91Ó°ÊÓ

Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a stable, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings: (a) \({ }^{235} \mathrm{U} \rightarrow{ }^{232} \mathrm{Th}+{ }^{3} \mathrm{He},\) (b) \({ }^{235} \mathrm{U} \rightarrow{ }^{231} \mathrm{Th}+{ }^{4} \mathrm{He}\) (c) \({ }^{235} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+{ }^{5} \mathrm{He}\) The needed atomic masses are $$ \begin{array}{llll} { }^{232} \mathrm{Th} & 232.0381 \mathrm{u} & { }^{3} \mathrm{He} & 3.0160 \mathrm{u} \\ { }^{231} \mathrm{Th} & 231.0363 \mathrm{u} & { }^{4} \mathrm{He} & 4.0026 \mathrm{u} \\ { }^{230} \mathrm{Th} & 230.0331 \mathrm{u} & { }^{5} \mathrm{He} & 5.0122 \mathrm{u} \\ { }^{235} \mathrm{U} & 235.0429 \mathrm{u} & & \end{array} $$

Step by step solution

01

Understanding the Problem

We are asked to calculate the disintegration energies for the decay of uranium-235 by emitting different nucleon combinations and discuss which is more stable or favorable. This involves calculating the mass defect and using it to find energy released.

02

Calculating Mass Defect for Process (a)

Process: \(^{235} \text{U} \rightarrow~ ^{232} \text{Th} + ^{3} \text{He}\).Calculate the mass difference: \[\Delta m_a = (m_{^{232}Th} + m_{^3He}) - m_{^{235}U} = (232.0381 \text{ u} + 3.0160 \text{ u}) - 235.0429 \text{ u} = -0.0112 \text{ u}\]

03

Calculating Energy for Process (a)

Convert the mass defect to energy using the relation \(E = \Delta m c^2\):\[E_a = -0.0112 \text{ u} \times 931.5 \text{ MeV/u} = -10.43 \text{ MeV}\]The negative sign indicates the process requires energy, rather than releasing it.

04

Calculating Mass Defect for Process (b)

Process: \(^{235} \text{U} \rightarrow~ ^{231} \text{Th} + ^{4} \text{He}\).Calculate the mass difference: \[\Delta m_b = (m_{^{231}Th} + m_{^4He}) - m_{^{235}U} = (231.0363 \text{ u} + 4.0026 \text{ u}) - 235.0429 \text{ u} = -0.0040 \text{ u}\]

05

Calculating Energy for Process (b)

Convert the mass defect to energy:\[E_b = -0.0040 \text{ u} \times 931.5 \text{ MeV/u} = -3.73 \text{ MeV}\]This process also requires energy input.

06

Calculating Mass Defect for Process (c)

Process: \(^{235} \text{U} \rightarrow~ ^{230} \text{Th} + ^{5} \text{He}\).Calculate the mass difference: \[\Delta m_c = (m_{^{230}Th} + m_{^5He}) - m_{^{235}U} = (230.0331 \text{ u} + 5.0122 \text{ u}) - 235.0429 \text{ u} = 0.0024 \text{ u}\]

07

Calculating Energy for Process (c)

Convert the mass defect to energy:\[E_c = 0.0024 \text{ u} \times 931.5 \text{ MeV/u} = 2.24 \text{ MeV}\]A positive energy implies energy release.

08

Discussion of Results

Based on the energy calculations, process (c) has a positive energy, indicating it releases energy, unlike processes (a) and (b) which require energy. However, among realistic processes, alpha decay (process b) is more common as it requires the least energy input compared to the others.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Defect

The concept of mass defect is central in understanding nuclear decay energetics. When a nucleus undergoes a reaction, there is often a small difference between the mass of the reactants and the products. This difference is known as the mass defect. It occurs because the mass of the constituent nucleons (protons and neutrons) in a bound nucleus is less than the total mass of those nucleons when separated. This mass difference is what powers a nuclear reaction, as it gets converted to energy.

Using Einstein's famous equation, the mass defect \(\Delta m\) is directly converted into energy through \(E = \Delta m c^2\), where \(c\) is the speed of light.

This energy conversion explains why even a small mass defect can result in a significant energy release, as seen in nuclear decay processes.

Disintegration Energy

Disintegration energy is the energy change associated with a nuclear decay reaction. It is calculated by evaluating the mass defect for a particular decay process and converting it into energy, usually in mega-electron volts (MeV).

In the context of uranium-235 decay, this energy tells us whether the decay process releases or requires energy. When we calculate the disintegration energy, a negative value indicates endothermic reactions, where energy is absorbed.
For instance, in process (a) and (b) of the exercise, \(^{235}\mathrm{U}\) decays require energy to proceed (energies of -10.43 MeV and -3.73 MeV respectively). Conversely, a positive energy, as seen in process (c) with 2.24 MeV, indicates that energy is released, making it an exothermic process.

This concept is crucial for understanding why certain nuclear decay processes occur naturally and which configurations of nucleons are likely to be stable or undergo decay.

Alpha Decay

Alpha decay is a common form of nuclear decay where an unstable nucleus emits an alpha particle, consisting of two protons and two neutrons. Because of its tightly bound structure, the alpha particle is a very stable configuration of nucleons.

In the given exercise, one of the scenarios (process b) involves \(^{235}\mathrm{U}\) decaying by emitting an alpha particle to form \([^4\mathrm{He}]\) and \(^{231}\mathrm{Th}\). This kind of decay is particularly likely for heavy nuclides because it achieves a large mass defect and, therefore, releases energy, making it energetically favorable.

Though other types of decay paths are possible, alpha decay stands out due to its lower energy requirement, as illustrated by the calculation yielding -3.73 MeV, which is more achievable than other combinations. This stability makes alpha decay a preferred path for heavy elements attempting to reach a more stable state.