Problem 27 - Chemistry: The Central Science

Question

Complete and balance the following nuclear equations by supplying the missing particle: (a) \({ }_{98}^{25} \mathrm{Cf}+{ }_{5}^{10} \mathrm{~B} \longrightarrow 3_{0}^{1} \mathrm{n}+?\) (b) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+?\)

Step-by-Step Solution

Answer

(a) \({ }_{98}^{25}\mathrm{Cf} + { }_{5}^{10}\mathrm{B} \longrightarrow 3_{0}^{1}\mathrm{n} + { }_{103}^{32}\mathrm{Lr}\) (b) \({ }_{1}^{2}\mathrm{H} + { }_{2}^{3}\mathrm{He} \longrightarrow { }_{2}^{4}\mathrm{He} + { }_{1}^{1}\mathrm{H}\)

Step 1: (a) Analyzing the given nuclear equation

In the first equation, we have Californium (Cf) reacting with Boron (B) to produce 3 neutrons (n) and an unknown particle. First, let's find the total atomic and mass numbers on both sides of the equation. Left side: Atomic number: 98 + 5 = 103 Mass number: 25 + 10 = 35 Right side: Atomic number: 3 * 0 = 0 (since neutrons have an atomic number of 0) Mass number: 3 * 1 = 3 (since neutrons have a mass number of 1)

Step 2: (a) Determining the missing particle

To find the missing particle, we need to ensure both the atomic and mass number are conserved. Therefore: Missing particle's atomic number: 103 - 0 = 103 Missing particle's mass number: 35 - 3 = 32 The resulting particle is an Isotope with an atomic number of 103 and a mass number of 32. This isotope is Lr, or Lawrencium. So the balanced nuclear equation is: \({ }_{98}^{25}\mathrm{Cf} + { }_{5}^{10}\mathrm{B} \longrightarrow 3_{0}^{1}\mathrm{n} + { }_{103}^{32}\mathrm{Lr}\)

Step 3: (b) Analyzing the given nuclear equation

In the second equation, we have Deuterium (H) reacting with Helium-3 (He) to produce Helium-4 (He) and an unknown particle. Let's find the total atomic and mass numbers on both sides of the equation. Left side: Atomic number: 1 + 2 = 3 Mass number: 2 + 3 = 5 Right side: Atomic number: 2 Mass number: 4

Step 4: (b) Determining the missing particle

To find the missing particle, we need to ensure both the atomic and mass number are conserved. Therefore: Missing particle's atomic number: 3 - 2 = 1 Missing particle's mass number: 5 - 4 = 1 The resulting particle is an Isotope with an atomic number of 1 and a mass number of 1. This isotope is a hydrogen atom H. So the balanced nuclear equation is: \({ }_{1}^{2}\mathrm{H} + { }_{2}^{3}\mathrm{He} \longrightarrow { }_{2}^{4}\mathrm{He} + { }_{1}^{1}\mathrm{H}\)

Key Concepts

Conservation of Mass Number

In nuclear reactions, the conservation of mass number is a fundamental principle. The mass number is the sum of protons and neutrons in an atomic nucleus. During nuclear reactions, the total mass number before and after the reaction should remain constant. This principle helps ensure that no nucleons are lost or gained in the process. For example, in part (a) of the given exercise, the initial total mass number is calculated by adding the mass numbers of Californium and Boron, which results in 35. On the product side, the mass numbers of all produced particles, including the three neutrons, should also sum to 35, ensuring the mass number is conserved. When balancing such equations, it's crucial to track the mass numbers of all involved particles meticulously. If there's a discrepancy, adjust the missing particle's mass number to maintain balance. This principle plays a vital role in understanding nuclear stability and transformations.

Conservation of Atomic Number

The conservation of atomic number is another key principle in balancing nuclear reactions. The atomic number represents the number of protons in an atom's nucleus, which determines the element's identity. In a nuclear reaction, this number must be conserved across the reactants and products. In exercise part (a), initially, Californium and Boron together have an atomic number sum of 103. As reactions progress, this total should remain the same on the product side, helping us deduce any unknown entities involved. By maintaining the continuity of the atomic number, we ensure the elemental composition of the system remains intact. Conservation of atomic number is crucial not just for equation balancing but also for comprehending the nature of radioactive decay and synthesis of new elements in nuclear chemistry.

Isotopes

Isotopes are variants of a given chemical element that exhibit the same number of protons but differ in the number of neutrons. This variation leads to different mass numbers, while the atomic number remains unchanged. Understanding isotopes is crucial for nuclear chemistry and reactions. In nuclear equations, isotopes can appear as both reactants and products, showcasing their involvement in nuclear transformations. For example, in the given exercise solution, the resulting isotopes like Lawrencium and hydrogen are crucial for balancing the equations. Recognizing isotopic identities helps grasp the changes that occur in nuclear reactions and assists in ensuring the conservation principles of mass and atomic numbers.

Balancing Nuclear Reactions

Balancing nuclear reactions involves ensuring both mass and atomic numbers are conserved on both sides of an equation. It is similar to balancing chemical equations but requires additional attention to nuclear particles. To balance a nuclear reaction:

• Sum up the mass numbers of the reactants and products and ensure they match.
• Do the same for atomic numbers, making sure they are equal on both sides.

In the original exercise, each nuclear equation is carefully balanced through these calculations. First, the known particles are analyzed, and then the unknowns are deduced by maintaining the equality of total mass and atomic numbers. Balancing nuclear reactions provides insight into nuclear decay, fission, fusion, and helps explain the formation of new isotopes, making it foundational for studies in nuclear physics and chemistry.