Solved NUCLEAR CHEMISTRY BALANCING NUCLEAR REACTIONS | Chegg.com - Free Printable

Problem Overview:

The task involves balancing nuclear reactions and identifying the type of nuclear reaction (e.g., α emission, β emission, γ emission, positron emission, artificial transmutation, fission, or fusion). Additionally, we need to write balanced nuclear equations for specific decays.

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Solution:

#### 1. Balancing Nuclear Reactions:

We will balance each nuclear reaction step by step and identify the type of reaction.

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##### 1.
$$
^{40}_{19} \text{K} \rightarrow { }_{-1}^{\circ} \text{e} + \_\_\_\_
$$

- Analysis: The reactant is potassium-40 ($^{40}_{19}\text{K}$). The product includes an electron ($\beta$ particle), indicating $\beta^-$ decay.
- Balancing: In $\beta^-$ decay, a neutron in the nucleus converts into a proton, emitting an electron and an antineutrino ($\bar{\nu}_e$). The atomic number increases by 1, and the mass number remains the same.
- Result: The missing product is calcium-40 ($^{40}_{20}\text{Ca}$).
- Type of Reaction: $\beta^-$ emission.

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

---

##### 2.
$$
^{236}_{94} \text{Pu} \rightarrow { }_{2}^{4} \text{He} + \_\_\_\_
$$

- Analysis: The reactant is plutonium-236 ($^{236}_{94}\text{Pu}$). The product includes an alpha particle ($\alpha$ particle = $^{4}_{2}\text{He}$), indicating $\alpha$ decay.
- Balancing: In $\alpha$ decay, the nucleus emits an alpha particle, reducing the atomic number by 2 and the mass number by 4. The resulting nucleus is uranium-232 ($^{232}_{92}\text{U}$).
- Result: The missing product is uranium-232 ($^{232}_{92}\text{U}$).
- Type of Reaction: $\alpha$ emission.

$$
^{236}_{94} \text{Pu} \rightarrow { }_{2}^{4} \text{He} + ^{232}_{92} \text{U}
$$

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##### 3.
$$
^{233}_{92} \text{U} \rightarrow \_\_\_\_ + { }_{82}^{209} \text{Pb}
$$

- Analysis: The reactant is uranium-233 ($^{233}_{92}\text{U}$). The product includes lead-209 ($^{209}_{82}\text{Pb}$), indicating multiple steps of decay (likely involving $\alpha$ and $\beta$ emissions).
- Balancing: To balance this, we need to determine how many $\alpha$ and $\beta$ emissions occur. The difference in atomic numbers is $92 - 82 = 10$, and the difference in mass numbers is $233 - 209 = 24$. Each $\alpha$ emission reduces the mass number by 4 and the atomic number by 2. Therefore, there are $24 / 4 = 6$ $\alpha$ emissions, which reduce the atomic number by $6 \times 2 = 12$. To compensate for the extra reduction in atomic number, there must be $12 - 10 = 2$ $\beta^-$ emissions.
- Result: The missing product is 6 $\alpha$ particles ($^{4}_{2}\text{He}$) and 2 $\beta^-$ particles ($^{0}_{-1}\text{e}$).
- Type of Reaction: Artificial transmutation (series of $\alpha$ and $\beta^-$ emissions).

$$
^{233}_{92} \text{U} \rightarrow 6 \cdot { }_{2}^{4} \text{He} + 2 \cdot { }_{-1}^{0} \text{e} + ^{209}_{82} \text{Pb}
$$

---

##### 4.
$$
^{1}_{1} \text{H} + { }_{1}^{1} \text{H} \rightarrow \_\_\_\_
$$

- Analysis: The reactants are two hydrogen-1 nuclei (protons). This indicates a fusion reaction.
- Balancing: In fusion, two light nuclei combine to form a heavier nucleus. Here, two protons fuse to form deuterium ($^{2}_{1}\text{H}$) and release a positron ($^{0}_{+1}\text{e}$) and a neutrino ($\nu_e$).
- Result: The missing product is deuterium ($^{2}_{1}\text{H}$) and a positron ($^{0}_{+1}\text{e}$).
- Type of Reaction: Fusion.

$$
^{1}_{1} \text{H} + { }_{1}^{1} \text{H} \rightarrow ^{2}_{1} \text{H} + { }_{+1}^{0} \text{e}
$$

---

##### 5.
$$
^{6}_{3} \text{Li} + { }_{0}^{1} \text{n} \rightarrow { }_{2}^{4} \text{He} + \_\_\_\_
$$

- Analysis: The reactants are lithium-6 ($^{6}_{3}\text{Li}$) and a neutron ($^{1}_{0}\text{n}$). This indicates a nuclear reaction involving a neutron.
- Balancing: When a neutron collides with lithium-6, it can cause the nucleus to split into two alpha particles ($^{4}_{2}\text{He}$). This is an example of a nuclear fission-like process.
- Result: The missing product is another alpha particle ($^{4}_{2}\text{He}$).
- Type of Reaction: Artificial transmutation (neutron-induced fission).

$$
^{6}_{3} \text{Li} + { }_{0}^{1} \text{n} \rightarrow 2 \cdot { }_{2}^{4} \text{He}
$$

---

##### 6.
$$
^{27}_{13} \text{Al} + { }_{2}^{4} \text{He} \rightarrow { }_{15}^{30} \text{P} + \_\_\_\_
$$

- Analysis: The reactants are aluminum-27 ($^{27}_{13}\text{Al}$) and an alpha particle ($^{4}_{2}\text{He}$). This indicates an artificial transmutation.
- Balancing: In this reaction, the alpha particle is absorbed, increasing the atomic number by 2 and the mass number by 4. However, the resulting phosphorus-30 nucleus is unstable and emits a neutron to reach stability.
- Result: The missing product is a neutron ($^{1}_{0}\text{n}$).
- Type of Reaction: Artificial transmutation.

$$
^{27}_{13} \text{Al} + { }_{2}^{4} \text{He} \rightarrow { }_{15}^{30} \text{P} + { }_{0}^{1} \text{n}
$$

---

##### 7.
$$
^{9}_{4} \text{Be} + { }_{1}^{1} \text{H} \rightarrow \_\_\_\_ + { }_{2}^{4} \text{He}
$$

- Analysis: The reactants are beryllium-9 ($^{9}_{4}\text{Be}$) and a proton ($^{1}_{1}\text{H}$). This indicates a nuclear reaction involving a proton.
- Balancing: When a proton collides with beryllium-9, it can cause the nucleus to split into an alpha particle ($^{4}_{2}\text{He}$) and a deuteron ($^{2}_{1}\text{H}$).
- Result: The missing product is a deuteron ($^{2}_{1}\text{H}$).
- Type of Reaction: Artificial transmutation.

$$
^{9}_{4} \text{Be} + { }_{1}^{1} \text{H} \rightarrow { }_{1}^{2} \text{H} + { }_{2}^{4} \text{He}
$$

---

##### 8.
$$
^{37}_{19} \text{K} \rightarrow { }_{-1}^{0} \text{e} + \_\_\_\_
$$

- Analysis: The reactant is potassium-37 ($^{37}_{19}\text{K}$). The product includes an electron ($\beta$ particle), indicating $\beta^-$ decay.
- Balancing: In $\beta^-$ decay, a neutron converts into a proton, emitting an electron and an antineutrino. The resulting nucleus is calcium-37 ($^{37}_{20}\text{Ca}$).
- Result: The missing product is calcium-37 ($^{37}_{20}\text{Ca}$).
- Type of Reaction: $\beta^-$ emission.

$$
^{37}_{19} \text{K} \rightarrow { }_{-1}^{0} \text{e} + ^{37}_{20} \text{Ca}
$$

---

##### 9.
$$
\_\_\_\_ + { }_{0}^{1} \text{n} \rightarrow { }_{56}^{142} \text{Ba} + { }_{36}^{89} \text{Kr} + 3 \cdot { }_{0}^{1} \text{n}
$$

- Analysis: The products include barium-142 ($^{142}_{56}\text{Ba}$), krypton-89 ($^{89}_{36}\text{Kr}$), and three neutrons. This indicates a fission reaction.
- Balancing: The total mass number on the right side is $142 + 89 + 3 = 234$. Therefore, the reactant must have a mass number of 234. The most likely reactant is uranium-235 ($^{235}_{92}\text{U}$), which undergoes fission when bombarded by a neutron.
- Result: The missing reactant is uranium-235 ($^{235}_{92}\text{U}$).
- Type of Reaction: Fission.

$$
^{235}_{92} \text{U} + { }_{0}^{1} \text{n} \rightarrow { }_{56}^{142} \text{Ba} + { }_{36}^{89} \text{Kr} + 3 \cdot { }_{0}^{1} \text{n}
$$

---

##### 10.
$$
^{238}_{92} \text{U} + { }_{2}^{4} \text{He} \rightarrow \_\_\_\_ + { }_{0}^{1} \text{n}
$$

- Analysis: The reactants are uranium-238 ($^{238}_{92}\text{U}$) and an alpha particle ($^{4}_{2}\text{He}$). This indicates an artificial transmutation.
- Balancing: When an alpha particle is absorbed, the atomic number increases by 2 and the mass number increases by 4. The resulting nucleus is thorium-242 ($^{242}_{94}\text{Th}$), but it emits a neutron to reach stability.
- Result: The missing product is thorium-242 ($^{242}_{94}\text{Th}$).
- Type of Reaction: Artificial transmutation.

$$
^{238}_{92} \text{U} + { }_{2}^{4} \text{He} \rightarrow ^{242}_{94} \text{Th} + { }_{0}^{1} \text{n}
$$

---

##### 11.
$$
^{14}_{6} \text{C} \rightarrow { }_{7}^{14} \text{N} + \_\_\_\_
$$

- Analysis: The reactant is carbon-14 ($^{14}_{6}\text{C}$). The product includes nitrogen-14 ($^{14}_{7}\text{N}$), indicating $\beta^-$ decay.
- Balancing: In $\beta^-$ decay, a neutron converts into a proton, emitting an electron and an antineutrino. The resulting nucleus is nitrogen-14 ($^{14}_{7}\text{N}$).
- Result: The missing product is an electron ($^{0}_{-1}\text{e}$).
- Type of Reaction: $\beta^-$ emission.

$$
^{14}_{6} \text{C} \rightarrow { }_{7}^{14} \text{N} + { }_{-1}^{0} \text{e}
$$

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##### 12.
$$
^{187}_{75} \text{Re} + \_\_\_\_ \rightarrow { }_{75}^{188} \text{Re} + { }_{1}^{1} \text{H}
$$

- Analysis: The reactant is rhenium-187 ($^{187}_{75}\text{Re}$). The product includes rhenium-188 ($^{188}_{75}\text{Re}$) and a proton ($^{1}_{1}\text{H}$), indicating a nuclear reaction involving a proton.
- Balancing: To balance this, the reactant must include a deuteron ($^{2}_{1}\text{D}$), which splits into a proton and a neutron. The neutron is absorbed by the rhenium-187 nucleus, increasing its mass number by 1.
- Result: The missing reactant is a deuteron ($^{2}_{1}\text{D}$).
- Type of Reaction: Artificial transmutation.

$$
^{187}_{75} \text{Re} + { }_{1}^{2} \text{D} \rightarrow { }_{75}^{188} \text{Re} + { }_{1}^{1} \text{H}
$$

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##### 13.
$$
^{21}_{11} \text{Na} + \_\_\_\_ \rightarrow { }_{10}^{22} \text{Ne}
$$

- Analysis: The reactant is sodium-21 ($^{21}_{11}\text{Na}$). The product includes neon-22 ($^{22}_{10}\text{Ne}$), indicating a nuclear reaction involving a neutron.
- Balancing: To balance this, the reactant must include a neutron ($^{1}_{0}\text{n}$), which is absorbed by the sodium-21 nucleus, increasing its mass number by 1. The resulting nucleus then emits a proton to reach stability.
- Result: The missing reactant is a neutron ($^{1}_{0}\text{n}$).
- Type of Reaction: Artificial transmutation.

$$
^{21}_{11} \text{Na} + { }_{0}^{1} \text{n} \rightarrow { }_{10}^{22} \text{Ne} + { }_{1}^{1} \text{H}
$$

---

##### 14.
$$
^{238}_{94} \text{Po} \rightarrow \_\_\_\_ + { }_{2}^{4} \text{He}
$$

- Analysis: The reactant is polonium-238 ($^{238}_{94}\text{Po}$). The product includes an alpha particle ($^{4}_{2}\text{He}$), indicating $\alpha$ decay.
- Balancing: In $\alpha$ decay, the nucleus emits an alpha particle, reducing the atomic number by 2 and the mass number by 4. The resulting nucleus is lead-234 ($^{234}_{92}\text{Pb}$).
- Result: The missing product is lead-234 ($^{234}_{92}\text{Pb}$).
- Type of Reaction: $\alpha$ emission.

$$
^{238}_{94} \text{Po} \rightarrow ^{234}_{92} \text{Pb} + { }_{2}^{4} \text{He}
$$

---

##### 15.
$$
^{235}_{92} \text{Es} + { }_{2}^{4} \text{He} \rightarrow \_\_\_\_ + { }_{0}^{1} \text{n}
$$

- Analysis: The reactants are einsteinium-235 ($^{235}_{92}\text{Es}$) and an alpha particle ($^{4}_{2}\text{He}$). This indicates an artificial transmutation.
- Balancing: When an alpha particle is absorbed, the atomic number increases by 2 and the mass number increases by 4. The resulting nucleus is fermium-239 ($^{239}_{94}\text{Fm}$), but it emits a neutron to reach stability.
- Result: The missing product is fermium-239 ($^{239}_{94}\text{Fm}$).
- Type of Reaction: Artificial transmutation.

$$
^{235}_{92} \text{Es} + { }_{2}^{4} \text{He} \rightarrow ^{239}_{94} \text{Fm} + { }_{0}^{1} \text{n}
$$

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#### 2. Writing Balanced Nuclear Equations:

##### a) Plutonium-234 ($^{234}_{94}\text{Pu}$): Alpha Decay
- Process: 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.
- Equation:
$$
^{234}_{94} \text{Pu} \rightarrow ^{230}_{92} \text{U} + { }_{2}^{4} \text{He}
$$

##### b) Strontium-90 ($^{90}_{38}\text{Sr}$): Beta Decay
- Process: In beta decay, a neutron converts into a proton, emitting an electron ($^{0}_{-1}\text{e}$) and an antineutrino ($\bar{\nu}_e$). The atomic number increases by 1, and the mass number remains the same.
- Equation:
$$
^{90}_{38} \text{Sr} \rightarrow ^{90}_{39} \text{Y} + { }_{-1}^{0} \text{e} + \bar{\nu}_e
$$

##### c) Radium-226 ($^{226}_{88}\text{Ra}$): Alpha, Beta, and Gamma Decay
- Alpha Decay:
$$
^{226}_{88} \text{Ra} \rightarrow ^{222}_{86} \text{Rn} + { }_{2}^{4} \text{He}
$$
- Beta Decay:
$$
^{222}_{86} \text{Rn} \rightarrow ^{222}_{87} \text{At} + { }_{-1}^{0} \text{e} + \bar{\nu}_e
$$
- Gamma Decay: Gamma emission does not change the atomic number or mass number but releases energy.
$$
^{222}_{87} \text{At} \rightarrow ^{222}_{87} \text{At}^* \rightarrow ^{222}_{87} \text{At} + \gamma
$$

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Final Answer:

$$
\boxed{
\begin{aligned}
1. & \quad ^{40}_{19} \text{K} \rightarrow { }_{-1}^{0} \text{e} + ^{40}_{20} \text{Ca} \quad (\beta^- \text{emission}) \\
2. & \quad ^{236}_{94} \text{Pu} \rightarrow { }_{2}^{4} \text{He} + ^{232}_{92} \text{U} \quad (\alpha \text{emission}) \\
3. & \quad ^{233}_{92} \text{U} \rightarrow 6 \cdot { }_{2}^{4} \text{He} + 2 \cdot { }_{-1}^{0} \text{e} + ^{209}_{82} \text{Pb} \quad (\text{Artificial transmutation}) \\
4. & \quad ^{1}_{1} \text{H} + { }_{1}^{1} \text{H} \rightarrow ^{2}_{1} \text{H} + { }_{+1}^{0} \text{e} \quad (\text{Fusion}) \\
5. & \quad ^{6}_{3} \text{Li} + { }_{0}^{1} \text{n} \rightarrow 2 \cdot { }_{2}^{4} \text{He} \quad (\text{Artificial transmutation}) \\
6. & \quad ^{27}_{13} \text{Al} + { }_{2}^{4} \text{He} \rightarrow { }_{15}^{30} \text{P} + { }_{0}^{1} \text{n} \quad (\text{Artificial transmutation}) \\
7. & \quad ^{9}_{4} \text{Be} + { }_{1}^{1} \text{H} \rightarrow { }_{1}^{2} \text{H} + { }_{2}^{4} \text{He} \quad (\text{Artificial transmutation}) \\
8. & \quad ^{37}_{19} \text{K} \rightarrow { }_{-1}^{0} \text{e} + ^{37}_{20} \text{Ca} \quad (\beta^- \text{emission}) \\
9. & \quad ^{235}_{92} \text{U} + { }_{0}^{1} \text{n} \rightarrow { }_{56}^{142} \text{Ba} + { }_{36}^{89} \text{Kr} + 3 \cdot { }_{0}^{1} \text{n} \quad (\text{Fission}) \\
10. & \quad ^{238}_{92} \text{U} + { }_{2}^{4} \text{He} \rightarrow ^{242}_{94} \text{Th} + { }_{0}^{1} \text{n} \quad (\text{Artificial transmutation}) \\
11. & \quad ^{14}_{6} \text{C} \rightarrow { }_{7}^{14} \text{N} + { }_{-1}^{0} \text{e} \quad (\beta^- \text{emission}) \\
12. & \quad ^{187}_{75} \text{Re} + { }_{1}^{2} \text{D} \rightarrow { }_{75}^{188} \text{Re} + { }_{1}^{1} \text{H} \quad (\text{Artificial transmutation}) \\
13. & \quad ^{21}_{11} \text{Na} + { }_{0}^{1} \text{n} \rightarrow { }_{10}^{22} \text{Ne} + { }_{1}^{1} \text{H} \quad (\text{Artificial transmutation}) \\
14. & \quad ^{238}_{94} \text{Po} \rightarrow ^{234}_{92} \text{Pb} + { }_{2}^{4} \text{He} \quad (\alpha \text{emission}) \\
15. & \quad ^{235}_{92} \text{Es} + { }_{2}^{4} \text{He} \rightarrow ^{239}_{94} \text{Fm} + { }_{0}^{1} \text{n} \quad (\text{Artificial transmutation}) \\
\end{aligned}
}
$$

$$
\boxed{
\begin{aligned}
& \text{Plutonium-234: } ^{234}_{94} \text{Pu} \rightarrow ^{230}_{92} \text{U} + { }_{2}^{4} \text{He} \\
& \text{Strontium-90: } ^{90}_{38} \text{Sr} \rightarrow ^{90}_{39} \text{Y} + { }_{-1}^{0} \text{e} + \bar{\nu}_e \\
& \text{Radium-226: } \\
& \quad \text{Alpha Decay: } ^{226}_{88} \text{Ra} \rightarrow ^{222}_{86} \text{Rn} + { }_{2}^{4} \text{He} \\
& \quad \text{Beta Decay: } ^{222}_{86} \text{Rn} \rightarrow ^{222}_{87} \text{At} + { }_{-1}^{0} \text{e} + \bar{\nu}_e \\
& \quad \text{Gamma Decay: } ^{222}_{87} \text{At} \rightarrow ^{222}_{87} \text{At}^* \rightarrow ^{222}_{87} \text{At} + \gamma \\
\end{aligned}
}
$$