atomic number decreases by 1 ($Z = 19 - 1 = 18$). The missing particle is a positron (${}_{1}^{0}e$).

The completed equation is:
$$
{}_{19}^{40}\text{K} \rightarrow {}_{1}^{0}e + {}_{18}^{40}\text{Ar}
$$

---

(vi) ${}_{4}^{7}\text{Be} \rightarrow {}_{4}^{7}\text{Be} + \_\_\_$

This is a gamma decay process. The mass number and atomic number remain the same, and the only change is the release of a gamma photon ($\gamma$).

The completed equation is:
$$
{}_{4}^{7}\text{Be} \rightarrow {}_{4}^{7}\text{Be} + \gamma
$$

---

Final Answers:

#### Part 1:
1. (i) ${}_{38}^{90}\text{Sr} \rightarrow {}_{39}^{90}\text{Y} + {}_{-1}^{0}e$
2. (ii) ${}_{92}^{238}\text{U} \rightarrow {}_{90}^{234}\text{Th} + {}_{2}^{4}\text{He}$
3. (iii) ${}_{19}^{40}\text{K} \rightarrow {}_{20}^{40}\text{Ca} + {}_{-1}^{0}e$
4. (iv) ${}_{43}^{99}\text{Tc} \rightarrow {}_{44}^{99}\text{Ru} + {}_{-1}^{0}e$
5. (v) ${}_{15}^{32}\text{P} \rightarrow {}_{16}^{32}\text{S} + {}_{-1}^{0}e$
6. (vi) ${}_{87}^{212}\text{Fr} \rightarrow {}_{85}^{208}\text{At} + {}_{2}^{4}\text{He}$

#### Part 2:
1. (i) ${}_{1}^{3}\text{H} \rightarrow {}_{2}^{3}\text{He} + {}_{-1}^{0}e$
2. (ii) ${}_{92}^{232}\text{U} \rightarrow {}_{2}^{4}\text{He} + {}_{90}^{238}\text{Th}$
3. (iii) ${}_{60}^{144}\text{Nd} \rightarrow {}_{-1}^{0}e + {}_{59}^{144}\text{Pr}$
4. (iv) ${}_{30}^{65}\text{Zn} \rightarrow {}_{-1}^{0}e + {}_{31}^{65}\text{Ga}$
5. (v) ${}_{19}^{40}\text{K} \rightarrow {}_{1}^{0}e + {}_{18}^{40}\text{Ar}$
6. (vi) ${}_{4}^{7}\text{Be} \rightarrow {}_{4}^{7}\text{Be} + \gamma$

Boxed Final Answer:
$$
\boxed{
\begin{aligned}
&\text{(i)} \; {}_{38}^{90}\text{Sr} \rightarrow {}_{39}^{90}\text{Y} + {}_{-1}^{0}e \\
&\text{(ii)} \; {}_{92}^{238}\text{U} \rightarrow {}_{90}^{234}\text{Th} + {}_{2}^{4}\text{He} \\
&\text{(iii)} \; {}_{19}^{40}\text{K} \rightarrow {}_{20}^{40}\text{Ca} + {}_{-1}^{0}e \\
&\text{(iv)} \; {}_{43}^{99}\text{Tc} \rightarrow {}_{44}^{99}\text{Ru} + {}_{-1}^{0}e \\
&\text{(v)} \; {}_{15}^{32}\text{P} \rightarrow {}_{16}^{32}\text{S} + {}_{-1}^{0}e \\
&\text{(vi)} \; {}_{87}^{212}\text{Fr} \rightarrow {}_{85}^{208}\text{At} + {}_{2}^{4}\text{He} \\
&\text{(i)} \; {}_{1}^{3}\text{H} \rightarrow {}_{2}^{3}\text{He} + {}_{-1}^{0}e \\
&\text{(ii)} \; {}_{92}^{232}\text{U} \rightarrow {}_{2}^{4}\text{He} + {}_{90}^{238}\text{Th} \\
&\text{(iii)} \; {}_{60}^{144}\text{Nd} \rightarrow {}_{-1}^{0}e + {}_{59}^{144}\text{Pr} \\
&\text{(iv)} \; {}_{30}^{65}\text{Zn} \rightarrow {}_{-1}^{0}e + {}_{31}^{65}\text{Ga} \\
&\text{(v)} \; {}_{19}^{40}\text{K} \rightarrow {}_{1}^{0}e + {}_{18}^{40}\text{Ar} \\
&\text{(vi)} \; {}_{4}^{7}\text{Be} \rightarrow {}_{4}^{7}\text{Be} + \gamma
\end{aligned}
}
$$