Problem 58 - Physics

Question

Concept Questions(a) How is the total released energy related to the decrease in mass that accompanies \(\beta^{-}\) decay? (b) When a parent nucleus undergoes \(\beta^{-}\) decay, a daughter nucleus, a \(\beta^{-}\) particle, and an antineutrino \(\bar{\gamma}\) are produced. The released energy is shared among these three particles. Assume that the kinetic energy of the \(\beta^{-}\) particle is known and that the antineutrino carries away the maximum possible energy. What then can you say about the kinetic energy of the recoiling daughter nucleus? Problem The \(\beta^{-}\) decay of phosphorus \({ }_{15}^{32} \mathrm{P}\) (atomic mass \(=31.973907 \mathrm{u}\) ) produces a daughter nucleus that is sulfur \(\underset{16}^{32} \mathrm{~S}\) (atomic mass \(=31.972070 \mathrm{u}\) ), a \(\beta^{-}\) particle, and an antineutrino. The kinetic energy of the \(\beta^{-}\) particle is \(0.90 \mathrm{MeV}\). Find the maximum possible energy that the antineutrino could carry away.

Step-by-Step Solution

Answer

The antineutrino can carry away a maximum of 0.81163 MeV of energy.

Step 1: Understanding Energy-Mass Equivalence

According to Einstein's principle of energy-mass equivalence, the energy released during a nuclear reaction is directly related to the decrease in mass. The energy (E) can be calculated using the equation \(E = \Delta m \cdot c^2\), where\(\Delta m\) is the change in mass and\(c\) is the speed of light.

Step 2: Calculating Mass Defect

To find the released energy, calculate the mass defect, \(\Delta m\). This is the difference between the mass of the parent nucleus and the sum of the masses of the decay products. So,\(\Delta m = m_{\mathrm{P}} - m_{\mathrm{S}}\), where\(m_{\mathrm{P}} = 31.973907\, \mathrm{u}\)and\(m_{\mathrm{S}} = 31.972070\, \mathrm{u}\).Calculate \(\Delta m = 31.973907 - 31.972070 = 0.001837\, \mathrm{u}\).

Step 3: Converting Mass Defect to Energy

One atomic mass unit (u) corresponds to an energy equivalence of 931.5 MeV. Thus, the energy released is\(E = 0.001837\, \mathrm{u} \times 931.5\, \mathrm{MeV/u} = 1.71163\, \mathrm{MeV}\).

Step 4: Distributing Released Energy

The total released energy, 1.71163 MeV, is distributed among the \(\beta^{-}\) particle, the antineutrino, and the recoiling daughter nucleus. Given the kinetic energy of the\(\beta^{-}\) particle is 0.90 MeV, the remaining energy is available to the antineutrino and the daughter nucleus.

Step 5: Calculating Antineutrino's Maximum Energy

Assume the antineutrino carries away as much energy as possible, meaning the daughter nucleus takes negligible energy (i.e., its kinetic energy is near zero). The maximum energy for the antineutrino is then \(E_{\bar{u}} = 1.71163\, \mathrm{MeV} - 0.90\, \mathrm{MeV} = 0.81163\, \mathrm{MeV}\).

Key Concepts

Energy-Mass Equivalence

The principle of energy-mass equivalence fundamentally connects energy and mass in physical processes. This concept, formulated by Albert Einstein, is famously expressed as \(E = mc^2\), where \(E\) represents energy, \(m\) is mass, and \(c\) is the speed of light. This equation shows that mass can be converted into energy and vice versa.
In the context of nuclear reactions, like beta decay, a small amount of mass is lost. This lost mass is transformed into energy. When a nucleus decays, the slight decrease in its mass results in a significant amount of energy being released, due to the large value of \(c^2\). Therefore, understanding the energy-mass equivalence helps in calculating how much energy results from a given nuclear reaction's mass changes.

Mass Defect

The concept of mass defect refers to the difference in mass between a parent nucleus before decay and the sum of its decay products. This difference arises because a small amount of mass is "missing" after the reaction.

• The parent nucleus is the isotope that undergoes decay, while the daughter nucleus is the isotope formed after decay.
• During the decay, some mass is converted into energy, which accounts for the decrease in mass.

The mass defect can be calculated using the equation \(\Delta m = m_{\text{parent}} - m_{\text{daughter}}\). In our example, for phosphorus \((_{15}^{32} \mathrm{P})\) and sulfur \((_{16}^{32} \mathrm{S})\), the mass defect is \(0.001837\, \mathrm{u}\), which leads to the energy calculation in beta decay.

Antineutrino Energy

Antineutrino energy comes into play when considering the distribution of energy among the particles produced in beta decay. The antineutrino is one of the decay products and can carry away a portion of the released energy.
In a beta decay process, the energy available is shared between the beta particle, the antineutrino, and the recoiling daughter nucleus. To find the maximum energy the antineutrino can carry, assume it takes most of the remaining energy after accounting for the known energy of the beta particle.

• In our problem, the total energy released is 1.71163 MeV.
• The beta particle carries away 0.90 MeV.
• The antineutrino can then potentially carry up to 0.81163 MeV of energy, assuming the daughter nucleus retains minimal kinetic energy.

Nuclear Reactions

Nuclear reactions involve transformations where nuclei of atoms change their structure through processes like decay or fusion. Beta decay is one common type of nuclear reaction.
In beta decay, a beta particle (electron or positron) is emitted from an atomic nucleus, transforming the original nucleus into a different element. This can also result in the emission of a neutral particle, such as an antineutrino.

• Beta decay leads to a change in the atomic number of the element involved, while the mass number remains the same.
• The energy released during nuclear reactions is substantial, especially compared to chemical reactions, due to the high energy-mass conversion rate.
• These reactions are core to many applications in energy production, medicine, and understanding fundamental physics.

Nuclear reactions are crucial in both natural processes, like stellar evolution, and technological applications such as nuclear power generation.