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The EXAFS function is as follows, k’N k x(k)= r. exp(-20k )sin(2kr, + 8, (k)) 2 However, oj (k) = 2ak + b. (a and b are
Posted: Tue Apr 26, 2022 7:04 pm
by answerhappygod
- The Exafs Function Is As Follows K N K X K R Exp 20k Sin 2kr 8 K 2 However Oj K 2ak B A And B Are 1 (82.94 KiB) Viewed 37 times
The EXAFS function is as follows, k’N k x(k)= r. exp(-20k )sin(2kr, + 8, (k)) 2 However, oj (k) = 2ak + b. (a and b are constants) Nj, rj, oj are jth Coordination number of the shell, interatomic distance, Debye-Waller factor, which is a constant. Calculate Fourier transform, 2 -2 ikr Φ(r) = VES_dkov e kºx(k) -2 ikr However, the basis vector (rotor) for the Fourier transform is e In addition, plot the absolute value o () against r.