Problem 48

Question

Element \(^{287} 114\) was made by firing a beam of \(^{48} \mathrm{Ca}\) ions at \(^{242}\mathrm{Pu.}\) Three neutrons were ejected in the reaction. Write a balanced nuclear equation for the synthesis of this super-heavy element.

Step-by-Step Solution

Verified
Answer
The balanced equation is: \(^{48}_{20}\text{Ca} + ^{242}_{94}\text{Pu} \rightarrow ^{287}_{114}114 + 3^1_0n\).
1Step 1: Identify the Reactants and Products
The reactants are the two elements combined: \(^{48}\text{Ca}\) and \(^{242}\text{Pu}\). The resulting product is \(^{287}114\), with three neutrons (\(3n\)) being ejected in the process.
2Step 2: Set Up the Nuclear Equation
The equation will be set up as: \(^{48}\text{Ca} + ^{242}\text{Pu} \rightarrow ^{287}114 + 3n\). Here, we must balance both the atomic and mass numbers.
3Step 3: Balance Mass Numbers
Calculate the total mass number on both sides. The mass numbers are 48 (Ca) + 242 (Pu) = 290 on the left side, and 287 (new element) + (3 neutrons x 1) = 290 on the right side. Mass numbers balance.
4Step 4: Balance Atomic Numbers
Calculate the total atomic number on both sides. The atomic numbers are 20 (Ca) + 94 (Pu) = 114 on the left side. The product has atomic number 114 and neutrons don't contribute to the atomic number, so the atomic numbers balance.
5Step 5: Write the Balanced Equation
The balanced nuclear equation is: \(^{48}_{20}\text{Ca} + ^{242}_{94}\text{Pu} \rightarrow ^{287}_{114}114 + 3^1_0n\).

Key Concepts

Super-Heavy ElementsNeutron EjectionAtomic Number Balancing
Super-Heavy Elements
Super-heavy elements are a fascinating part of the periodic table due to their unique nuclear properties and the challenges associated with their production. These elements usually have an atomic number greater than 104 and are not found naturally on Earth. They are created artificially by using particle accelerators to collide lighter nuclei. This process requires a precise combination of nuclear physics techniques and is typically conducted in specialized laboratories.

Producing super-heavy elements can help scientists understand more about the forces and decays involved in heavy nuclei, as well as the "island of stability", a hypothesized cluster of these elements that remain relatively stable compared to other heavy elements. The creation of each new super-heavy element also pushes the boundaries of chemistry and expands our understanding of atomic interactions and nuclear structure.
Neutron Ejection
In nuclear reactions, neutron ejection refers to the process where neutrons are expelled from the nucleus of an atom. This is a common occurrence in the synthesis of super-heavy elements, as seen in the creation of element 114 from the collisions of calcium-48 and plutonium-242.

During the reaction, it's not only about fusing the two atomic nuclei together. Often, the excess energy released from the collision results in the ejection of neutrons. For example, in this reaction, three neutrons are ejected.
  • This helps dissipate the excess energy, allowing the remaining nucleus to reach a more stable configuration.
  • The ejection does not affect the atomic number but does reduce the mass number of the new element.
Understanding neutron ejection aids in balancing nuclear equations and predicting the outcomes of such nuclear synthesis reactions.
Atomic Number Balancing
Atomic number balancing is a key step in the process of ensuring nuclear equations are correct. The atomic number reflects the number of protons in an atom's nucleus and defines the element. For any nuclear reaction, the total atomic number before and after the reaction must be equal.

For the reaction involving calcium and plutonium, the atomic numbers are balanced as follows:
The atomic numbers of the reactants are:
  • Calcium: 20
  • Plutonium: 94
Total on the left side: 20 + 94 = 114.
The product formed, element 114, has an atomic number of 114. As neutrons have an atomic number of 0, they don't affect this balance. Therefore, the equation is balanced with respect to the atomic number. This balance ensures that the reaction adheres to the law of conservation of charge, making it essential for writing accurate nuclear equations.