Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Energy from Removal of Proton/Neutron

  1. Apr 23, 2010 #1
    1. The problem statement, all variables and given/known data

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

    i. 1 neutron
    ii. 1 proton

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

    2. Relevant 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]

    3. The attempt at a solution

    I'm not sure how to do this.
  2. jcsd
  3. Apr 23, 2010 #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.
  4. Apr 23, 2010 #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]


    [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]

  5. Apr 23, 2010 #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.
  6. Apr 23, 2010 #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?
  7. Apr 23, 2010 #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.
  8. Apr 23, 2010 #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:


    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!
  9. Apr 23, 2010 #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.
  10. Apr 23, 2010 #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.
  11. Apr 23, 2010 #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.
  12. Apr 23, 2010 #11
    So basically the initial energy has to be greater than the final energy? Whichever way I do it.
  13. Apr 23, 2010 #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.
  14. Apr 24, 2010 #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.
  15. Apr 24, 2010 #14
    Yes, but make your final answer positive. Since they ask how much energy is required to complete this process.
  16. Apr 24, 2010 #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: Apr 24, 2010
  17. Apr 24, 2010 #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.
  18. Apr 25, 2010 #17
    Righto, so should be this:

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


    [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]


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



  19. Apr 25, 2010 #18
    Seems good.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook