Nuclear Interactions: Inelastic and elastic scattering

In summary, the conversation discusses a graduate level Medical Physics course and a homework problem about choosing the proper neutron interaction type. The instructor's answer suggests that both elastic and inelastic scattering are possible, but inelastic is more probable due to the large amount of excess energy available in the compound nucleus. However, there is a disagreement about the exact amount of excess energy and the determining factor for whether a neutron is ejected or not. The conversation also mentions the use of a table for mass excess calculations and the possibility of photons being unlikely due to the strong interaction.
  • #1
physmurf
27
0
I originally posted this in the homework section until I realized the homework section only covers through undergraduate courses. The course I am currently in is a graduate level Medical Physics course. So if you don't mind, here is the thread I posted in the homework section.

I am preparing for a Nuclear Physics test. One of the homework problems asks the following: Choose the proper neutron interaction type for each of the following scenarios and explain why. Interaction types: elastic, inelastic, (n,[tex] \gamma[/tex]),(n,2n),(n,[tex] \alpha[/tex]).

A. 10-MeV neutrons interacting with lead.
B. Thermal neutrons interacting with gold.
C. 1-MeV neutrons interacting with hydrogen in water.
D. Thermal Neutrons interacting with boron-10.
E. 6-MeV neutrons interacting with beryllium.



Instructors answer to part A: Both elastic and inelastic scattering are possible, but inelastic is more probable; This is because there is a large amount of "excess" energy (~17.5 Mev) available in the compound nucleus, and it takes little time ([tex]<10^{-14} [/tex] sec) for a neutron to gain enough energy (~7 MeV to escape. This most likely would leave the residual nucleus at an excited state as there are so many low-lying excited states available in a Pb nucleus.

First off, I am unsure of where he gets the 17.5 MeV. Every calculation I use gives an excess energy of about 22 MeV. This was obtained using the following formula:
I used a table in the back of my book which gives the mass excess for different nuclei: I decided to use [tex]^{208}Pb [/tex] for the target nucleus since it is the most abundant form of Lead.

[tex]_\Delta Q = (m_A+m_a-m_b-m_B)c^2[/tex]

Mass excesses are:
[tex]^{208}Pb [/tex]: -23364 [tex]_\mu[/tex]u
Neutron: 8665 [tex]_\mu[/tex]u
[tex]^{209}Pb [/tex]: -18926 [tex]_\mu[/tex]u

This gave me a result close to 12 MeV. When this is added to 10 Mev I get approximately 22 Mev.

In any event, even if this is 22 MeV, what will determine weather or not a Neutron is ejected from the compound nucleus as opposed to just some elastic scattering or [tex]\gamma[/tex] gamma decay? This is perhaps my biggest "hang up" with this. What do any of you think?

Thanks.
 
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  • #2
Photons are generally unlikely if there is a process that happens via the strong interaction.

The 17.5 MeV might be an approximation from the average binding energy per nucleon, which is roughly 7.5 MeV for heavy nuclei (+10 for the incoming neutron). That doesn't take into account specific nuclides and even small variations in the binding energy per nucleon can lead to large differences in the overall binding energy, so this is a very poor approximation method.
 

1. What is the difference between inelastic and elastic scattering in nuclear interactions?

Inelastic scattering occurs when the incident particle interacts with the nucleus and transfers energy to it, causing it to become excited or even break apart. Elastic scattering, on the other hand, is when the incident particle interacts with the nucleus without transferring energy, resulting in a change in direction but no change in the nucleus itself.

2. How is the scattering angle related to the incident energy in nuclear interactions?

In general, the higher the incident energy, the smaller the scattering angle will be for both inelastic and elastic scattering. This is because at higher energies, the particles have more momentum and are less likely to be deflected by the target nucleus.

3. What types of particles can be involved in nuclear interactions?

Any type of particle can be involved in nuclear interactions, but the most common ones are protons, neutrons, and alpha particles. Particles with higher energies, such as gamma rays and cosmic rays, can also interact with nuclei.

4. How are inelastic and elastic scattering used in nuclear physics research?

Inelastic and elastic scattering are used to study the structure and properties of nuclei. By analyzing the energies and scattering angles of the particles involved, scientists can gain insight into the internal structure of nuclei and the forces that hold them together.

5. Can nuclear interactions be dangerous?

Nuclear interactions can be dangerous if they occur in an uncontrolled environment, such as in a nuclear reactor or during a nuclear explosion. However, in controlled settings, such as in particle accelerators, nuclear interactions are used for research and do not pose a significant risk to health or safety.

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