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Lithium scattering on iron plate

  1. May 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Every second ##N_{Li}=2\cdot 10^8## atoms of ##_3^7Li## with kinetic energy ##30 MeV## is Coulomb scattering on ##_{26}^{56}Fe## plate ##d_{Fe}=0.3mm## thick with density ##\rho =7800kg/m^3##. Under angle of ##30^°## and ##r=0.2m## away from the plate we have a target with surface ##S_n=1cm^2## that contains ##N_n=6\cdot 10^{22}## neutrons.

    a) How many Lithium (N) falls on the neutron plate each second?
    b) How many Lithium (N') per second deviate due to the scattering on neutron plate? You can say that ##\sigma _n=0.2\cdot 10^{-28} m^2## for one of the Li.


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    2. Relevant equations



    3. The attempt at a solution

    Can I ask before I show my work? :))

    Is it ok if I calculate how much of the original Lithium flux overcomes the Fe plate, where it Coulomb scatters? This should than be the answer to part a) or is it not?

    All I am trying to say is that I don't get it why the angle of neutron plate and the distance is important? (not even for part b))
     
  2. jcsd
  3. Jun 2, 2014 #2

    BvU

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    Would that be in the forward direction ? None of those reach te neutron plate ! What relevant equations do you have available to tackle this problem ?
     
  4. Jun 2, 2014 #3
    Well I somehow have to find out how many are scattered in ##d\omega =\frac{dS}{r^2}## I just don't know how.

    Equations:

    Coulomb

    ##\frac{\mathrm{d} \sigma }{\mathrm{d} \Omega }=(\frac{e_1e_2}{16\pi \epsilon _0T})^2\frac{1}{sin^4(\theta /2)}##

    For scattering on one particle: ##d \Omega (\frac{\mathrm{d} \sigma }{\mathrm{d} \Omega })=\frac{N}{jdt}## where N is number of particles scattered in ##d\theta ## and ##j## initial flux.

    also

    ##N=\frac{N_pN_t\sigma}{S}## where N is number of reactions, ##N_p## number of projectiles and ##N_t## number of targets.

    That's more or less all I have.
     
  5. Jun 3, 2014 #4

    BvU

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    Coulomb expression: Well, there is a theta in there ! So the angle is needed. Now about the distance. Where is that hiding ? Or is it absent ? Once that is answered, your conditions for showing your work are fulfilled!
     
  6. Jun 3, 2014 #5
    ##d\Omega =\frac{dS}{r^2}##

    BUT

    ##N_td\Omega (\frac{d\sigma }{d\Omega })=\frac{N_{on sensor}}{jdt}##

    My question: How do I find out the number of Fe in a plate, ##N_t## ??? I know nothing about area size of flux.
     
  7. Jun 4, 2014 #6

    BvU

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    What about looking around in the given data ? ##\rho## for example ? Area size of flux divides out (you may assume the beam is like a pencil ray). I don't see it in your worked out thingy anyway (?)
     
  8. Jun 4, 2014 #7
    I haven't shown anything yet because I don't think that size of flux divides out.

    Let's take a look:

    Let me use notation ##N_x## instead of ##N_{on sensor}##.

    ##N_td\Omega (\frac{d\sigma }{d\Omega })=\frac{N_{x}}{jdt}##

    ##\frac{N_x}{dt}=N_tjd\Omega (\frac{d\sigma }{d\Omega })=N_tj\frac{dS}{r^2} (\frac{e_1e_2}{16\pi \epsilon _0T})^2\frac{1}{sin^4(\theta /2)}##

    Now ##N_t=\frac{mN_a}{M}=\frac{\rho S d N_a}{M}##. Now here is where it all stops. What am I supposed to do with that S in the numerator?
     
  9. Jun 4, 2014 #8
    Ahhh ok, I see it now. Why does this happen to me? o_O Why do i see it now and not days ago?

    ##j=N/tS##

    So ##\frac{N_x}{dt}=N_tjd\Omega (\frac{d\sigma }{d\Omega })=(\frac{\rho S d N_a}{M})\frac{N_0}{tS}\frac{dS}{r^2} (\frac{e_1e_2}{16\pi \epsilon _0T})^2\frac{1}{sin^4(\theta /2)}##

    So the answer should be:

    ##\frac{N_x}{dt}=\frac{\rho d N_a}{M} \frac{N_0}{t}\frac{dS}{r^2} (\frac{e_1e_2}{16\pi \epsilon _0T})^2\frac{1}{sin^4(\theta /2)}##
     
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