Beta Decay Processes of 5525Mn & 5526Fe

In summary, the atomic mass of 55 25Mn is 54.938047u and that of 55 26Fe is 54.938296u. The three possible β-decay processes involving these two nuclides are determined by the energy difference between the initial and final states. If it is less than 2mec2, β+ decay is not energetically possible and electron capture is the sole decay mode. However, in cases where β+ decay is allowed, it can be accompanied by the electron capture process. The energy spectrum of the products in these decays is determined by the conservation laws of energy and momentum, with multiple particles being emitted in processes such as beta minus and beta plus decay
  • #1
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Homework Statement


The atomic mass of 55 25Mn = 54.938047u and that of 55 26Fe=54.938296u. Explain which of
the three possible β-decay processes involving these two nuclides are forbidden, and which
can occur. For decays which can occur, describe the energy spectrum of the products.


Homework Equations


In all the cases where β+
decay is allowed energetically (and the proton is a part of a nucleus with electron shells), it is accompanied by the electron capture process) .
However, in proton-rich nuclei where the energy difference between initial and final states is less than 2mec2, then β+
decay is not energetically possible, and electron capture is the sole decay mode.



The Attempt at a Solution



β+ decay can only happen inside nuclei when the value of the binding energy of the mother nucleus is less than that of the daughter nucleus.
B.E of Fe = 481.050477 Mev B.E of Mn = 482.033 Mev
So β+ decay seems possible.
Also there is no reason to exclude b- and electron capture processes .
Im also confused about the energy spectrum of created nuclei .
How would be possible to know their K.E ??
 
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  • #2
Remember that positron decay competes with electron capture, so if the positron decay is possible, you have to look to see if if the positron can be emitted (remember its rest mass energy is 511keV). If nuclide A decays by positron (or electron capture capture) to nuclide B, then B cannot decay by beta minus to nuclide A.

As for the energy spectrum, in any of the processes, if you have multiple particles emitted, Conservation of Energy along with Conservation of Momentum dictates that the energy is shared between the particles, each particle with a minimum decay energy and a maximum decay energy, so which modes emit multiple particles?
 
  • #3
multiple particles are emitted in b minus and beta plus decays .
Since the positron can be emitted from energy point of view as well

it should be a beta plus decay.
 
  • #4
i don't know if you had this in your mind when you say devay modes emit multiple partivles but thanks for the answer
 
  • #5


I would approach this problem by first looking at the energy difference between the initial and final states of the nuclei. In this case, the atomic masses of 55 25Mn and 55 26Fe are very similar, indicating that there may be a small energy difference between the two states. This means that all three β-decay processes (β-, β+, and electron capture) are possible.

To determine which process is most likely to occur, I would look at the binding energies of the two nuclei. The fact that the binding energy of 55 25Mn is slightly higher than that of 55 26Fe suggests that β+ decay is more likely to occur, as this would result in a more stable nucleus with a higher binding energy.

As for the energy spectrum of the products, this would depend on the specific decay process that occurs. In β+ decay, the energy spectrum would consist of a continuous range of energies, with a maximum energy equal to the difference between the initial and final states. In β- decay, the energy spectrum would also be continuous, but with a maximum energy that is slightly lower than in β+ decay. In electron capture, the energy spectrum would consist of discrete energy levels, with the highest energy level being equal to the difference between the initial and final states.

To determine the kinetic energy of the created nuclei, one would need to measure the energy of the emitted particles (positrons, electrons, or photons) and subtract this from the maximum energy available for the decay process. This would give the kinetic energy of the nucleus.
 

1. What is beta decay and how does it work?

Beta decay is a type of nuclear decay where an unstable atom releases a beta particle (either an electron or a positron) in order to become more stable. This process involves changes in the nucleus of the atom, specifically in the number of protons and neutrons present. Beta decay can occur in three ways: beta minus decay, beta plus decay, and electron capture.

2. What is the difference between beta minus and beta plus decay?

Beta minus decay occurs when a neutron in the nucleus of an atom changes into a proton, releasing a beta particle (electron) and an antineutrino in the process. This results in an increase in the atomic number of the atom. On the other hand, beta plus decay occurs when a proton in the nucleus changes into a neutron, releasing a beta particle (positron) and a neutrino. This results in a decrease in the atomic number of the atom.

3. How do the beta decay processes of 5525Mn and 5526Fe differ?

The beta decay process of 5525Mn involves beta minus decay, where a neutron in the nucleus changes into a proton, resulting in an increase in the atomic number from 25 to 26. On the other hand, the beta decay process of 5526Fe involves beta plus decay, where a proton in the nucleus changes into a neutron, resulting in a decrease in the atomic number from 26 to 25. Both processes release a beta particle and a neutrino, but the type of particle emitted and the resulting change in atomic number differ.

4. What are the applications of studying beta decay processes?

Studying beta decay processes is important in the field of nuclear physics and has various applications. For example, it helps us understand the stability of isotopes and the process of radioactive decay. It also has practical applications in fields such as nuclear energy and medicine, where radioactive isotopes are used for imaging and cancer treatment.

5. Is beta decay a random process?

Yes, beta decay is a random process. The exact timing of when an atom will undergo beta decay cannot be predicted, but the probability of decay can be calculated using the half-life of the isotope. This means that while we cannot predict when a specific atom will decay, we can determine how many atoms in a given sample will decay over a certain period of time.

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