Explaining Alpha Decay Energies in 232Th

In summary, when 232Th undergoes alpha emission, the alpha particle can have two different energies, 4.01 Mev and 3.95 Mev. This is due to the different probability weightings of the decay products, with the majority of cases resulting in two different excited states of Ra. If the decay results in the higher energy state of Ra, there is less energy available for the alpha particle. This can be seen in the equation Th232 = Ra228 + He4, where the alpha particle is a product of the decay.
  • #1
jackxxny
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Homework Statement



I got ask the following question, and i don't know the answer.

Why in a rdioactive decay of 232Th by alpha emission, the alpha particle can have energy 4.01 Mev and 3.95 Mev?

Homework Equations





The Attempt at a Solution




I look up this website and i found it.

http://atom.kaeri.re.kr/cgi-bin/decay?Th-232 A

But i don't understand why.

Is because of the probility?
 
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  • #2
Yes, there are different possible decay products with different probability weightings.
 
  • #3
the part that i don't get is how Th232 -->decay--> Ra228

why does the alpha particle can have 2 different energies??
 
  • #4
Judging from your picture it looks like Th will decay in the majority of cases to two different excited states of Ra (which can subsequently emit a gamma to get to the ground state of Ra). If it decays to the higher energy state of Ra then there is less energy left for the alpha.
 
  • #5
well if Th232 decays by alpha emission it loses a He atom
thus Th232 = Ra228 + He4
hope this helps
 

1. What is alpha decay and how does it occur in 232Th?

Alpha decay is a type of radioactive decay in which an unstable nucleus emits an alpha particle (consisting of two protons and two neutrons) to become a more stable nucleus. In the case of 232Th, the nucleus spontaneously emits an alpha particle to become 228Ra, which is a more stable isotope.

2. What causes the release of energy in alpha decay of 232Th?

The release of energy in alpha decay of 232Th is due to the difference in binding energy between the initial and final nuclei. The nucleus of 232Th has a higher binding energy than the nucleus of 228Ra, so when the alpha particle is emitted, the excess energy is released in the form of kinetic energy.

3. How is the energy released in alpha decay calculated for 232Th?

The energy released in alpha decay of 232Th can be calculated using the equation E = mc^2, where E is the energy released, m is the mass defect (difference in mass between the initial and final nuclei), and c is the speed of light. The mass defect can be calculated using the masses of the initial and final nuclei.

4. Are there any other factors that can affect the energy released in alpha decay of 232Th?

Yes, there are other factors that can affect the energy released in alpha decay of 232Th. These include the nuclear structure of the initial and final nuclei, the spin and parity of the particles involved, and the quantum numbers of the particles.

5. How is the energy released in alpha decay of 232Th used in practical applications?

The energy released in alpha decay of 232Th can be used in a variety of practical applications, such as in nuclear energy production and medical treatments for cancer. It is also used in radiometric dating to determine the age of rocks and other materials.

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