00:01
For this question, we have an x -ray wavelength lambda of 0 .1 nanometers, which is 0 .1 times 10 to the minus 9 meters.
00:08
The sample is located 12 centimeters from the photographic film.
00:12
So l is 12 centimeters or 0 .12 meters.
00:16
And we're also told that the atomic spacing, d, is 0 .222 nanometers or 0 .22 times 10 to the minus 9 meters.
00:23
So always make sure that if you're dividing or anything or multiplying anything together that they have the same.
00:30
Units.
00:31
So if one has nanometers and one has centimeters, you need to convert both the centimeters or both meters, however you see fit.
00:38
Just be consistent.
00:40
So we're asked to figure out the radii diffraction rings corresponding to the first and second order.
00:47
Look at figure 28 and you see that deflected, the x -ray beam is deflected through an angle of 2 .5.
00:53
Okay.
00:54
So first things first, from bragg's equation, we find the angle of incident for the first and second order come from the equation that says 2 times d times the sign of the angle phi is equal to m times lambda.
01:12
Or in other words, because we're asked to first find the first thing we need to do is find the angles phi to find the two radii of curvature, we find that phi is equal to the inverse sign, sine to the minus one of m times lambda divided by two times d.
01:32
We're asked to find the first two...