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Calculation of Intensity of of AX crystals

  1. Mar 3, 2015 #1
    I'm still confuse on using the equation about atomic scattering factor, f = a*e^(-b+((sin(theta)/lamda)^2) + c (general equation),

    where sin(theta)/lamda = 1/2d = 1/(2*(sqrt(a^2/(h^2 + k^2 +l^2)))), so this term is dependent on the plane.
    which you will used to find the Intensity of crystals in a certain plane. I = F^2 (disregard multiplicity factor, absorption and etc.), F = f*cos(2pi(hx + ky + lz)) (i omitted the isine part)

    so given a compound NaCl where a, b and c are also given, i try to so solve I in 111 plane.

    first i solve fNa and fCl using the equation above, this gives 26.30, and 23.71 respectively, and substitute to the equation of intensity i got I(NaCl)=106.44, but when i'm try to check using related literature, some website (http://pd.chem.ucl.ac.uk/pdnn/diff2/structf.htm) and its jcpds, my answer was wrong so please enlighten me about this matter.

    constant i used:
    for Na:
    for Cl:
    a1Cl=18.2915;b1Cl=0.0066; a2Cl=17.604; b2Cl=1.1717;a3Cl=6.5337;b3Cl=19.5424;a4Cl=2.3386;b4Cl=60.4486;c1Cl=-16.378;
  2. jcsd
  3. Mar 3, 2015 #2
    The formula for the atomic (or ionic) scattering factor given here



    f(s) = \sum\limits_{i=1}^{4}a_i \exp(-b_i s^2) +c \\
    s = \sin(\theta)/\lambda = Q/4\pi = 1/2d

    (you have a "+" which is probably a typo)

    You should use the scattering factors for the ions, Na+ and Cl-

    The constants given in the paper look a bit different from yours, but I have not checked every one of them.
  4. Mar 3, 2015 #3
    Last edited: Mar 3, 2015
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