91Ó°ÊÓ

A nucleus \({ }_{n}^{M} \mathrm{P}\), the parent nucleus, decays to a daughter nucleus D by positron decay: $$ { }_{n}^{n} \rightarrow D+{ }_{+1}^{0} e+{ }_{0}^{0} v $$ where \(v\) is a neutrino, a particle that has nearly zero mass and zero charge. ( \(a\) ) What are the subscript and superscript for D? (b) Prove that the mass loss in the reaction is \(M_{p}-M_{d}-2 m_{e}\), where \(M_{p}\) and \(M_{d}\) are the atomic masses of the parent and daughter. (a) To balance the subscripts and superscripts, we must have \({ }_{n-1}^{M} \mathrm{D}\). (b) The table of masses for the nuclei involved is Subtraction gives the mass loss: $$ \left(M_{p}-n m_{e}\right)-\left(M_{d}-n m_{e}+2 m e\right)=M_{p}-M_{d}-2 m_{e} $$ Notice how important it is to keep track of the electron masses in this and the previous problem.

Step by step solution

01

Determine the Subscripts and Superscripts for D

In a positron decay, a proton in the parent nucleus is converted into a neutron. The atomic number, which is the subscript, decreases by one because a proton is lost. The mass number, which is the superscript, remains the same because the change is between a proton and a neutron. Thus, for nucleus D, the subscript should be \(n-1\) and the superscript should remain \(M\). Therefore, nucleus D is represented as \({}_{n-1}^{M}\text{D}\).

02

Define Mass Components and Neutron-Electron Mass Subtraction

Initially, consider the atomic masses given: \(M_p\) for the parent and \(M_d\) for the daughter nucleus. The presence of electrons means the mass of electrons must be considered in calculations. The parent with subtracted electron mass can be represented as \(M_p - n m_e\) and the daughter nucleus includes electron subtraction but also accounts for the two produced positrons \(2m_e\), which is \(M_d - n m_e + 2m_e\).

03

Calculate Mass Loss in Reaction

Subtract the daughter's mass configuration from that of the parent: \[(M_p - n m_e) - (M_d - n m_e + 2m_e) \].Simplifying gives:\[M_p - n m_e - M_d + n m_e - 2m_e = M_p - M_d - 2m_e\].This confirms the mass loss is \(M_{p} - M_{d} - 2 m_{e}\).

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Key Concepts

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

Positron Decay

Positron decay is a type of radioactive decay in which a proton in the nucleus is transformed into a neutron. This process results in the emission of a positron, which is the antiparticle of the electron, and a neutrino. A positron has the same mass as an electron but carries a positive charge.
This decay process reduces the atomic number of the element because it converts a proton into a neutron, leading to the formation of a new element. Despite this change, the mass number of the nucleus remains the same since a proton and a neutron have approximately the same mass.
Positron decay is a significant process that occurs in certain unstable isotopes, often used in medical imaging techniques, like PET scans.

Atomic Mass

Atomic mass is the mass of an atom, usually expressed in atomic mass units (amu). It is approximately equivalent to the sum of the number of protons and neutrons in the atom, since electrons have negligible mass compared to the nucleus.
In nuclear reactions, it's crucial to consider the atomic masses of the nuclei involved. In positron decay, we deal with the atomic mass of the parent nucleus, denoted as \(M_p\), and the atomic mass of the daughter nucleus, \(M_d\).
Atomic masses can be measured using mass spectrometry and are essential for calculating the energy changes in nuclear reactions.

Subscripts and Superscripts

Subscripts and superscripts in nuclear chemistry represent important properties of nuclei. The subscript indicates the atomic number, which is the number of protons in the nucleus. The superscript is the mass number, representing the total count of protons and neutrons.
When evaluating nuclear decay equations, it's essential to balance these numbers on both sides of the reaction. In positron decay, the subscript decreases by one as a proton changes into a neutron, but the superscript remains unchanged as the overall nucleon count does not change.
Understanding subscripts and superscripts helps visualize transformations within the atom's nucleus, preserving certain properties while altering others.

Electron Mass

The mass of an electron is about \(9.11 \times 10^{-31}\) kilograms, but in atomic mass units, it is approximately \(0.0005485799\, ext{amu}\). Although this seems small, electron mass becomes significant in precise nuclear mass calculations.
During positron decay, the mass loss includes the accounting of electron-like particles, namely the emitted positrons. Even though electronic mass is minor, in calculating the mass difference resulting from decay, these values must be carefully considered.
In mass calculations, each electron's mass should be subtracted from atomic masses to provide a more accurate result, especially when assessing the net mass change in reactions.

Mass Calculations in Decay Reactions

In nuclear reactions, determining the mass change is crucial for understanding the process. The mass calculated often translates into energy changes due to the mass-energy equivalence principle \(E=mc^2\). For positron decay, accurate mass calculations involve the initial atomic mass of the parent nucleus minus the mass of the daughter nucleus and electron considerations.

• The atomic mass of the parent nucleus is adjusted by subtracting the combined mass of its electrons.
• The daughter's atomic mass accounts for electron mass as well as the mass of the produced positrons.
• This method ensures that the calculation considers all mass losses, including emissions and transformations during decay.

Performing these calculations correctly ensures the conservation laws are upheld, providing insight into energy distribution post-decay.