Problem 57 - University Physics with Modern Physics

Question

The atomic mass of \(_{12}^{25} \mathrm{Mg}\) is 24.985837 \(\mathrm{u},\) and the atomic mass of \(^{25}_{13}\) Al is 24.990429 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?

Step-by-Step Solution

Answer

(a) \(^{25}_{12} \text{Mg}\) decays into \(^{25}_{13} \text{Al}\); (b) beta-minus decay occurs; (c) 4.278 MeV is released.

Step 1: Understanding the Given Problem

We have two isotopes, \(^{25}_{12} \text{Mg}\) and \(^{25}_{13} \text{Al}\), with their respective atomic masses. We need to determine (a) which nucleus will decay into the other, (b) the type of decay, and (c) the energy released during this decay.

Step 2: Identifying Possible Decay

Since \( \text{Mg} \) has 12 protons and \( \text{Al} \) has 13 protons, if decay were to occur, \(^{25}_{12} \text{Mg}\) would need to increase its proton number to match \(^{25}_{13} \text{Al}\). This suggests that \(^{25}_{12} \text{Mg}\) may undergo decay to form \(^{25}_{13} \text{Al}\).

Step 3: Determining the Type of Decay

For \(^{25}_{12} \text{Mg}\) to become \(^{25}_{13} \text{Al}\), one of its neutrons must convert into a proton through beta-minus (\(\beta^-\)) decay, emitting an electron and an anti-neutrino. Thus, \(^{25}_{12} \text{Mg}\) decays to \(^{25}_{13} \text{Al}\) via \(\beta^-\) decay.

Step 4: Calculating Energy Released

The energy released is the difference in mass energy between the initial and final states. Calculate this using:\[ E = (m_1 - m_2) \times 931.5 \text{ MeV/u} \]where \(m_1 = 24.985837 \text{ u} \) (mass of \(^{25}_{12}\text{Mg}\)) and \(m_2 = 24.990429 \text{ u}\) (mass of \(^{25}_{13}\text{Al}\)).

Step 5: Performing the Calculation

Substitute the masses into the formula:\[ E = (24.985837 - 24.990429) \times 931.5\]\[ E = (-0.004592) \times 931.5 \text{ MeV/u}\]\[ E = -4.278444 \text{ MeV} \]Since we look for magnitude of energy, the energy released is \(4.278 \text{ MeV}\).

Key Concepts

Beta Decay

In nuclear physics, beta decay is a type of radioactive decay that involves the transformation of a neutron in an unstable nucleus into a proton, or vice versa. This process results in the emission of an electron (beta particle) and an anti-neutrino from the nucleus, specifically in beta-minus decay, as seen in the transformation of

• eutrons into protons, which increases the atomic number by one.

Beta decay is a crucial mechanism for changes in the identity of an element due to its alterations in the number of protons in a nucleus.
Beta decay occurs when a nucleus is unstable due to an excess number of neutrons compared to protons, and it seeks stability by undergoing this transformation.
This type of decay is essential in the operation of certain isotopes, like

• magnesium-25 (\(^{25}_{12}\) Mg)

decaying into aluminum-25 (\(^{25}_{13}\) Al).

Energy Release

When nuclear decay processes occur, such as beta decay, there is a release of energy. This energy release happens as a result of the difference in mass between the original and final nuclei. In the example of magnesium-25 decaying into aluminum-25, the energy released can be calculated using the masses provided:

• The initial mass of magnesium-25 is 24.985837 u
• The final mass of aluminum-25 is 24.990429 u

The energy release can be calculated using the formula:

• \[ E = (m_{initial} - m_{final}) \times 931.5 ext{ MeV/u} \]

Substituting the masses:

• \[ E = (24.985837 - 24.990429) \times 931.5 \]

The negative result of \(-4.278 ext{ MeV} \) indicates a release of 4.278 MeV of energy.
This negative sign represents the energy given off by the decay process, which corresponds to the excess energy being expelled as the nucleus transitions from a higher to a lower energy state.

Nuclear Physics

Nuclear physics is the branch of physics that studies atomic nuclei and their constituents, as well as their behavior in different contexts, such as decay processes or interactions. It seeks to understand the forces holding nuclei together and the reactions they undergo.
Key topics within nuclear physics include:

• The study of nuclear decay mechanisms, such as \(beta decay\) ,
• The interactions between particles in a nucleus, and
• The energy changes and emissions that occur during these reactions.

In our example involving magnesium-25 and aluminum-25, nuclear physics principles explain how and why magnesium-25 undergoes beta decay.
This field also investigates the fundamental forces involved in interactions, like the nuclear force and weak force, which are crucial for processes like beta decay and others.

Atomic Mass

The atomic mass of an isotope is a weighted average of the masses of an element's isotopes, influenced by both its isotopic abundance and mass number. This property is crucial for understanding nuclear reactions and stability, as it provides insights into the energy involved in nuclear transitions.
In the context of media example, the atomic mass of

• magnesium-25 \(^{25}_{12} ext{Mg}\)
• is 24.985837 u,
• while aluminum-25 \(^{25}_{13} ext{Al}\)
• has an atomic mass of 24.990429 u.

Understanding these atomic masses helped in determining which isotope is more stable, thereby guiding the analysis of the probable direction of decay and the calculation of energy released during such decay processes. The atomic mass is a pivotal factor in understanding the comparison and outcome of nuclear reactions like beta decay.