Question

Balance the following equations by filling in the blanks. (a) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} n \longrightarrow{ }_{54}^{137}---+2{ }_{0}^{1} n+\)--- (b) \({ }_{90}^{232} \mathrm{Th}+{ }_{6}^{12}\)--- \(-\)---- \({ }_{0}^{1} n+{ }_{96}^{240} \mathrm{Cm}\) (c) \({ }_{2}^{4} \mathrm{He}+{ }_{42}^{96} \mathrm{Mo} \longrightarrow{ }_{43}^{100}\)----+----- (d)_____ \(+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{84}^{210}---- \longrightarrow+{ }_{0}^{1} n\)

Short answer

}_{38} \mathrm{Y}\) (b) \(^{232}_{90} \mathrm{Th} + ^{?}_{?} \mathrm{X} \longrightarrow\ ^{1}_0 n + ^{240}_{96} \mathrm{Cm}\) (c) \(^{4}_2 \mathrm{He} + ^{96}_{42} \mathrm{Mo} \longrightarrow\ ^{100}_{43} \mathrm{X} + ^{?}_{?} \mathrm{Y}\) (d) \(^?_? \mathrm{X} + ^2_1 \mathrm{H} \longrightarrow\ ^{210}_{84} \mathrm{Y} + ^{1}_0 n\) Answer: (a) The balanced nuclear reaction is \(^{235}_92 \mathrm{U} + ^{1}_0 n \longrightarrow\ ^{137}_{54} \mathrm{Xe} + 2(^{1}_0 n) + ^{97}_{38} \mathrm{Sr}\), with X as Xenon (Xe) and Y as Strontium (Sr). (b) The balanced nuclear reaction is \(^{232}_{90} \mathrm{Th} + ^{12}_{6} \mathrm{C} \longrightarrow\ ^{1}_0 n + ^{240}_{96} \mathrm{Cm}\), with X as Carbon-12 nucleus (\(^{12}_6 \mathrm{C}\)). (c) The balanced nuclear reaction is \(^{4}_2 \mathrm{He} + ^{96}_{42} \mathrm{Mo} \longrightarrow\ ^{100}_{43} \mathrm{Tc} + ^0_{-1} e\), with X as Technetium (Tc) and Y as an electron (\(^0_{-1} e\)). (d) The balanced nuclear reaction is \(^{208}_{83} \mathrm{Bi} + ^2_1 \mathrm{H} \longrightarrow\ ^{210}_{84} \mathrm{Po} + ^1_0 n\), with X as Bismuth (Bi) and Y as Polonium (Po).

Step-by-step solution

Identify missing element

Based on the given atomic number 54, the missing element is Xenon (Xe).

Balance the atomic mass and number

To balance the atomic mass and number, \(^{235}_92 \mathrm{U} + ^{1}_0 n \longrightarrow\ ^{137}_{54} \mathrm{Xe} + 2(^{1}_0 n) + ^{?}_{38} \mathrm{Y}\), we have Y as Sr (Strontium).

Final Equation (a)

\(^{235}_92 \mathrm{U} + ^{1}_0 n \longrightarrow\ ^{137}_{54} \mathrm{Xe} + 2(^{1}_0 n) + ^{97}_{38} \mathrm{Sr}\) (b) Balancing the second reaction

Identify missing particles

From the given equation, we need to find a particle with atomic mass 12 and atomic number 6. This particle is the Carbon-12 nucleus \(^{12}_6 \mathrm{C}\).

Balance the atomic mass and number

The equation is already balanced, and there are no other missing particles.

Final Equation (b)

\(^{232}_{90} \mathrm{Th} + ^{12}_{6} \mathrm{C} \longrightarrow\ ^{1}_0 n + ^{240}_{96} \mathrm{Cm}\) (c) Balancing the third reaction

Identify missing element

Based on the given atomic number 43, the missing element is Tc (Technetium).

Balance the atomic mass and number

To balance the atomic mass and number, \(^{4}_2 \mathrm{He} + ^{96}_{42} \mathrm{Mo} \longrightarrow\ ^{100}_{43} \mathrm{Tc} + ^{?}_{?} \mathrm{X}\), we have X as \(^0_{-1} e\), an electron.

Final Equation (c)

\(^{4}_2 \mathrm{He} + ^{96}_{42} \mathrm{Mo} \longrightarrow\ ^{100}_{43} \mathrm{Tc} + ^0_{-1} e\) (d) Balancing the fourth reaction

Identify missing element

Based on the given atomic mass 210 and atomic number 84, the missing element is Po (Polonium).

Identify missing particles

We have a missing particle in the first half of the reaction. Using the balanced atomic mass and number, \(^?_? \mathrm{X} + ^2_1 \mathrm{H} \longrightarrow\ ^{210}_{84} \mathrm{Po} + ^{1}_0 n\), we find the particle as \(^{208}_{83} \mathrm{Bi}\) (Bismuth).

Final Equation (d)

\(^{208}_{83} \mathrm{Bi} + ^2_1 \mathrm{H} \longrightarrow\ ^{210}_{84} \mathrm{Po} + ^1_0 n\)

Key concepts

Nuclear Chemistry

Nuclear chemistry delves into the reactions and processes involving the nucleus of atoms. It differs from ordinary chemical reactions, which involve electrons and other atomic particles, by focusing on phenomena such as radioactivity, nuclear transmutation, and nuclear properties. An essential aspect of nuclear chemistry is understanding how unstable nuclei undergo transformations to achieve stability, releasing energy in the process.

These transformations, captured in nuclear equations much like the ones seen in the exercise, are fundamental to multiple applications, ranging from medical treatments using radioisotopes to the generation of electricity in nuclear reactors. Mastering the art of balancing nuclear reactions is not only key to comprehending how elements can be converted from one form to another but is also crucial for harnessing nuclear energy safely and effectively.

Radioactive Decay

Radioactive decay is a random but predictable process where an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. There are several types of radioactive decay, such as alpha decay, beta decay, and gamma radiation. For instance, in the nuclear reaction from the exercise, the emission of an electron (beta particle) accompanies the creation of Technetium-100.

This process, along with others, contributes to the transformation of elements and the generation of energy or other particles. Understanding radioactive decay is integral to predicting how certain materials will behave over time, whether it's for dating archaeological finds via carbon-14 decay or managing nuclear waste.

Nuclide Notation

Nuclide notation is a system used to denote specific isotopes of elements, providing information about both the number of protons and neutrons in a nucleus. The notation includes the element's symbol surrounded by two numbers: the atomic number (number of protons) is shown as a subscript to the left of the symbol, and the mass number (total number of protons and neutrons) as a superscript to the left.

For example, Uranium-235 is written as \( ^{235}_{92} \mathrm{U} \). This notation is essential for understanding which nuclide is involved in a nuclear reaction, as seen in the exercise where the identification of nuclides is the first step to balancing the equations. Properly identifying these symbols ensures correct application of the laws of conservation of mass and atomic number when balancing nuclear reactions.

Atomic Mass and Number

The atomic mass and number are vital for identifying elements and their isotopes. The atomic number, represented by the symbol 'Z', is the number of protons found in the nucleus, and it defines the chemical identity of the element. The atomic mass number, denoted by 'A', is the sum of protons and neutrons, giving us a glimpse into the isotope's version of the element.

In practice, like in the provided exercise, these numbers must be conserved in nuclear equations. When balancing a nuclear reaction, the sum of the atomic mass numbers and the sum of the atomic numbers on both sides of the equation must be equal. This is a reflection of the conservation laws that govern all physical processes, and it is through these principles that we can systematically solve for unknown isotopes or particles produced in nuclear reactions.