InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A beam of X-rays with wavelength lambda = 40pm falls normally on aplne rectangular array of scattering centres and produces a system of diffraction maxima (Fig.) on a plane screen removed from the array by a distance l = 10cm. Find the array periods a and b along the x and y axes if the distance between symmetrically located maxima of second order are equal to Deltax = 60mm (along the x axis) and Deltay = 40mm (along the yaxis). |
|
Answer» Solution :We given here a simple derivation of the condition for diffraction maxima, KNOWS as Laue equations. It is easy to see from the above figure that the path DIFFERENCE between waves scattered by nearby scattering centres `P_(1)` and `P_(2)` is `P_(2)A - P_(1)B = OVERSET rarr(r ).overset rarr(s_(0)) - overset rarr(e ). overset rarr(s)` `=overset rarr(r ). (overset rarr(s_(0))-overset rarr(s)) =overset rarr(r ). overset rarr(S)`. Here `overset rarr(r)` is the radius vector `overset rarr(P_(1))P_(2))`. For maxima this path difference must be an inyeger multiple of `lambda` for any two neighbouring atoms. In the present case of two dunensional lattice with `X`-rays incident normally `oversetrarr(r).oversetrarr(s) = 0`. Taking successively. nearest neighbours in the x-&y- directions We get equations `a cos alpha = h lambda` `b cos alpha = k lambda` Here `cos alpha` and `cos beta` ar the direction consines of the RAY with respect to the `x` & `y` axes of the two dimensional crystal. `cos alpha = (Deltax)/(sqrt((Deltax)^(2) + 4l^(2))) = sin (tan^(-1)(Deltax)/(2l)) = 0.28735` so using `h = k = 2` we get `a = (40 xx 2)/(28735)pm = 0.278nm` Similarly `cos beta = (Deltay)/(sqrt((Deltay)^(2) + 4l^(2))) = sin (tan^(-1)(Deltay)/(2l)) = 0.19612` `b = (80)/(cos beta)pm = 0.408nm` 
|
|