91Ó°ÊÓ

Understanding atomic mass comparison is crucial in determining which nucleus will decay.
When looking at nuclear decay, the atomic mass gives hints about the stability of nuclei.
In general, a nucleus will decay to a more stable state, usually signified by a lower atomic mass. Let's consider the example given in the exercise with magnesium-25 (\(^{25}_{12} \text{Mg}\)) and aluminum-13 (\(^{13}_{12} \text{Al}\)). Here are the key points:

• Magnesium-25 has an atomic mass of 24.985837 u.
• Aluminum-13 has an atomic mass of 24.990429 u.

The lower atomic mass for magnesium means it is in a more stable state compared to aluminum.
In situations where mass is the determining factor, the nucleus with the higher atomic mass (aluminum in this case) will undergo decay to achieve a more stable configuration.
Thus, aluminum-13 will naturally decay into magnesium-25.

Beta-plus decay, often referred to as positron emission, is a type of radioactive decay.
In beta-plus decay, a proton within the nucleus is transformed into a neutron, resulting in the emission of a positron and a neutrino. In the given exercise, we observe the transformation of\(^{13}_{12} \text{Al}\) to\(^{25}_{12} \text{Mg}\):

• Aluminum has an atomic number of 13, meaning it contains 13 protons.
• Magnesium has an atomic number of 12, consisting of 12 protons.

This change indicates that one proton is lost, confirming the occurrence of beta-plus decay.

What Happens in the Nucleus?

During the decay:

• One proton in the aluminum nucleus is converted to a neutron.
• This conversion results in the emission of a positron (\(e^+\)).

The atomic number decreases by one while the mass number remains unchanged, consistent with the properties of beta-plus decay. This process is utilized in various applications including medicine.

Calculating the energy released during nuclear decay can be determined by evaluating the mass defect.
The mass defect refers to the difference in atomic masses before and after the decay event. In the case of the decay from\(^{13}_{12} \text{Al}\) to\(^{25}_{12} \text{Mg}\), we observe the following steps:

• Determine the difference in atomic masses (mass defect):\[\Delta m = 24.990429 \, \text{u} - 24.985837 \, \text{u} = 0.004592 \, \text{u}\]
• The energy released is calculated using the formula:\[E = \Delta m \times 931.5 \, \text{MeV/u}\]
• Substituting the values gives us:\[E = 0.004592 \, \text{u} \times 931.5 \, \text{MeV/u} \approx 4.276 \, \text{MeV}\]

Understanding the Energy Conversion

This conversion of mass defect into energy release reflects Einstein’s mass-energy equivalence principle given by:\[E = mc^2\]where energy \(E\) is directly proportional to the mass defect \(m\). Thus, the 4.276 MeV energy released highlights how even small amounts of mass can convert into significant energy. This principle is foundational in nuclear chemistry.