Energy from Removal of Proton/Neutron

  • Thread starter Lissajoux
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In summary: O} \rightarrow {}^{15}_{8}\text{O} + \text{p}So just need to rearrange and input the relevant energy values given.This gives:Q = 2.9 \times 10^{-3}u = 2.701MeVRight?Yes, this is the equation for the proton removal.
  • #1
Lissajoux
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Homework Statement



How much energy (in MeV) is required to remove:

i. 1 neutron
ii. 1 proton

From [itex]{}^{16}_{8}O[/itex]?

Homework Equations



The masses of relevant neutral atoms:

[tex]M\left({}^{15}_{7}N \right)=15.0001u[/tex]

[tex]M\left({}^{15}_{8}O \right)=15.0030u[/tex]

[tex]M\left({}^{16}_{8}O \right)=15.9949u[/tex]

The Attempt at a Solution



I'm not sure how to do this.
 
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  • #2
Just set up the equations. For example, the first one is:

[tex]{}^{16}_{8}\text{O} \rightarrow {}^{15}_{8}\text{O} + \text{n}[/tex]

So just subtract the masses from both sides and convert to energy.
 
  • #3
Right, so for the removal of the neutron:

[tex]Q = M\left({}^{16}_{8}O \right) - M\left({}^{15}_{8}O \right) = 15.9949u - 15.0030u = 0.9919u[/tex]

Then need to convert this to MeV, so use that [itex]u=931.494MeV[/itex]

Hence:

[tex]Q = 923.949MeV[/tex]

.. correct?

Then do similar for the removal of proton case:

The equation is:

[tex]{}^{15}_{8}\text{O} \rightarrow {}^{15}_{7}\text{O} + \text{p}[/tex]

Then just need to rearrange and input the relevant energy values given.

This gives:

[tex]Q = 2.9 \times 10^{-3}u = 2.701MeV[/tex]

Right?
 
  • #4
Your first equation seems to have a little latex trouble. But I don't see the mass of the neutron in there. The neutron doesn't disappear, so it carries mass as well.
 
  • #5
I've edited the equations now so it should look fine, it does look much better to me now.

So I have to minus the value of the neutron mass as well then? Is that what you mean?
 
  • #6
Yes, you should also subtract off the mass of the neutron. Also, you really should switch the order around and do final-initial. That way you can see how much greater the final energy is over the initial state. So you will know you need that much extra energy to get the final state.
 
  • #7
I think I see what you mean about switching it around, but it seemed more simple and intuitive to do it the way I did.

I see though that when I account for the subtraction of the neutron mass, this would give me a negative result presumably due to the method.

So if I rearrange it to have:

[tex]Q = M\left({}^{15}_{8}O \right) - m_{n} - M\left({}^{16}_{8}O \right) = 15.0030u - m_{n} - 15.9949u = -0.9919u - m_{n}[/tex]

So I just need to correct this by taking away [itex]m_{n}=939.566MeV[/itex], and convert 0.9919u to MeV as well.

Which will therefore give [tex]Q=-940.558MeV[/tex]

But doing it the original way, simply taking off the value for the neutron mass gives:

[tex]Q=-936.864MeV[/tex]

Both of these methods seem to give me a negative value of Q, but I think this is OK though? I can't remember why, something like it shows it isn't spontaneous process.

Is this along the right lines now? Hopes so!
 
  • #8
Ok, if you switched the order to final-initial, then you need to remember that the isolated neutron is a final product. So now you add the mass of the neutron into the equation and not subtract.
 
  • #9
Was there anything wrong with this method:

[tex]Q = M\left({}^{16}_{8}O \right) - M\left({}^{15}_{8}O \right) - m_{n}[/tex]

??

This is the way I find easiest, for some reason switching things around is more confusing.
 
  • #10
Alright, keep it that way. That will get you an answer as well, but it should be negative meaning the final energy is greater than the initial energy. So you will need add energy to the initial state.
 
  • #11
So basically the initial energy has to be greater than the final energy? Whichever way I do it.
 
  • #12
You want the initial energy to equal the final energy. That means you will need to add in a specific amount of energy to the initial state to get the final state. You are trying to solve for that energy.

If you add in extra energy, then that will just go to kinetic energy for the final products.
 
  • #13
SO if I have that:

[tex]Q(energy)=(total mass of initial state)-(total mass of final state)[/tex]

Which is:

[tex]Q = M\left({}^{16}_{8}O \right) - M\left({}^{15}_{8}O \right)-m_{n} = 15.9949u - 15.0030u -m_{n}= 0.9919u-m_{n} [/tex]

Then whatever this answer is is the correct value of Q?

Then just repeat but change n with p.
 
  • #14
Yes, but make your final answer positive. Since they ask how much energy is required to complete this process.
 
  • #15
Swtiching to final-initial I have:

[tex]Q = M\left({}^{15}_{8}O \right) + m_{n} - M\left({}^{16}_{8}O \right) = 15.0030u - 15.9949u + m_{n} = -0.9919u + m_{n}[/tex]

Then need to convert u to MeV, and input that [itex]m_{n} = 939.565560 MeV[/itex]

This gives:

[tex]Q=15.617 MeV[/tex]

Which is positive.. hooray! So this is the energy to remove a neutron, right?

Then need to do same but with proton, so is it just:

[tex]{}^{15}_{8}\text{O} \rightarrow {}^{15}_{7}\text{O} + \text{p}[/tex]

correct equation? Then just repeat as with steps from before, just change n for p?
 
Last edited:
  • #16
Yes, that is the energy required to remove the neutron. Do the same for the proton, but pay attention to the number of protons. If you change the number of protons you will change the element type. There is no such thing as Oxygen with 7 protons.
 
  • #17
Righto, so should be this:

[tex]{}^{15}_{8}\text{O} \rightarrow {}^{15}_{7}\text{N} + \text{p}[/tex]

Hence:

[tex]Q = M\left({}^{15}_{7}N\right) + m_{p} - M\left({}^{15}_{8}O \right) = 15.0001u -15.0030u + m_{p} = -2.9\times 10^{-3}u + m_{p}[/tex]

Then:

[itex]-2.9\times 10^{-3}u=-2.701MeV[/itex] and [itex]m_{p}=938.3MeV[/itex]

Therefore:

[tex]Q=935.6MeV[/tex]

..correct?
 
  • #18
Seems good.
 

1. How is energy obtained from the removal of a proton or neutron?

Energy can be obtained from the removal of a proton or neutron through a process called nuclear fission. This involves splitting an atom's nucleus into smaller fragments, which releases a large amount of energy in the form of heat and radiation.

2. Can energy from the removal of a proton or neutron be used as a source of electricity?

Yes, the energy released from nuclear fission can be converted into electricity through a nuclear power plant. The heat generated from the fission process is used to produce steam, which then turns a turbine and generates electricity.

3. What are the benefits of using energy from the removal of a proton or neutron?

One of the main benefits of using nuclear energy is that it does not produce air pollution or greenhouse gas emissions. It is also a reliable source of energy and can provide a large amount of electricity without the need for a lot of resources.

4. Are there any risks associated with using energy from the removal of a proton or neutron?

Like any form of energy production, there are risks associated with using nuclear energy. The main risk is the potential for accidents, such as a nuclear meltdown, which can have severe environmental and health consequences. However, the industry has implemented strict safety measures to minimize these risks.

5. How is the waste from nuclear energy stored and disposed of?

The waste from nuclear energy is highly radioactive and must be stored safely to prevent harm to humans and the environment. It is typically stored in special containers and kept in secure facilities such as underground repositories. Long-term solutions for the disposal of nuclear waste are still being developed and researched.

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