Binding Energy & Nuclear Reactions

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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
 

Answers and Replies

  • #2
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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
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okay, i sorta 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
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pavadrin said:
okay, i sorta 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
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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
nrqed
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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).
 
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  • #7
nrqed
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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 travelling 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 travelling into, etc (so I don't know where the figure of 10^(-8) s comes from)

Just to make this clear.

Regards

Patrick
 
  • #8
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many thanks for the replies, however im still unsure what to do about the gamma radiation problem. what is mean't by the four-momentum?
thanks
Pavadrin
 
  • #9
nrqed
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pavadrin said:
many thanks for the replies, however im 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
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ooooo i c now, thanks Patrick :smile:
 

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