Question

Complete these nuclear equations and identify \(X\) in each case: (a) \({ }^{135}{ }_{53} \mathrm{I} \longrightarrow{ }_{54}^{135} \mathrm{Xe}+\mathrm{X}\) (b) \({ }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{-1}^{0} \beta+\mathrm{X}\) (c) \({ }_{27}^{59} \mathrm{Co}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{25}^{56} \mathrm{Mn}+\mathrm{X}\) (d) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{40}^{99} \mathrm{Zr}+{ }_{52}^{135} \mathrm{Te}+2 \mathrm{X}\)

Short answer

The identified unknown \(X\) in each case is: (a) Electron \(e^{-}\), (b) Calcium \(^{40}_{20}Ca\), (c) Alpha particle \(^{4}_{2}He\), (d) Neutron \(^{1}_{0}n\)

Step-by-step solution

Analyze the equations and fill missing values

For each equation, the aim should be to check that the numbers on both sides of the reaction equation add up. This means that the sum of atomic numbers (subscripts) and the sum of mass numbers (superscripts) should be the same on both sides of the equation.

Solve for X in first equation

For the equation \(^{135}_{53}I \rightarrow ^{135}_{54}Xe + X\), by comparing the superscripts, we notice that \(135=135 + mass~number~of~X\). This means that the mass number of X is 0. Comparing the subscripts, \(53 = 54 + atomic~number~of~X\), so the atomic number of X is -1. An atomic number of -1 and mass number of 0 corresponds to an electron. Hence, \(X\) is an electron.

Solve for X in second equation

For the equation \(^{40}_{19}K \rightarrow ^{0}_{-1}\beta + X\), we again similarly get mass number of \(X\) as \(40\) and atomic number as \(20\). A mass number of \(40\) and atomic number \(20\) corresponds to calcium (Ca). Hence, \(X\) is \(^{40}_{20}Ca\).

Solve for X in third equation

For \(^{59}_{27}Co + ^{1}_{0}n \rightarrow ^{56}_{25}Mn + X\), to solve for \(X\), we can see that the mass number of \(X\) will be \(59+1-56=4\), and atomic number will be \(27+0-25=2\). So, \(X\) will be an alpha particle \(^{4}_{2}He\).

Solve for X in fourth equation

The final equation is \(^{235}_{92}U + ^{1}_{0}n \rightarrow ^{99}_{40}Zr + ^{135}_{52}Te + 2X\). Here, the mass number of \(X\) will be \((235+1-99-135)/2=1\), and the atomic number will be \((92+0-40-52)/2=0\). Therefore, \(X\) will be a neutron \(^{1}_{0}n\).

Key concepts

Atomic Number

The atomic number is fundamental to understanding nuclear equations. It represents the number of protons in an atom's nucleus and determines the element's identity. Each element has a unique atomic number. For instance, the atomic number for iodine (I) is 53, meaning it has 53 protons.
In nuclear reactions, the atomic numbers must balance on both sides of the equation to obey the law of conservation of charge. This ensures that the total number of protons does not change during the reaction. When analyzing nuclear equations:

• Sum up the atomic numbers on one side.
• Make sure they equal the sum on the other side.
• If there's a discrepancy, it indicates a change in element identity or the presence of particles like electrons or positrons, which can alter the atomic number without affecting mass.

Understanding atomic numbers helps you balance reactions and identify unknowns like "X" in given nuclear equations.

Mass Number

The mass number is crucial for analyzing nuclear equations, representing the total number of protons and neutrons in a nucleus. Unlike the atomic number, it is not unique to each element due to isotopes, which have the same atomic number but different mass numbers because of varying neutron counts.
In nuclear processes, the sum of mass numbers on both sides of the reaction equation should also be equal. This conservation ensures that the total count of nucleons (protons and neutrons) remains constant. When working with nuclear equations:

• Calculate the total mass number on each side of the equation.
• Ensure that these totals are identical.
• Discrepancies can help identify particles like neutrons or alpha particles, which influence mass but not charge.

In this way, comprehending mass numbers aids in solving nuclear reactions and detecting missing components through equations.

Alpha Particle

An alpha particle plays a significant role in nuclear equations. It consists of two protons and two neutrons and is identical to a helium-4 nucleus, hence symbolized as dhc . In nuclear reactions, alpha particles often emerge as products, indicating the emission of this particle from a nucleus.
Since alpha particles have both mass and charge, they affect the equation's balance. They have an atomic number of 2 and a mass number of 4. Thus, their addition or subtraction in a reaction:

• Reduces the original nucleus's mass number by 4.
• Decreases the atomic number by 2.
• Is common in decay processes like alpha decay.

Identifying alpha particles in nuclear equations is crucial because they significantly alter the composition of the remaining nucleus and affect the elements produced.

Neutron

Neutrons are neutral particles found in an atom's nucleus. Unlike protons, they don't influence the atomic number, but they do contribute to the mass number. This is why different isotopes of the same element have the same atomic number but different mass numbers. In nuclear reactions, neutrons can be either reactants or products. They are involved in processes such as nuclear fission and fusion, affecting the equation as:

• Neutrons have a mass number of 1 and an atomic number of 0.
• They increase or decrease the total mass number.
• Despite having no charge, their presence is critical, especially in balancing nuclear equations.

Understanding the role of neutrons enables you to complete nuclear equations effectively, ensuring that both mass and identity conservation laws are upheld in processes involving changes in atomic nuclei.