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What is the kinetic energy of a proton when neutron decays?

  1. Mar 12, 2017 #1
    1. The problem statement, all variables and given/known data
    What is the kinetic energy given to the proton in the decay of a neutron when:
    a) The electron has negligibly small kinetic energy
    b) The neutrino has negligibly small kinetic energy

    2. Relevant equations
    Q = (mn - mp - me - mv ) c2 = .782MeV

    Where T is kinetic energy, and the neutron is at rest:
    Q = Tp + Te + Tv

    3. The attempt at a solution
    Now my thoughts are similar for both part a) and b):
    For a,
    if the electron has zero kinetic energy, then by conservation of momentum, the proton and the neutrino must have equal and opposite momentums. But from lecture, my professor explicitly wrote:
    Since mp >> me, and mp >> mv,
    Tp << Te, and Tp << Tv, so the Q value is practically
    shared between the electron and the neutrino.
    and when Tv = 0, Te = Q

    So is this a trick question then? Is the kinetic energy given to the proton always going to be essentially zero in part a) and b) , since it is so massive?
     
  2. jcsd
  3. Mar 12, 2017 #2

    haruspex

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    Perhaps, but you have enough information to find out, so try it.
     
  4. Mar 13, 2017 #3
    Are you sure? I have 2 unknowns. By conservation of momentum I have
    mp vp = ∂v mv vv
    But i don't know vp or vv. But wait! Could I maybe say:
    P=momentum
    Pp = -Pv = P , and then say
    Q=Tp + Tv
    Q=sqrt(P2c2 + mp2c4) + sqrt(P2c2 + mv2c4) = .782MeV
    and then solve for P!? which i can then use to find just the kinetic energy of the proton!? :D
     
  5. Mar 13, 2017 #4

    haruspex

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    Yes. If the electron has negligible KE then it certainly has negligible momentum.
     
  6. Mar 13, 2017 #5
    Awesome! I appreciate your help very much! I couldn't have done it without you!!
     
  7. Mar 13, 2017 #6
    Actually, im getting THIS so i want to convert my 0.782MeV into 8.395x10^-4 amu*c2 by dividing by 931.5. If i do this, my units of momentum would be mass and speed, which is good, right? Or do you think it changes things with all the exponents to the 8th, 4th, and 2nd mixed in?

    wolfram.PNG
     
  8. Mar 13, 2017 #7

    haruspex

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    I am no quantum theorist so I may have this wrong, but it seems to me you already subtracted out rest mass energies to arrive at Q.
    Shouldn't it be something like
    ##\Sigma \sqrt{(p_i c)^2+(m_i c^2)^2}## is constant, where the sum is over the particles, and the mi are their rest masses?
     
  9. Mar 13, 2017 #8
    Well wolfram is saying P equals a momentum of 4.89x1036 amu*(m/s)2 = 8.13x109 kg*(m/s)2 which does not seem reasonable because the proton is heavy and should not act relativistic... But i could be wrong about that. Im not very certain of these things either since im in a class 2 semesters ahead of all the other classes im in and i skipped the pre-requisites lol.
    You are right, your equation
    ##\Sigma \sqrt{(p_i c)^2+(m_i c^2)^2}##
    is correct, but in this case, there are only 2 particles, and their momentums must be equal so my equation that i asked wolfram to solve actually fits your generalized equation, and should be correct too.

    But its 4:30am here so im just about done with this problem anyway lol. Unless you have a breakthrough epiphany type of solution idea then i think i should be fine haha. I have the concept down much better now and I owe it all to you for helping me get to the realization of
    and just a small numerical error shouldn't have me losing too many points
     
  10. Mar 13, 2017 #9

    haruspex

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    I still don't see that. What follows is for case (b).
    I am saying
    ##m_Nc^2=\sqrt{(pc)^2+(m_Pc^2)^2}+\sqrt{(pc)^2+(m_ec^2)^2}+m_{\nu}c^2##
    Not sure because I thought it was unclear whether neutrinos have mass.

    Edit:
    Since pc is to be found, we can write that in the form ##E=\sqrt{x+a}+\sqrt{x+b}##.
    From that I get ##4E^2x=E^4-2E^2(a+b)+(a-b)^2##.
     
    Last edited: Mar 13, 2017
  11. Mar 13, 2017 #10
    Oh thats interesting! so this equation says the rest energy of the neutrino equals the total energy of the proton plus the total energy of the electron?
     
  12. Mar 13, 2017 #11

    haruspex

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    No, sorry, I made a mistake.
    Correcting...

    Ok, I've corrected it. mN is the rest mass of the neutron. But on the right I left out the rest mass of the neutrino... does it have any?
     
  13. Mar 13, 2017 #12
    Oh no worries at all. Maybe the N stood for neutron? But I have to get some sleep now. But please do continue i am very interested! I will be up in about 3 hours
     
  14. Mar 13, 2017 #13

    PeroK

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    A suggestion about how to interpret this problem. The decay into three particles has too many variables to give a unique solution, hence the assumptions. In part a) you assume that the electron has no KE or momentum and, I suggest, assume that the neutrino has a velocity close to ##c##. This means that the energy and momentum of the neutrino are: ##E_{\nu} = \gamma mc^2## and ##p_{\nu} = \gamma mc##., hence ##E_{\nu} = cp_{\nu}##

    This means that the neutrinos act like a photon in the energy-momentum equations and you don't need to know its small rest mass.

    Part b) effectively means that no neutrino is produced, given that its rest mass, KE and momentum are negligible.
     
  15. Mar 13, 2017 #14

    PeroK

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    Another suggestion. In these problems, it's best to work with the energy and momentum for each particle and calculate the overall energy of the particle you want. If you split the energy into mass-energy and kinetic energy too early, it tends to get messy. Note that once you have got the energy of a particle, then you can get the KE simply by: ##T = E - mc^2##.

    Finally, there is actually a fairly simple general solution to the problem of decay of a single particle into two. So, the answer here should not be too complicated (although it's not so easy to get to that simple answer). The trick here is see that in part a) the electron is at rest, so its mass can effectively simply be subtracted from that of the neutron. So, I would set the problem up as follows:

    Energy conservation: ##E_n = E_e + E_p + E_{\nu}##, hence ##E_{\nu} = (E_n - E_e) - E_p = E_0 - E_p##

    Where, for part a), I've used ##E_0 = E_n - E_e = (m_n - m_e)c^2## to keep the equations as simple as possible.

    Then, I would square this equation to get:

    ##E_{\nu}^2 = E_0^2 + E_p^2 - 2E_0E_p##

    Now, use conservation of momentum to aim for an equation for ##E_p##.
     
    Last edited: Mar 13, 2017
  16. Mar 13, 2017 #15
    Okay I hear you out and I agree that the neutrino should have a velocity close to ##c##, but if we treat it like a photon, then
    ##E_{\nu} = cp_{\nu}## would become
    ##E_{\nu} = chf## where ##f## is the frequency of the photon.
    How could we possibly know the frequency of the photon/neutrino?
     
  17. Mar 14, 2017 #16

    PeroK

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    The energy and momentum of a particle are related by:

    ##E^2 = p^2c^2 + mc^4##

    In the case of a truly massless particle or a particle of negligible mass, this reduces to:

    ##E^2 = p^2c^2##

    This gives the energy of a photon (or a neutrino). There is absolutely no reason to start introducing the frequency of the photon (or neutrino) into the equation.
     
  18. Mar 14, 2017 #17
    Brilliant! Your math checks out
    You're right, I found that out. I already got my answers and im confident in them. Thank you very much :)
     
  19. Mar 14, 2017 #18
    Summary for future readers looking for the guidance more consice:
    Using these equations in the order they are written would be helpful for both parts but part A specifically
    ##Q=({\Delta}m)c^2 = .782##MeV
    ##Q = E_n = E_p + E_{e^-} + E_{\nu}##
    ##Q = E = \sqrt{x+a} + \sqrt{x+b} + d##
    where ##x=(Pc)^2##, ##a=(m_pc^2)^2##, ##b=(m_{\nu}c^2)^2=0##, ##d=(m_{e^-}c^2)^2=.5114##MeV
    The electron has only rest mass energy because its kinetic energy is said to be zero. ##Pc## is the same for both the proton and the neutrino because momentum must be conserved and equal if the electron has zero momentum.
    Finally,
    ##(Pc)^2 = x = \frac{(a-d^2+2dE-E^2)^2}{4(d-E)^2}## and using momentum, one can find kinetic energy
    All credit goes to haruspex and to PeroK.
    PeroK has an excellent explanation on the final post on page 2 as well!
     
    Last edited: Mar 14, 2017
  20. Mar 14, 2017 #19

    haruspex

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    No, that is not right. The total energy is conserved.
    ##E_N = E_p + E_{e^-} + E_{\nu}##
     
  21. Mar 14, 2017 #20
    Oh, thank you. Fixed.
    Again though, thank you for all your help, you're a lifesaver. I understand how to use these equations so much better now
     
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