Problem 21 - Chemistry and Chemical Reactivity

Question

Complete the following nuclear equations. Write the mass number and atomic number for the remaining particle, as well as its symbol. (a) \(^{54}_{26} \mathrm{Fe}+^{4}_{2} \mathrm{He} \longrightarrow 2 \frac{1}{0} \mathrm{n}+?\) (b) \(^{27}_{13} \mathrm{Al}+^{4}_{2} \mathrm{He} \longrightarrow_{15}^{30} \mathrm{P}+?\) (c) \(^{32}_{16} \mathrm{S}+^{1}_{0} \mathrm{n} \longrightarrow_{1}^{1} \mathrm{H}+?\) (d) \(_{42}^{96} \mathrm{Mo}+_{1}^{2} \mathrm{H} \longrightarrow_{0}^{1} \mathrm{n}+?\) (e) \(^{98}_{42} \mathrm{Mo}+^{1}_{0} \mathrm{n} \longrightarrow _{99}^{43} \mathrm{Tc}+?\) (f) \(^{18}_{9} \mathrm{F} \longrightarrow_{18}^{8} \mathrm{O}+?\)

Step-by-Step Solution

Answer

(a) \(^{56}_{28} \mathrm{Ni}\) (b) \(^{1}_{0} \mathrm{n}\) (c) \(^{32}_{15} \mathrm{P}\) (d) \(^{97}_{43} \mathrm{Tc}\) (e) \(^{0}_{-1} \mathrm{e}\) (f) \(^{0}_{+1} \mathrm{e}\)

Step 1: Analyze Equation (a)

We have the nuclear reaction: \(^{54}_{26} \mathrm{Fe} + ^{4}_{2} \mathrm{He} \rightarrow 2 \frac{1}{0} \mathrm{n} + ?\). To find the missing particle, start by balancing the mass numbers and atomic numbers separately. The left side has a mass number total of \(54 + 4 = 58\) and an atomic number total of \(26 + 2 = 28\). On the right side, the 2 neutrons contribute \(2(1) = 2\) to the mass number and \(2(0) = 0\) to the atomic number. Therefore, the missing particle must have a mass number of \(58 - 2 = 56\) and an atomic number of \(28\). This corresponds to \(^{56}_{28} \mathrm{Ni}\).

Step 2: Analyze Equation (b)

The equation is \(^{27}_{13} \mathrm{Al} + ^{4}_{2} \mathrm{He} \rightarrow^{30}_{15} \mathrm{P} + ?\). Calculate the mass and atomic numbers on each side. Initially, the left side totals \(27 + 4 = 31\) for mass number and \(13 + 2 = 15\) for atomic number. The right side shows \(^{30}_{15} \mathrm{P}\), thus a mass number of \(31 - 30 = 1\) and an atomic number of \(15 - 15 = 0\). The missing particle is a neutron: \(^{1}_{0} \mathrm{n}\).

Step 3: Analyze Equation (c)

We have \(^{32}_{16} \mathrm{S} + ^{1}_{0} \mathrm{n} \rightarrow^{1}_{1} \mathrm{H} + ?\). Total the mass numbers, beginning with the left side: \(32 + 1 = 33\). The atomic number is \(16 + 0 = 16\). The right side includes \(^{1}_{1} \mathrm{H}\) out of \(33 - 1 = 32\) for the mass number and an atomic number of \(16 - 1 = 15\). The remaining particle is \(^{32}_{15} \mathrm{P}\).

Step 4: Analyze Equation (d)

Given \(_{42}^{96} \mathrm{Mo} + _{1}^{2} \mathrm{H} \rightarrow_{0}^{1} \mathrm{n} + ?\), we add mass numbers on the left: \(96 + 2 = 98\), and atomic numbers: \(42 + 1 = 43\). On the right, a neutron \(^{1}_{0} \mathrm{n}\) has a mass number of \(1\). Therefore, the missing particle accounts for: mass \(98 - 1 = 97\), and atomic number \(43\). Corresponds to \(^{97}_{43} \mathrm{Tc}\).

Step 5: Analyze Equation (e)

We work with \(^{98}_{42} \mathrm{Mo} + ^{1}_{0} \mathrm{n} \rightarrow _{99}^{43} \mathrm{Tc} + ?\). Begin by summing on the left: mass \(98 + 1 = 99\), atomic \(42 + 0 = 42\). Then, on the right, \(^{99}_{43} \mathrm{Tc}\). The mismatch is accounted by a beta particle: \(^{0}_{-1} \mathrm{e}\), allowing the conservation of charge with an electron ejection.

Step 6: Analyze Equation (f)

Consider \(^{18}_{9} \mathrm{F} \rightarrow_{18}^{8} \mathrm{O} + ?\). Mass number checks (remains 18), but the atomic number decreases by 1, from 9 to 8, indicating positron emission (beta-plus): \(^{0}_{+1} \mathrm{e}\), where positrons carry the opposite sign.

Key Concepts

Nuclear Equations

Nuclear equations are representations of nuclear reactions written in equation form, similar to chemical equations. Unlike chemical reactions that involve the exchange or sharing of electrons, nuclear reactions involve the core of an atom—its nucleus. In a nuclear equation, different particles are involved, including protons, neutrons, and even electrons, which may be emitted as beta particles. The goal of writing and balancing a nuclear equation is to ensure that both the mass number and the atomic number are conserved on both sides of the reaction. This means:

• The total sum of mass numbers (a count of protons and neutrons) must be equal on both sides of the equation.
• The total atomic numbers (the number of protons) must also balance out.

To solve such equations, you identify missing particles using the conservation laws mentioned above. Understanding how to balance nuclear equations is crucial in fields like nuclear chemistry and physics, where they describe processes such as nuclear decay, fission, and fusion.

Mass Number

The mass number is a crucial part of understanding nuclear chemistry. It represents the total number of protons and neutrons in the nucleus of an atom. The mass number is always a whole number and is denoted by the letter A in scientific notation.In nuclear equations, this number helps us track the mass balance, ensuring the reaction obeys the conservation of mass. For example, when balancing the equation for a reaction involving iron (\( ^{54}_{26} \text{Fe} \)) and a helium nucleus (\( ^{4}_{2} \text{He} \)), we start by calculating the total mass numbers on each side of the equation to make sure they match.Remember that the mass number is different from the atomic weight you might encounter in chemistry, as atomic weight is the average mass of all isotopes of an element. In nuclear chemistry, however, it's the mass number that's more relevant, as it helps us track the specific isotopes undergoing the reactions.

Atomic Number

The atomic number is another foundational concept in nuclear chemistry. It represents the number of protons in the nucleus of an atom, which determines the element's identity. The atomic number is indicated by the letter Z and is usually written as a subscript in nuclear notations.When balancing nuclear equations, the atomic number ensures that the identity of the nuclear species is preserved across the reaction. For instance, even if a neutron changes into a proton (like in beta decay), the atomic number will change accordingly so as to keep both sides of the equation balanced.In equations like \( ^{27}_{13} \text{Al} + ^{4}_{2} \text{He} \rightarrow ^{30}_{15} \text{P} + ? \), we see the importance of balancing these numbers to ensure that the same number of protons exists on both sides of the reaction, which determines the correct identity for the missing particle.

Particle Symbols

Particle symbols are shorthand notations used in nuclear chemistry to represent various particles involved in nuclear reactions. These symbols are essential for correctly writing nuclear equations and include:

• Protons, represented as \( ^{1}_{1} \text{H} \)
• Neutrons, which are \( ^{1}_{0} \text{n} \)
• Alpha particles (or helium nuclei), denoted by \( ^{4}_{2} \text{He} \)
• Beta particles, which can be negative, \( ^{0}_{-1} \text{e} \) (for beta-minus decay) or positive, \( ^{0}_{+1} \text{e} \) (for beta-plus decay)

The correct usage of particle symbols in an equation allows for accurate depiction and understanding of the changes happening at a nuclear level. For instance, if you see an alpha particle symbol in a nuclear reaction, you know that the process involves the emission or absorption of a helium nucleus, which can greatly affect both the mass and atomic numbers.