Problem 28 - Chemistry The Central Science

Question

Complete and balance the following nuclear equations by supplying the missing particle: (a) \({ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \longrightarrow ?+{ }_{1}^{1} \mathrm{H}\) (b) \({ }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{e}\) (orbital electron) \(\longrightarrow\) (c) ? \(+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{1}^{1} \mathrm{H}\) (d) \({ }_{26}^{58} \mathrm{Fe}+2{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{27}^{60} \mathrm{Co}+?\) (e) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{54}^{135} \mathrm{Xe}+2{ }_{0}^{1} \mathrm{n}+?\)

Step-by-Step Solution

Answer

(a) \( { }_{7}^{14} \mathrm{N} + { }_{2}^{4} \mathrm{He} \longrightarrow { }_{8}^{17} \mathrm{O} + { }_{1}^{1} \mathrm{H}\) (b) \( { }_{19}^{40} \mathrm{K} + { }_{-1}^{0} \mathrm{e} \longrightarrow { }_{18}^{40} \mathrm{Ar}\) (c) \( { }_{13}^{27} \mathrm{Al} + { }_{2}^{4} \mathrm{He} \longrightarrow { }_{14}^{30} \mathrm{Si} + { }_{1}^{1} \mathrm{H}\) (d) \({ }_{26}^{58} \mathrm{Fe} + 2{ }_{0}^{1} \mathrm{n} \longrightarrow { }_{27}^{60} \mathrm{Co} + 3{ }_{0}^{1} \mathrm{n}\) (e) \({ }_{92}^{235} \mathrm{U} + { }_{0}^{1} \mathrm{n} \longrightarrow { }_{54}^{135} \mathrm{Xe} + 2{ }_{0}^{1} \mathrm{n} + { }_{38}^{100} \mathrm{Sr}\)

Step 1: Identify the numbers of protons and neutrons in the given particles

In the nuclear equation, nitrogen-14 (\( { }_{7}^{14} \mathrm{N}\)) has 7 protons and 7 neutrons (14-7), and helium-4 (\( { }_{2}^{4} \mathrm{He}\)) has 2 protons and 2 neutrons (4-2). We also know that the hydrogen-1 (\( { }_{1}^{1} \mathrm{H}\)) has 1 proton and 0 neutrons.

Step 2: Determine the numbers of protons and neutrons in the missing particle

Since we need to balance both the number of protons and neutrons in the nuclear reaction, we can add the protons and neutrons of the given particles and subtract the protons and neutrons of hydrogen-1 from it. This gives us: Protons: 7 + 2 - 1 = 8 Neutrons: 7 + 2 - 0 = 9 Thus, the missing particle has 8 protons and 9 neutrons.

Step 3: Identify the missing particle

Using the periodic table, we can find that the element with 8 protons is oxygen (O). Since the total number of nucleons in the missing particle is 17 (8+9), the missing particle is oxygen-17 (\( { }_{8}^{17} \mathrm{O}\)). Thus, the complete balanced nuclear equation is: \({ }_{7}^{14} \mathrm{N} + { }_{2}^{4} \mathrm{He} \longrightarrow { }_{8}^{17} \mathrm{O} + { }_{1}^{1} \mathrm{H}\) (b) Balance the charge:

Step 1: Identify the charge of the given particles

In the nuclear equation, potassium-40 (\( { }_{19}^{40} \mathrm{K}\)) has a charge of +19, and orbital electron (\( { }_{-1}^{0} \mathrm{e}\)) has a charge of -1.

Step 2: Determine the charge of the missing particle

Since we need to balance the charge in the nuclear reaction, we can simply add the charges of the given particles. This gives us a total charge of +18.

Step 3: Identify the missing particle

Using the periodic table, we can find that the element with an atomic number of 18 is argon (Ar). Thus, the missing particle is argon-40 (\( { }_{18}^{40} \mathrm{Ar}\)). The complete balanced nuclear equation is: \({ }_{19}^{40} \mathrm{K} + { }_{-1}^{0} \mathrm{e} \longrightarrow { }_{18}^{40} \mathrm{Ar}\) (c) Balance protons and neutrons: Follow the same steps as in (a):

Step 1: Identify the number of protons and neutrons in other particles

Silicon-30 (\( { }_{14}^{30} \mathrm{Si}\)) has 14 protons and 16 neutrons (30-14), and helium-4 (\( { }_{2}^{4} \mathrm{He}\)) has 2 protons and 2 neutrons (4-2). We also know that the hydrogen-1 (\( { }_{1}^{1} \mathrm{H}\)) has 1 proton and 0 neutrons.

Step 2: Determine the number of protons and neutrons in the missing particle

We subtract the protons and neutrons of Silicon-30 and helium-4 and add the protons and neutrons of hydrogen-1. This gives us: Protons: 14 - 2 + 1 = 13 Neutrons: 16 - 2 + 0 = 14 Thus, the missing particle has 13 protons and 14 neutrons.

Step 3: Identify the missing particle

Using the periodic table, we find that the element with 13 protons is aluminum (Al). Since the total number of nucleons in the missing particle is 27 (13+14), the missing particle is aluminum-27 (\( { }_{13}^{27} \mathrm{Al}\)). The complete balanced nuclear equation is: \({ }_{13}^{27} \mathrm{Al} + { }_{2}^{4} \mathrm{He} \longrightarrow { }_{14}^{30} \mathrm{Si} + { }_{1}^{1} \mathrm{H}\) (d) Determine the missing neutron count:

Step 1: Identify the neutron count of each given particle

Iron-58 (\( { }_{26}^{58} \mathrm{Fe}\)) has 32 neutrons (58-26), cobalt-60 (\( { }_{27}^{60} \mathrm{Co}\)) has 33 neutrons (60-27), and a neutron (\( { }_{0}^{1} \mathrm{n}\)) has 1 neutron.

Step 2: Determine the number of missing neutrons

We need to balance the neutron count in this nuclear equation. There are 2 neutrons on both the reactant and product side. We can calculate the total neutron count on both sides and find the difference: Reactant side: 32 (Fe) + 2(2 neutrons) = 36 Product side: 33 (Co) So, the missing neutrons are: 36 - 33 = 3

Step 3: Write the complete balanced nuclear equation

The complete balanced nuclear equation is: \({ }_{26}^{58} \mathrm{Fe} + 2{ }_{0}^{1} \mathrm{n} \longrightarrow { }_{27}^{60} \mathrm{Co} + 3{ }_{0}^{1} \mathrm{n}\) (e) Identify the missing particle:

Step 1: Identify the number of protons and neutrons in the given particles

Uranium-235 (\( { }_{92}^{235} \mathrm{U}\)) has 92 protons and 143 neutrons (235-92), xenon-135 (\( { }_{54}^{135} \mathrm{Xe}\)) has 54 protons and 81 neutrons (135-54), and a neutron (\( { }_{0}^{1} \mathrm{n}\)) has 1 neutron.

Step 2: Determine the number of protons and neutrons in the missing particle

Since we need to balance both the number of protons and neutrons in the nuclear reaction, we can add the total nucleons of Uranium-235 and one neutron and then subtract the nucleons of Xenon-135 and the remaining neutron. This gives us: Protons: 92 (U) + 0 (single n) - 54 (Xe) = 38 Neutrons: 143 (U) + 1 (single n) - 81 (Xe) - 1 (remaining n) = 62 Thus, the missing particle has 38 protons and 62 neutrons.

Step 3: Identify the missing particle

Using the periodic table, we find that the element with 38 protons is strontium (Sr). Since the total number of nucleons in the missing particle is 100 (38+62), the missing particle is strontium-100 (\( { }_{38}^{100} \mathrm{Sr}\)). The complete balanced nuclear equation is: \({ }_{92}^{235} \mathrm{U} + { }_{0}^{1} \mathrm{n} \longrightarrow { }_{54}^{135} \mathrm{Xe} + 2{ }_{0}^{1} \mathrm{n} + { }_{38}^{100} \mathrm{Sr}\)

Key Concepts

Nuclear Chemistry

Nuclear chemistry involves the study of the chemical and physical properties of elements as influenced by changes in the structure of the atomic nucleus. This field addresses various topics such as radioactive decay, nuclear fission and fusion, and the synthesis of new elements.

Understanding nuclear chemistry is crucial when dealing with nuclear equations, as these reactions involve the transformation of one element into another through processes like alpha and beta decay, or neutron absorption. Unlike chemical reactions which only involve changes in the electron configurations of atoms, nuclear reactions alter the composition of nuclei, thus leading to the formation of different elements or isotopes.

In the context of the exercises provided, nuclear chemistry helps us understand how to identify and balance nuclear reactions, an essential skill in predicting the products of a given nuclear process and understanding natural phenomena such as radioactive decay in the environment or artificial processes in nuclear reactors.

Balancing Nuclear Reactions

Balancing nuclear reactions is analogous to balancing chemical equations but involves ensuring both mass and atomic numbers are conserved. Each nuclide is represented by its nuclear symbol, where the superscript indicates the mass number (protons plus neutrons) and the subscript indicates the atomic number (the number of protons).

To balance a nuclear equation, follow these steps: First, ensure that the sum of the mass numbers on both sides of the equation is equal. Then, do the same for the atomic numbers. This confirms that both nucleon (proton plus neutron) and charge conservation laws hold. In the example of nitrogen-14 reacting with helium-4 generating a particle and hydrogen-1, we applied these principles to deduce the other particle's identity. These skills are crucial for solving nuclear balances and reinforce the concept that nuclear reactions must respect the fundamental laws of conservation.

Nuclear Reaction Calculations

Calculating the details of a nuclear reaction requires understanding the changes in nucleus composition during the reaction. This involves identifying the number of protons and neutrons before and after the reaction to ensure the reaction abides by the law of conservation of mass and atomic numbers.

When a missing particle or isotope is involved, as seen in the exercises here, we calculate the difference in protons and neutrons to find the identity of the unknown component. Consulting the periodic table provides guidance in deducing the right element or isotope. For heavy elements like uranium, extensive nuclear reaction calculations might relate to fission processes, which are pivotal in nuclear power generation and the study of radiological materials. By mastering these calculations, students can predict the outcomes of nuclear reactions and grasp the complexities of nuclear transformations.