Binding Energy & Nuclear Reactions

In summary: Uranium - 235 can be calculated by finding the difference in mass between the individual nucleons and the combined mass of the nucleus, and then multiplying it by the speed of light squared.For the gamma ray photon to change into an electron and positron, it needs to interact with a nearby nucleus which can absorb some of its energy and momentum. The minimum energy required for this to happen is at least 2 times the rest mass energy of the electron, but it may be slightly higher depending on the reaction taking place. This process is known as "pair production" and the positron has a very short lifetime.
  • #1
pavadrin
156
0
Hey
I was studying for an upcoming test on Nuclear Technology, and I am unable to do the calculations for this unit. Ill post 2 of the many problems which I can’t do.



Calculate the binding energy for an atom of Uranium – 235.

My working:
Number of:
Protons – 92
Neutrons – 143
Therefore, mass of protons = [tex]92 \times 1.67265\times 10^-27 = 1.538838\times 10^-25[/tex]kg
Mass of neutrons = [tex]143 \times 1.67496\times 10^-27 = 2.3951928\times 10^-25[/tex]kg
Combined mass of nucleons = [tex]3.93440308\times 10^-25[/tex]kg
From here I do not know where to go.



Under certain circumstances, a gamma ray photon may suddenly change into an electron and positron as in the following equation:
[tex]\gamma \rightarrow e^0_-1 + e^0_+1[/tex]
Calculate the minimum energy of the photon.

This problem has me completely lost, as I don not understand how a gamma ray with no mass can all of a sudden be broken down into the two parts, and electron and a positron which have a mass.



Many thanks for any replies with the slightest help, as I am completely lost on these equations plus similar ones.
Thanks,
Pavadrin
 
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  • #2
The concept of binding energy is that the nucleus of a certain atom of an element is different from when the individual nucleons are separated from its own. The difference in this mass is related to energy by the equation : difference in mass = Energy/c^2 where c is speed of light. So all you have to do is find the difference in mass and multiply by c^2.

I think someone can help you better with the gamma rays I'm not very sure either :smile:
 
  • #3
okay, i sort of understand now, thanks al_201314. so basically i need to have the mass of the uranium - 235, so that i can subtract that form the mass of the products to find the mass defect/mass difference?
thanks,
Pavadrin
 
  • #4
pavadrin said:
okay, i sort of understand now, thanks al_201314. so basically i need to have the mass of the uranium - 235, so that i can subtract that form the mass of the products to find the mass defect/mass difference?
thanks,
Pavadrin

No worries. That's right. You already have the combined mass of the nucleons in your initial post so now you need the mass of the uranium nucleus and just find the difference in mass.
 
  • #5
pavadrin said:
Under certain circumstances, a gamma ray photon may suddenly change into an electron and positron as in the following equation:
[tex]\gamma \rightarrow e^0_-1 + e^0_+1[/tex]
Calculate the minimum energy of the photon.

This problem has me completely lost, as I don not understand how a gamma ray with no mass can all of a sudden be broken down into the two parts, and electron and a positron which have a mass.
I believe this process is called "Pair production" if I am not wrong.

By interaction via the Coulomb force, in the vicinity of the nucleus, the energy of the incident photon is spontaneously converted into the mass of an electron-positron pair. A positron is the anti-matter equivalent of an electron; it has the same mass as an electron, but it has a positive charge equal in strength to the negative charge of an electron. Energy in excess of the equivalent rest mass of the two particles appears as the kinetic energy of the pair and the recoil nucleus. The positron has a very short lifetime (about 10-8 seconds). At the end of its range, it combines with a free electron.
 
  • #6
pavadrin said:
Hey



Under certain circumstances, a gamma ray photon may suddenly change into an electron and positron as in the following equation:
[tex]\gamma \rightarrow e^0_-1 + e^0_+1[/tex]
Calculate the minimum energy of the photon.

This problem has me completely lost, as I don not understand how a gamma ray with no mass can all of a sudden be broken down into the two parts, and electron and a positron which have a mass.



Many thanks for any replies with the slightest help, as I am completely lost on these equations plus similar ones.
Thanks,
Pavadrin

Strictly speaking, this reaction is not possible (it violates conservation of four-momentum). But in the presence of something else (like a nearby nucleus) which can absorb some four-momentum, the reaction can take place. So the question is a bad one, because to answer one really needs to know what is the complete reaction (what else is around to absorb the four-momentum). But a quick (but not completely correct) answer is that one needs at least the rest mass energy of the electron and positron so one needs at least [itex] 2 m_e c^2 [/itex]. Again, the *real* answer is that the minimum energy is slightly above this because there must be another constituent around that will absorb some of the energy (and momentum).
 
Last edited:
  • #7
Reshma said:
I believe this process is called "Pair production" if I am not wrong.

By interaction via the Coulomb force, in the vicinity of the nucleus, the energy of the incident photon is spontaneously converted into the mass of an electron-positron pair. A positron is the anti-matter equivalent of an electron; it has the same mass as an electron, but it has a positive charge equal in strength to the negative charge of an electron. Energy in excess of the equivalent rest mass of the two particles appears as the kinetic energy of the pair and the recoil nucleus. The positron has a very short lifetime (about 10-8 seconds). At the end of its range, it combines with a free electron.

One correction. The lifetime of the positron is infinite! It is a stable particle! (lifetime is defined in terms of what happens to a particle in a vacuum..whether it decays or not). Of course, if a positron is not in a vacuum but is traveling in matter (in a gas, say), then it will eventually combine with an electron. But that is not a *decay* reaction, it's annihilation. So one does not talk about "lifetime" at all in that context. And the average time the positron will survive depends on many factors, including its energy, the type and density of matter it is traveling into, etc (so I don't know where the figure of 10^(-8) s comes from)

Just to make this clear.

Regards

Patrick
 
  • #8
many thanks for the replies, however I am still unsure what to do about the gamma radiation problem. what is mean't by the four-momentum?
thanks
Pavadrin
 
  • #9
pavadrin said:
many thanks for the replies, however I am still unsure what to do about the gamma radiation problem. what is mean't by the four-momentum?
thanks
Pavadrin
You haven't done some special relativity?
In any case, don't worry about my comments concerning four-momentum (it is just a way to combine energy and momentum in a way that makes it easier to discuss special relativity and Lorentz transformations). The point is that energy and momentum cannot be conserved in the reaction "photon goes to electron plus positron". There must be another constituent nearby (like a nucleus or another particle) to interact with.
Since your class seems to cover things at an introductory level, I would say that the answer they expect is simply that the photon must have at least an energy equal to the rest-mass energy of the electron plus the rest-mass energy of the positron (so that they are created!). So [itex] E_{\gamma, minimum} \approx 2 m_e c^2 [/itex].

Patrick
 
  • #10
ooooo i c now, thanks Patrick :smile:
 

1. What is binding energy?

Binding energy refers to the energy that holds the nucleus of an atom together. It is the energy required to break apart the nucleus into its individual protons and neutrons. This energy is released during nuclear reactions and is responsible for the stability of atoms.

2. How is binding energy related to nuclear reactions?

Nuclear reactions involve the transformation of one or more atomic nuclei. These reactions release or absorb energy, which is the result of changes in the binding energy of the nuclei involved. The difference in binding energy between the reactants and products determines the amount of energy released or absorbed during the reaction.

3. What factors affect the binding energy of a nucleus?

The binding energy of a nucleus is affected by the number of protons and neutrons in the nucleus, known as the mass number. The stronger the nuclear force between the protons and neutrons, the higher the binding energy. Additionally, the arrangement of protons and neutrons in the nucleus also affects the binding energy.

4. How is binding energy calculated?

The binding energy of a nucleus can be calculated using Einstein's famous equation, E=mc². In this equation, E represents energy, m represents mass, and c represents the speed of light. By calculating the difference in mass between the reactants and products of a nuclear reaction, the corresponding change in binding energy can be determined.

5. What are the applications of binding energy and nuclear reactions?

Nuclear reactions and binding energy have various applications, including nuclear power generation, medical imaging and treatments, and nuclear weapons. They are also used in research and development of new materials and technologies, such as the production of isotopes for medical use and the study of nuclear fusion for potential energy production.

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