Nuclear Reactions & Binding Energy

In summary, a proton with a kinetic energy of 0.42 x 10-11 J is fired at a stationary oxygen-18 nucleus with a binding energy of 2.35 x 10-11 J. This results in the production of fluorine-18 with a binding energy of 2.30 x 10-11 J, a neutron, and a gamma ray. A high energy proton is required to overcome the electrostatic repulsion of the oxygen-18 nucleus. Using conservation of energy, it can be estimated that the energy of the gamma photon is 0.37 x 10-11 J. However, there are still unknowns in this calculation, such as the kinetic energy of the neutron and the fluor
  • #1
Jimmy87
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Homework Statement


A proton of kinetic energy 0.42 x 10-11 J is fired at a stationary oxygen-18 which has a binding energy of 2.35 x 10-11 J. This reaction produces flourine-18 of binding energy 2.30 x 10-11 J with the production of a neutron and gamma ray.

Q1) Explain why the high energy proton is required.
Q2) Estimate the energy of the gamma photon

Homework Equations


None

The Attempt at a Solution


Q1) - is it because energy is required to turn a neutron into a proton inside the nucleus? Or would I have to talk about the bidning energy?
Q2) I am unsure if it is 0.05 x 10-11 J (difference in binding energy) or 0.37 x 10-11 J (the remaining kinetic energy of the proton (I am thinking this is more likely)

Some guidance/confirmation would be greatly appreciated!
 
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  • #2
Jimmy87 said:
Q1) - is it because energy is required to turn a neutron into a proton inside the nucleus? Or would I have to talk about the bidning energy?
Why would a low-energy proton not induce the reaction?

Jimmy87 said:
Q2) I am unsure if it is 0.05 x 10-11 J (difference in binding energy) or 0.37 x 10-11 J (the remaining kinetic energy of the proton (I am thinking this is more likely)
What does conservation of momentum tell you?
 
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  • #3
DrClaude said:
Why would a low-energy proton not induce the reaction?What does conservation of momentum tell you?

Is it because the binding energy is less so work has to be done to get a nucleus of lower binding energy?

I am unsure on how conservation of momentum plays out here. I thought mass or energy are not separately conserved in nuclear reactions only mass-energy together? This is why I am unsure. Wouldn’t the fluorine nucleus have less mass as it is less tightly bound?
 
  • #4
Jimmy87 said:
Is it because the binding energy is less so work has to be done to get a nucleus of lower binding energy?
No, it's not related to the binding energy. What is stopping a low-energy proton from hitting the nucleus?

Jimmy87 said:
I am unsure on how conservation of momentum plays out here. I thought mass or energy are not separately conserved in nuclear reactions only mass-energy together? This is why I am unsure. Wouldn’t the fluorine nucleus have less mass as it is less tightly bound?
Yes, they are separately conserved. But in the conservation of energy, you have to consider that you initially have kinetic energy of the proton + binding energy of the nucleus, and after you have kinetic energy of the neutron and nucleus + binding energy of the nucleus + energy of the photon. You have to estimate the final kinetic energy if you want to calculate the photon energy.
 
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  • #5
Thanks. So the high energy proton is needed to overcome the electrostatic repulsion of the oxygen nucleus?

So is it just as simple as energy before equals energy after? Would it therefore be 0.37 x 10-11 J for the photon? How do we even estimate if we don’t know anything about the neutron? Do we assume it has zero energy? What specifically happens to the 0.05 x 10-11 J? I know the binding energy has gone down but what is specifically happening in this example inside the oxygen-18 nucleus?
 
  • #6
Jimmy87 said:
Thanks. So the high energy proton is needed to overcome the electrostatic repulsion of the oxygen nucleus?
Yes.

Jimmy87 said:
So is it just as simple as energy before equals energy after?
Yes.

Jimmy87 said:
Would it therefore be 0.37 x 10-11 J for the photon?
How did you get that number? Using conservation of energy, would everything balance out?

Jimmy87 said:
How do we even estimate if we don’t know anything about the neutron? Do we assume it has zero energy?
That wouldn't be compatible with conservation of momentum, would it?

Jimmy87 said:
What specifically happens to the 0.05 x 10-11 J?
You simply need to consider the overall energy balance. That difference in energy will appear naturally in the equation.

Jimmy87 said:
I know the binding energy has gone down but what is specifically happening in this example inside the oxygen-18 nucleus?
The proton took the place of a neutron, so the element transmuted from O to F.
 
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  • #7
DrClaude said:
Yes.Yes.How did you get that number? Using conservation of energy, would everything balance out?That wouldn't be compatible with conservation of momentum, would it?You simply need to consider the overall energy balance. That difference in energy will appear naturally in the equation.The proton took the place of a neutron, so the element transmuted from O to F.

Thanks. For the energy part I was thinking 0.05 of the protons kinetic energy is used for the binding energy so the rest must have gone into the photon and neutron? I’m really not sure on the second part of the question - the more I think about it the more confused I get. I don’t even know what a lower binding energy of flourine means - has it taken some of the protons KE? or is this energy released and therefore the neutron and photon have the energy of the original proton plus this binding energy so 0.47?
 
  • #8
Ok, so been doing some more reading online. The physics websites I have been looking at say that if the binding energy of the products is lower (like this example) then the nuclear reaction is endothermic and requires an input energy. So the binding energy of the reactant minus the product is the energy required - 0.05. This is the input energy required to make the reaction work and this must come from the incoming proton. Therefore, the proton energy minus 0.05 should be the remaining energy left since the proton deposited all of its energy into the oxygen nucleus which equals 0.21 like I said unless I am missing something.
 
  • #9
Jimmy87 said:
the proton deposited all of its energy into the oxygen nucleus
That is an assumption. Both the neutron and the fluorine nucleus may have KE afterwards.
But that seems to leave too many unknowns to solve for using just energy and momentum ... @DrClaude ?
 
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  • #10
haruspex said:
That is an assumption. Both the neutron and the fluorine nucleus may have KE afterwards.
But that seems to leave too many unknowns to solve for using just energy and momentum ... @DrClaude ?

If the neutron and flourine nucleus did not have any kinetic energy please could you confirm the photon would have 0.21 x 10-11J? Thanks.
 
  • #11
haruspex said:
That is an assumption. Both the neutron and the fluorine nucleus may have KE afterwards.
But that seems to leave too many unknowns to solve for using just energy and momentum ... @DrClaude ?

Sorry 0.37 x 10-11J I meant to say!
 
  • #12
Jimmy87 said:
If the neutron and flourine nucleus did not have any kinetic energy please could you confirm the photon would have 0.21 x 10-11J? Thanks.
But by conservation of momentum they cannot both be at rest.
 
  • #13
haruspex said:
But by conservation of momentum they cannot both be at rest.

I will get the actual full question tomorrow and upload it as it was on a practice exam paper but I am almost certain the OP gives all the info.
 
  • #14
haruspex said:
But by conservation of momentum they cannot both be at rest.

Hi, so the only other info in additional to my OP is that it says the oxygen nucleus is stationary (says this in bold) and that the neutron and proton have zero binding energy.
 
  • #15
haruspex said:
But by conservation of momentum they cannot both be at rest.

And the exact question is “estimate the minimum energy of the photon”
 
  • #16
Jimmy87 said:
And the exact question is “estimate the minimum energy of the photon”
That's vital. As I surmised, there is not enough information otherwise.
What equations can you write using conservation laws?
How do you find a minimum?
 
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  • #17
haruspex said:
That's vital. As I surmised, there is not enough information otherwise.
What equations can you write using conservation laws?
How do you find a minimum?

Sorry, I meant maximum energy of the photon as it then says for the next part to calculate the minimum wavelength.

So would maximum energy of photon assume the neutron and flourine nucleus have negligible/very little energy? And therefore this is why they say estimate I assume. So the maximum photon energy would be 0.37 x 10-11 J then?
 
  • #18
Jimmy87 said:
would maximum energy of photon assume the neutron and flourine nucleus have negligible/very little energy?
The least energy consistent with all conservation laws. I keep pointing this out but you persist in ignoring it.
 
  • #19
haruspex said:
The least energy consistent with all conservation laws. I keep pointing this out but you persist in ignoring it.

Sorry. I didn’t ignore it I was saying that I got the question wrong. The questions asks for the maximum photon energy and minimum wavelength.
 
  • #20
Jimmy87 said:
Sorry. I didn’t ignore it I was saying that I got the question wrong. The questions asks for the maximum photon energy and minimum wavelength.
So did you take it into account in arriving at
Jimmy87 said:
0.37 x 10-11 J
?
You say you got that figure by assuming the neutron and F nucleus are both at rest, but by conservation of momentum they cannot be. So you are ignoring it.
 
  • #21
haruspex said:
So did you take it into account in arriving at

?
You say you got that figure by assuming the neutron and F nucleus are both at rest, but by conservation of momentum they cannot be. So you are ignoring it.

I really don’t know what else I can do? I don’t know the velocities of the neutron or F nucleus? Let’s say some of the remaining 0.37 went into the kinetic energy of the neutron and F nucleus (which I fully appreciate it would for momentum to be conserved) how would I calculate this not knowing the velocity of either one? Thanks.
 
  • #22
Jimmy87 said:
I really don’t know what else I can do? I don’t know the velocities of the neutron or F nucleus? Let’s say some of the remaining 0.37 went into the kinetic energy of the neutron and F nucleus (which I fully appreciate it would for momentum to be conserved) how would I calculate this not knowing the velocity of either one? Thanks.
You have three unknowns, the three energies after the reaction.
You have two equations: conservation of energy and of momentum. Write out those equations using the unknowns.
Maximising the photon energy effectively gives you your third equation.
 

Related to Nuclear Reactions & Binding Energy

1. What is a nuclear reaction?

A nuclear reaction is a process in which the nucleus of an atom is altered by either splitting into smaller fragments (fission) or combining with other nuclei (fusion). This process releases a large amount of energy.

2. What is binding energy?

Binding energy is the amount of energy required to hold the nucleus of an atom together. This energy is released during a nuclear reaction and is what powers nuclear reactors and bombs.

3. How is binding energy related to the stability of an atom?

The higher the binding energy of an atom, the more stable it is. This is because the energy required to break apart the nucleus is greater, making it more difficult for the atom to undergo a nuclear reaction.

4. What is the difference between fission and fusion reactions?

Fission is the process of splitting a heavy nucleus into smaller fragments, while fusion is the process of combining two or more lighter nuclei to form a heavier nucleus. Fission releases energy by breaking apart atoms, while fusion releases energy by combining atoms.

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

Nuclear reactions and binding energy have many practical applications, including generating electricity in nuclear power plants, producing medical isotopes for diagnosis and treatment, and powering spacecraft and submarines. However, they can also be used for destructive purposes, such as nuclear weapons.

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