Question

The atomic mass of \(^{25}_{12}Mg\) is 24.985837 u, and the atomic mass of \(^{25}_{13}Al\) is 24.990428 u. (a) Which of these nuclei will decay into the other? (b) What type of decay will occur? Explain how you determined this. (c) How much energy (in MeV) is released in the decay?

Short answer

(a) \(^{25}_{13}Al\) decays into \(^{25}_{12}Mg\). (b) \(\beta^+\) decay. (c) 4.276 MeV is released.

Step-by-step solution

Determine the Decay Process

Compare the atomic masses of \(^{25}_{12}Mg\) and \(^{25}_{13}Al\). Since \(^{25}_{12}Mg\) has a lower atomic mass than \(^{25}_{13}Al\), the nucleus will likely decay into the nucleus with a higher atomic mass. Hence, \(^{25}_{13}Al\) will decay into \(^{25}_{12}Mg\).

Identify the Type of Decay

Since an aluminum atom is decaying to a magnesium atom, the process involves a change in atomic number from 13 to 12. This indicates a positive beta decay (\(\beta^+\) decay), as a proton transforms into a neutron.

Calculate the Energy Released

The energy released \(Q\) in a decay is given by the formula: \[ Q = (m_i - m_f) \cdot c^2 \] where \(m_i\) is the initial mass (24.990428 u), \(m_f\) is the final mass (24.985837 u), and \(c^2 = 931.5 \text{ MeV}/\text{u}\). Calculate \(Q\):\[ Q = (24.990428 - 24.985837) \cdot 931.5 \]Resulting:\[ Q \approx 4.276 \text{ MeV} \]

Verify the Steps

Ensuring that the atomic number change matches the observed decay process and the calculated energy is plausible confirms consistency in the answer provided.

Key concepts

Beta Decay

Beta decay is a type of nuclear decay where a nucleon (protons or neutrons) in an unstable atom's nucleus transforms into a different type of nucleon. In our exercise, the transformation from aluminum to magnesium involves a special form known as positive beta decay, or \(\beta^+\) decay. This particular transformation occurs when a proton is converted into a neutron, accompanied by the emission of a positron (the antimatter equivalent of an electron) and a neutrino. Understanding the way beta decay changes atomic numbers is crucial:

• In \(\beta^+\) decay, the atomic number decreases by one as a proton turns into a neutron.
• The mass number remains constant because the total number of nucleons does not change during the decay process.

This nuanced transformation explains why in the original exercise, \(^{25}_{13}\text{Al}\) becomes \(^{25}_{12}\text{Mg}\) as the aluminum loses one proton, turning into magnesium.

Atomic Mass

Atomic mass is a critical factor in nuclear reactions as it contributes to determining the stability and eventual decay paths of isotopes. In the exercise, we compared the atomic masses of \(^{25}_{12}\text{Mg}\) and \(^{25}_{13}\text{Al}\) to ascertain which nucleus would undergo decay. Atomic mass generally reflects the total number of protons and neutrons in an atom.A key observation is:

• \(^{25}_{12}\text{Mg}\) has an atomic mass of 24.985837 u.
• \(^{25}_{13}\text{Al}\) has an atomic mass of 24.990428 u.

By examining these values, we see that \(^{25}_{13}\text{Al}\) has a slightly higher atomic mass. In the context of nuclear decay, such minor differences can signal a propensity for one isotope to transform into another more stable form, releasing energy in the process.

Energy Release

During nuclear decay, energy release is a significant aspect. Energy is released because the initial golden rule of physics applies—total energy conservation—which means the excess binding energy is emitted. This is articulated via the formula:\[ Q = (m_i - m_f) \cdot c^2 \]Where:

• \(m_i\) represents the initial atomic mass, here \(24.990428\) u for \(^{25}_{13}\text{Al}\).
• \(m_f\) the final atomic mass, here \(24.985837\) u for \(^{25}_{12}\text{Mg}\).
• \(c^2\) is the speed of light squared, converted for atomic units as \(931.5 \, \text{MeV/u}\).

From this calculation in the exercise:\[ Q \approx 4.276 \, \text{MeV} \]This surplus energy released in MeV is what powers the transformations in nuclear reactions, making them an intriguing subject of study.

Nuclear Reactions

Nuclear reactions are processes where the structure of an atom's nucleus changes, resulting in transformations that release or absorb energy. They are fundamental to understanding phenomena like decay, fusion, and fission.In the exercise, the specific nuclear reaction involves a \(\beta^+\) decay of an aluminum nucleus into a magnesium nucleus. The primary elements of nuclear reactions involve:

• Changes in the atomic and mass numbers.
• Emission or absorption of particles like neutrons, protons, or subatomic particles such as positrons.
• Energy release, which can be harnessed in various ways, such as in nuclear power industries.

This process illustrates the profound capability of nuclear reactions to transform substances by altering their core atomic structures—transitioning between isotopes—through atomic-scale energy exchanges.