Saved Bookmarks
| 1. |
A beam of X-rays with wavelength lambda = 174pm falls on the surface of a single crystal rotating about its axis which is parallel let to its surface and perpendicular to the direction of the incident beam. In this case the direction to the maxima of second and third order from the system of planes parallel to the surface of the single crystal from an angle alpha = 60^(@) between them. Find the corresponding interplanar disatnce. |
|
Answer» Solution :When the CRYSTAL is roated, the incident monochromatic beam is diffracted from a given crystal plane of interplanar spacing `d` whenever in the course of ROTATION the value of `theta` satisfies the Bragg equation. We have the EQUATIONS `2d sin theta_(1) = k_(1) lambda` and `2d sin theta_(2) = k_(2) lambda` But `pi-2theta_(1) = pi- 2 theta_(2) + alpha` or `2theta_(1) = 2 theta_(2) - alpha` so `theta_(2) = theta_(1) + (alpha)/(2)`. Thus `2d {sin .theta_(1) cos.(alpha)/(2) + cos theta_(1) sin.(alpha)/(2)} = k_(2) lambda` Hence `2d sin .(alpha)/(2)cos .theta_(1) = (k_(2) - k_(1)cos .(alpha)/(2))lambda` also `2d sin.(alpha)/(2)sin .theta_(1) = k_(1)lambda sin .(alpha)/(2)` SQUARING and adding `2dsin .(alpha)/(2) = (k_(1)^(2) + k_(2)^(2) - 2k_(1)k_(2)cos.(alpha)/(2))^(1//2)lambda` Hence `d = (lambda)/(2sin.(alpha)/(2)) [k_(1)^(2) + k_(2)^(2) - 2k_(1)k_(2)cos.(alpha)/(2)]^(1//2)` Substituting `alpha = 60^(@), k_(1) = 2, k_(2) = 3, lambda = 174pm` we get `d = 128pm = 2.81Å` (and not `0.281pm` as given in the BOOK) (Lattice parameters are typically in `Å's` and not in fractions of a pm.) 
|
|