Problem 5 - Chemistry: Principles and Reactions

Question

Write balanced nuclear equations for (a) the alpha emission resulting in the formation of \(\mathrm{Pa}-233\). (b) the loss of a positron by \(\mathrm{Y}-85\). (c) the fusion of two C-12 nuclei to give sodium-23 and another particle. (d) the fission of Pu-239 to give tin-130, another nucleus, and an excess of two neutrons.

Step-by-Step Solution

Answer

Question: Write balanced nuclear equations for the following processes: a) Alpha emission resulting in the formation of Pa-233 b) Loss of a positron by Y-85 c) Fusion of two C-12 nuclei to give sodium-23 and another particle d) Fission of Pu-239 to give Sn-130, another nucleus, and an excess of two neutrons Answer: a) ^{237}_{93}Np → ^{233}_{91}Pa + ^4_2He b) ^{85}_{39}Y → ^{85}_{38}Sr + ^0_{-1}β c) ^{12}_6C + ^{12}_6C → ^{23}_{11}Na + ^1_1p d) ^{239}_{94}Pu → ^{130}_{50}Sn + ^{107}_{44}Ru + 2×^1_0n

Step 1: Identify the reactant nucleus and the emission particle

The original reactant nucleus is the one which undergoes alpha emission to produce Pa-233. An alpha particle consists of 2 protons and 2 neutrons, represented as ^4_2He.

Step 2: Determine the reactant nucleus and write the balanced nuclear equation

To find the reactant nucleus, we need to consider the conservation of mass number and atomic number. Let's denote Reactant nucleus as X. The mass number and atomic number of Pa (Protactinium) are 91 and 233 respectively. So, the equation can be written as: \[^A_ZX \rightarrow ^{233}_{91}\mathrm{Pa} + ^4_2\mathrm{He}\] By applying conservation of mass number: \[A = 233 + 4 \Rightarrow A = 237\] By applying conservation of atomic number: \[Z = 91 + 2 \Rightarrow Z = 93\] So, the reactant nucleus is ^237_93X, which is an isotope of Neptunium (Np). The balanced nuclear equation is: \[^{237}_{93}\mathrm{Np} \rightarrow ^{233}_{91}\mathrm{Pa} + ^4_2\mathrm{He}\] #b) Loss of a positron by Y-85#

Step 1: Identify the original nucleus and the emission particle

The original nucleus is Y-85 (Yttrium), and the particle emitted is a positron, denoted as ^0_1e or ^0_{-1}\beta.

Step 2: Determine the product nucleus and write the balanced nuclear equation

To find the product nucleus, we need to consider the conservation of mass number and atomic number. The mass number and the atomic number of Y are 39 and 85 respectively. So, the equation can be written as: \[^{85}_{39}\mathrm{Y} \rightarrow ^A_ZX + ^0_{-1}\beta\] By applying conservation of mass number: \[85 = A\] By applying conservation of atomic number: \[39 = Z - (-1) \Rightarrow Z = 38\] So, the product nucleus is ^{85}_{38}X, which is an isotope of Sr (Strontium). The balanced nuclear equation is: \[^{85}_{39}\mathrm{Y} \rightarrow ^{85}_{38}\mathrm{Sr} + ^0_{-1}\beta\] #c) Fusion of two C-12 nuclei to give sodium-23 and another particle#

Step 1: Write the reactants and known product in the nuclear equation

The reactants are two C-12 nuclei, and the known product is Na-23 (Sodium). The equation can be written as: \[^{12}_6\mathrm{C} + ^{12}_6\mathrm{C} \rightarrow ^{23}_{11}\mathrm{Na} + ^A_ZX\]

Step 2: Determine the missing particle and write the balanced nuclear equation

To find the missing particle, we need to consider the conservation of mass number and atomic number. By applying conservation of mass number: \[12 + 12 = 23 + A \Rightarrow A = 1\] By applying conservation of atomic number: \[6 + 6 = 11 + Z \Rightarrow Z = 1\] So, the missing particle is ^1_1X, which is a proton (p). The balanced nuclear equation is: \[^{12}_6\mathrm{C} + ^{12}_6\mathrm{C} \rightarrow ^{23}_{11}\mathrm{Na} + ^1_1\mathrm{p}\] #d) Fission of Pu-239 to give Sn-130, another nucleus, and an excess of two neutrons#

Step 1: Write the reactants and known products in the nuclear equation

The reactant is Pu-239 (Plutonium), and the known products are Sn-130 (Tin) and two neutrons. The equation can be written as: \[^{239}_{94}\mathrm{Pu} \rightarrow ^{130}_{50}\mathrm{Sn} + ^A_ZX + 2\times^1_0\mathrm{n}\]

Step 2: Determine the other nucleus and write the balanced nuclear equation

To find the other nucleus, we need to consider the conservation of mass number and atomic number. By applying conservation of mass number: \[239 = 130 + A + 2 \Rightarrow A = 107\] By applying conservation of atomic number: \[94 = 50 + Z \Rightarrow Z = 44\] So, the other nucleus is ^{107}_{44}X, which is an isotope of Ru (Ruthenium). The balanced nuclear equation is: \[^{239}_{94}\mathrm{Pu} \rightarrow ^{130}_{50}\mathrm{Sn} + ^{107}_{44}\mathrm{Ru} + 2\times^1_0\mathrm{n}\]

Key Concepts

Alpha Emission

Alpha emission is a type of radioactive decay where an unstable nucleus releases an alpha particle to become more stable. An alpha particle is composed of 2 protons and 2 neutrons, similar to a helium nucleus, and is represented as

• \(^4_2\text{He}\)

When an alpha particle is emitted from a nucleus, the atomic number (number of protons) decreases by 2, and the mass number (total number of protons and neutrons) decreases by 4.
This process transforms the original element into a different one located two places back in the periodic table. For example, when Neptunium-237 undergoes alpha emission, it forms Protactinium-233. This process helps reduce the instability of heavy nuclei by emitting particles to transform into a lighter nucleus.

Positron Emission

Positron emission is a form of beta decay where a proton in a nucleus transforms into a neutron while emitting a positron. A positron is the antimatter equivalent of an electron, carrying the same mass but with a positive charge. It is symbolized by

• \(^0_{+1}\beta\) or \(^0_{+1}\text{e}\)

In positron emission, the atomic number of the nucleus decreases by 1, but the mass number remains unchanged, due to the transformation of a proton into a neutron.
This results in the formation of a different element, located one place to the left in the periodic table.
For instance, when Yttrium-85 emits a positron, it becomes Strontium-85. Positron emission often occurs in proton-rich isotopes, where a higher neutron-to-proton balance is more stable.

Nuclear Fusion

Nuclear fusion is a reaction where two light atomic nuclei combine to form a heavier nucleus. This process releases a tremendous amount of energy, much more than energy released by other forms of nuclear reactions.
During fusion, the nuclei must overcome their electrostatic repulsion to come close enough for the strong nuclear force to bond them together. In the exercise, two Carbon-12 nuclei fuse to create Sodium-23 and a proton.

• \(^{12}_6\text{C} + ^{12}_6\text{C} \rightarrow ^{23}_{11}\text{Na} + ^1_1\text{p}\)

Fusion reactions, such as those occurring in the sun, require extremely high temperatures and pressures to initiate. They are the most promising source of clean energy, as they produce minimal nuclear waste compared to fission.

Nuclear Fission

Nuclear fission is a process where a heavy nucleus splits into smaller nuclei, along with the release of energy and, typically, free neutrons. It is the opposite of nuclear fusion and occurs spontaneously or can be induced by neutron bombardment.
Fission can trigger a chain reaction if the released neutrons initiate further fission in nearby nuclei. This is the principle behind nuclear reactors and atomic bombs.
For instance, when Plutonium-239 undergoes fission, it can produce Tin-130, Ruthenium-107, and two extra neutrons:

• \(^{239}_{94}\text{Pu} \rightarrow ^{130}_{50}\text{Sn} + ^{107}_{44}\text{Ru} + 2\times^1_0\text{n}\)

Fission reactions are highly exothermic, meaning they release a large amount of heat energy, used to generate electricity in nuclear power plants. Understanding the balance and conservation of mass and atomic numbers is essential to predict the products of a fission reaction.