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Absorption of X-rays

  1. Nov 16, 2008 #1

    N8

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    1. The problem statement, all variables and given/known data

    If the ratio [tex]I(K\alpha)/I(K\beta)[/tex] before filtering is 7.5:1 for a copper target, then compute the thickness of a nickel filter, that would increase this ratio of 500:1.

    Mass Absorption Coefficients (cm[tex]^{2}[/tex]/g)

    Cu: [tex]\rho[/tex] = 8.93 [tex]g/cm^{3}[/tex]

    [tex]K\alpha[/tex] (0.1542nm)
    [tex]\mu/\rho[/tex] = 51.54

    [tex]K\beta[/tex] (0.1392nm)
    [tex]\mu/\rho[/tex] = 38.74

    Ni: [tex]\rho[/tex] = 8.91 [tex]g/cm^{3}[/tex]

    [tex]K\alpha[/tex] (0.1542nm)
    tex]\mu/\rho[/tex] = 48.83

    [tex]K\beta[/tex] (0.1392nm)
    [tex]\mu / \rho[/tex] = 282.8


    2. Relevant equations

    [tex]I_{x} = I_{0} e^{-\mu x}[/tex]

    3. The attempt at a solution

    The answer is 20 microns.

    Trying to align these equations up and solve has not yielded the correct answer. Also, the effect of filter on the mass absorption coeffient can be modified by weight fractions [tex](\mu / \rho ) = x(\mu / \rho)_{1} + (1-x)(\mu / \rho)_{2} [/tex]where x is the weight fraction of one element.

    But I am not sure how to incorporate this into my calculations.


    Also, unfortunately, my book Structure of Materials (Graff) does not provide any worked out problems.

    Any help would be much appreciated.
     
  2. jcsd
  3. Nov 16, 2008 #2

    N8

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    nevermind I was ably to solve this with the relevant equation and calculations for the linear absorption coeffecients of Kalpha and Kbeta for Nickel under copper radiation
     
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