Worksheet titled "Balancing Nuclear Equations Worksheet" featuring exercises on nuclear reactions, decay processes, and particle identification.

Balancing Nuclear Equations Worksheet Solution

#### 1. Consider the following two nuclear reactions and answer the given questions.
The given reactions are:
1. $^{220}_{87}\text{Fr} \rightarrow ^{4}_{2}\text{He} + ^{216}_{85}\text{At}$
2. $^{16}_{7}\text{N} \rightarrow ^{0}_{-1}\text{e} + ^{16}_{8}\text{O}$

##### a) Write the names of the products of francium-220 decay.
The reaction is:
$$
^{220}_{87}\text{Fr} \rightarrow ^{4}_{2}\text{He} + ^{216}_{85}\text{At}
$$
- The product $^{4}_{2}\text{He}$ is an alpha particle (or helium-4 nucleus).
- The product $^{216}_{85}\text{At}$ is astatine-216.

Thus, the products are:
- Alpha particle (helium-4 nucleus)
- Astatine-216

##### b) Write the names of the products of nitrogen-16 decay.
The reaction is:
$$
^{16}_{7}\text{N} \rightarrow ^{0}_{-1}\text{e} + ^{16}_{8}\text{O}
$$
- The product $^{0}_{-1}\text{e}$ is a beta particle (or electron).
- The product $^{16}_{8}\text{O}$ is oxygen-16.

Thus, the products are:
- Beta particle (electron)
- Oxygen-16

---

#### 2. Given the following nuclear reaction:
$$
^{27}_{13}\text{Al} + ^{4}_{2}\text{H} \rightarrow ^{30}_{15}\text{P} + ^{1}_{0}\text{n}
$$

##### a) What particle is bombarded with aluminum-27?
The reactants on the left side of the equation are $^{27}_{13}\text{Al}$ and $^{4}_{2}\text{H}$. The particle bombarded with aluminum-27 is $^{4}_{2}\text{H}$, which is a deuteron (a nucleus of deuterium).

##### b) What particle is released from the reaction?
The products on the right side of the equation are $^{30}_{15}\text{P}$ and $^{1}_{0}\text{n}$. The particle released from the reaction is $^{1}_{0}\text{n}$, which is a neutron.

---

#### 3. Fill in the blanks to complete the following nuclear reactions.

##### a) $^{42}_{19}\text{K} \rightarrow ^{0}_{-1}\text{e} + \_\_\_\_\_$
This is a beta decay reaction. In beta decay, a neutron in the nucleus converts into a proton, releasing a beta particle ($^{0}_{-1}\text{e}$). The atomic number increases by 1, and the mass number remains the same.

- Initial nucleus: $^{42}_{19}\text{K}$ (potassium-42)
- Final nucleus: $^{42}_{20}\text{Ca}$ (calcium-42)

Thus, the completed reaction is:
$$
^{42}_{19}\text{K} \rightarrow ^{0}_{-1}\text{e} + ^{42}_{20}\text{Ca}
$$

##### b) $^{239}_{94}\text{Pu} \rightarrow ^{4}_{2}\text{He} + \_\_\_\_\_$
This is an alpha decay reaction. In alpha decay, the nucleus emits an alpha particle ($^{4}_{2}\text{He}$), reducing the atomic number by 2 and the mass number by 4.

- Initial nucleus: $^{239}_{94}\text{Pu}$ (plutonium-239)
- Final nucleus: $^{235}_{92}\text{U}$ (uranium-235)

Thus, the completed reaction is:
$$
^{239}_{94}\text{Pu} \rightarrow ^{4}_{2}\text{He} + ^{235}_{92}\text{U}
$$

##### c) $^{235}_{92}\text{U} \rightarrow ^{4}_{2}\text{He} + \_\_\_\_\_$
This is another alpha decay reaction.

- Initial nucleus: $^{235}_{92}\text{U}$ (uranium-235)
- Final nucleus: $^{231}_{90}\text{Th}$ (thorium-231)

Thus, the completed reaction is:
$$
^{235}_{92}\text{U} \rightarrow ^{4}_{2}\text{He} + ^{231}_{90}\text{Th}
$$

##### d) $^{1}_{1}\text{H} + ^{3}_{1}\text{H} \rightarrow \_\_\_\_\_$
This is a nuclear fusion reaction. When a proton ($^{1}_{1}\text{H}$) and a triton ($^{3}_{1}\text{H}$) fuse, they form a helium-4 nucleus ($^{4}_{2}\text{He}$) and release energy.

Thus, the completed reaction is:
$$
^{1}_{1}\text{H} + ^{3}_{1}\text{H} \rightarrow ^{4}_{2}\text{He}
$$

##### e) $^{6}_{3}\text{Li} + ^{1}_{0}\text{n} \rightarrow ^{0}_{-1}\text{e} + \_\_\_\_\_ + ^{3}_{2}\text{He}$
This is a nuclear reaction involving lithium-6 and a neutron. The total mass number and atomic number must be conserved.

- Initial total mass number: $6 + 1 = 7$
- Initial total atomic number: $3 + 0 = 3$

- Final total mass number: $0 + x + 3 = 7 \implies x = 4$
- Final total atomic number: $-1 + y + 2 = 3 \implies y = 2$

Thus, the missing nucleus has a mass number of 4 and an atomic number of 2, which is beryllium-7 ($^{7}_{3}\text{Be}$).

The completed reaction is:
$$
^{6}_{3}\text{Li} + ^{1}_{0}\text{n} \rightarrow ^{0}_{-1}\text{e} + ^{7}_{3}\text{Be} + ^{3}_{2}\text{He}
$$

---

#### 4. Write a balanced nuclear equation for each decay process indicated.

##### a) The isotope Th-234 decays by an alpha emission.
In alpha decay, the nucleus emits an alpha particle ($^{4}_{2}\text{He}$), reducing the atomic number by 2 and the mass number by 4.

- Initial nucleus: $^{234}_{90}\text{Th}$
- Final nucleus: $^{230}_{88}\text{Ra}$ (radium-230)

The balanced equation is:
$$
^{234}_{90}\text{Th} \rightarrow ^{4}_{2}\text{He} + ^{230}_{88}\text{Ra}
$$

##### b) The isotope Fe-59 decays by beta emission.
In beta decay, a neutron in the nucleus converts into a proton, releasing a beta particle ($^{0}_{-1}\text{e}$). The atomic number increases by 1, and the mass number remains the same.

- Initial nucleus: $^{59}_{26}\text{Fe}$
- Final nucleus: $^{59}_{27}\text{Co}$ (cobalt-59)

The balanced equation is:
$$
^{59}_{26}\text{Fe} \rightarrow ^{0}_{-1}\text{e} + ^{59}_{27}\text{Co}
$$

##### c) The isotope Tc-99 decays by gamma emission.
In gamma decay, the nucleus releases a gamma photon ($\gamma$), but the atomic number and mass number remain unchanged. This typically occurs after an initial decay (e.g., beta decay) when the nucleus is in an excited state.

- Initial nucleus: $^{99}_{43}\text{Tc}$
- Final nucleus: $^{99}_{43}\text{Tc}$ (in its ground state)

The balanced equation is:
$$
^{99}_{43}\text{Tc}^* \rightarrow ^{99}_{43}\text{Tc} + \gamma
$$
Here, $^{99}_{43}\text{Tc}^*$ represents the excited state of technetium-99.

---

Final Answers:

1. a) Products of francium-220 decay: Alpha particle (helium-4 nucleus), astatine-216
b) Products of nitrogen-16 decay: Beta particle (electron), oxygen-16

2. a) Particle bombarded with aluminum-27: Deuteron ($^{4}_{2}\text{H}$)
b) Particle released from the reaction: Neutron ($^{1}_{0}\text{n}$)

3. a) $^{42}_{19}\text{K} \rightarrow ^{0}_{-1}\text{e} + ^{42}_{20}\text{Ca}$
b) $^{239}_{94}\text{Pu} \rightarrow ^{4}_{2}\text{He} + ^{235}_{92}\text{U}$
c) $^{235}_{92}\text{U} \rightarrow ^{4}_{2}\text{He} + ^{231}_{90}\text{Th}$
d) $^{1}_{1}\text{H} + ^{3}_{1}\text{H} \rightarrow ^{4}_{2}\text{He}$
e) $^{6}_{3}\text{Li} + ^{1}_{0}\text{n} \rightarrow ^{0}_{-1}\text{e} + ^{7}_{3}\text{Be} + ^{3}_{2}\text{He}$

4. a) $^{234}_{90}\text{Th} \rightarrow ^{4}_{2}\text{He} + ^{230}_{88}\text{Ra}$
b) $^{59}_{26}\text{Fe} \rightarrow ^{0}_{-1}\text{e} + ^{59}_{27}\text{Co}$
c) $^{99}_{43}\text{Tc}^* \rightarrow ^{99}_{43}\text{Tc} + \gamma$

Boxed Final Answer:
$$
\boxed{
\begin{aligned}
&\text{1. a) Alpha particle, astatine-216; b) Beta particle, oxygen-16} \\
&\text{2. a) Deuteron; b) Neutron} \\
&\text{3. a) } ^{42}_{20}\text{Ca}; \text{ b) } ^{235}_{92}\text{U}; \text{ c) } ^{231}_{90}\text{Th}; \text{ d) } ^{4}_{2}\text{He}; \text{ e) } ^{7}_{3}\text{Be} \\
&\text{4. a) } ^{234}_{90}\text{Th} \rightarrow ^{4}_{2}\text{He} + ^{230}_{88}\text{Ra}; \text{ b) } ^{59}_{26}\text{Fe} \rightarrow ^{0}_{-1}\text{e} + ^{59}_{27}\text{Co}; \text{ c) } ^{99}_{43}\text{Tc}^* \rightarrow ^{99}_{43}\text{Tc} + \gamma
\end{aligned}
}
$$