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Photon collision with neutron

  1. Feb 15, 2006 #1
    I am to calculate everything in a non-relativistic state:

    a 15MeV photon is completely absorbed in colliding with a neutron initially at rest. Determine the speed of the neutron:

    conservation of momentum:

    [tex]E=\frac{hc}{\lambda}[/tex]
    [tex]15MeV=\frac{1240MeV nm}{\lambda}[/tex]
    [tex]\lambda=8.2667x10^{-5}nm[/tex]

    [tex]p_{photon}=\frac{h}{\lambda}=m_{neutron}v_{post collision}[/tex]
    where m is the mass of the neutron and v is its velocity
    [tex]v= \frac{h}{ (8.2667x10^{-5}nm )(1.674927 x10^{-27} kg)}[/tex]
    [tex]v=4785505.59 m/s[/tex]

    after collisions, the neutron slows down and comes to rest, it emits a photon using up all the energy stored internally, what is the wavelength of this photon?

    [tex]KE=.5mv^2[/tex]
    [tex]KE=.5(1.674927 x 10^{-27}kg)(4785505.59 m/s)^2[/tex]
    [tex]KE=0.11MeV[/tex]

    this is the kinetic energy of the neutron, I was going to subtract that number from 15MeV to get the potential energy and then find the wavelength of the photon using E=hc/lambda, but I think I have already made a mistake somewhere since the KE should be a lot bigger.

    I've checked my work several times and I dont know where i went wrong.

    also, how do i know how much energy converts to KE and how much converts to PE?
     
    Last edited: Feb 15, 2006
  2. jcsd
  3. Feb 18, 2006 #2

    Astronuc

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    Staff: Mentor

    In an inelastic collision, momentum is conserved, but kinetic energy is lost or rather some of it is transformed.

    pphoton = E/c = pneutron = mv (nonrelativistic) and from that get the kinetic energy, and the remaining energy is absorbed as internal energy.

    Your work seems fine - the photon has very little momentum, so little of the photon energy becomes kinetic energy of the neutron (0.11 MeV), and the remaining energy becomes internal energy of the neutron.

    If the neutron is stopped and photon emitted, don't forget that the neutron recoils with some kinetic energy (again based on conservation of momentum) and the photon will have less than (15 - 0.11) MeV.
     
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