Nuclear Physics Scattering: Elastic, Inelastic etc

In summary, nuclear physics scattering is the study of interactions between particles, specifically the scattering of a projectile particle off a target nucleus. This can provide valuable information about the structure and properties of the nuclei involved. Elastic scattering refers to a process where the kinetic energy of the particles is conserved, while inelastic scattering results in a loss or gain of energy. Nuclear physics scattering is commonly used in research to study nuclear properties and reactions, as well as in various applications such as particle accelerators, medical imaging, and material development.
  • #1
physmurf
27
0
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?
 
Last edited:
Physics news on Phys.org
  • #2


The excess energy available in a compound nucleus after a neutron interaction plays a significant role in determining the type of interaction that will occur. In the case of 10-MeV neutrons interacting with lead, both elastic and inelastic scattering are possible, but inelastic scattering is more probable due to the large amount of excess energy available (~22 MeV). This excess energy can easily be transferred to the compound nucleus, causing it to become excited and eventually decay through the emission of a gamma ray or the ejection of a neutron.

The instructor's calculation of 17.5 MeV may be based on a different target nucleus or may have used a different method to calculate the mass excess. However, the important point to note is that there is a significant amount of excess energy available in this scenario, making inelastic scattering the most likely outcome.

In the case of thermal neutrons interacting with gold (B), the excess energy available is much smaller (~0.8 eV), making elastic scattering the most probable outcome. Similarly, for 1-MeV neutrons interacting with hydrogen in water (C), the excess energy is even smaller (~0.02 MeV), making elastic scattering the most likely interaction.

In contrast, thermal neutrons interacting with boron-10 (D) have a larger excess energy (~2.4 MeV), making (n, \alpha) reactions more probable. This is because boron-10 has a low binding energy for neutrons, so the excess energy can easily be transferred to the compound nucleus, causing it to decay through the emission of an alpha particle.

Finally, for 6-MeV neutrons interacting with beryllium (E), the excess energy available (~4 MeV) is enough to overcome the binding energy of the neutron, leading to (n,2n) reactions. This is because beryllium has a high binding energy for neutrons, so a larger amount of excess energy is needed to cause the compound nucleus to decay through the emission of two neutrons.

In summary, the excess energy available in a compound nucleus after a neutron interaction plays a crucial role in determining the type of interaction that will occur. Inelastic scattering is more probable when there is a large excess energy, while elastic scattering, (n, \alpha), and (n,2n) reactions are more likely for smaller excess energies.
 

1. What is nuclear physics scattering?

Nuclear physics scattering is the study of interactions between particles, typically involving the scattering of a projectile particle off a target nucleus. This can provide valuable information about the structure and properties of the nuclei involved.

2. What is elastic scattering in nuclear physics?

Elastic scattering refers to a type of scattering process in which the kinetic energy of the particles involved is conserved. This means that the particles involved in the collision do not lose or gain energy, but simply change direction. In nuclear physics, elastic scattering can provide information about the size and shape of the target nucleus.

3. What is inelastic scattering in nuclear physics?

Inelastic scattering refers to a type of scattering process in which the kinetic energy of the particles involved is not conserved. This means that the particles involved in the collision lose or gain energy, resulting in a change in their internal structure or state. In nuclear physics, inelastic scattering can provide information about the excited states of the target nucleus.

4. How is nuclear physics scattering used in research?

Nuclear physics scattering is used in research to study the properties of nuclei and the fundamental forces that govern their interactions. It can also be used to investigate nuclear reactions and the behavior of particles in extreme conditions, such as in high-energy collisions.

5. What are some applications of nuclear physics scattering?

Nuclear physics scattering has a wide range of applications, including in particle accelerators, nuclear reactors, and medical imaging techniques such as positron emission tomography (PET). It is also used in the development of new materials for various industries, such as nuclear power and defense.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
20
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • Nuclear Engineering
Replies
4
Views
3K
  • High Energy, Nuclear, Particle Physics
Replies
3
Views
4K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
4K
  • Advanced Physics Homework Help
Replies
5
Views
3K
  • High Energy, Nuclear, Particle Physics
Replies
6
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Beyond the Standard Models
2
Replies
39
Views
5K
Back
Top