I have SAED pattern for my sample which is Zn-Ni coating. Based on XRD results, it is a mixture of pure Zn and some intermetallic Zn-Ni phases. How can I index rings? The attachment is SAED of my sample. I wanted to try using equation Rd=lambda*L to calculate d spacing, but found it too complicated for a sample, having a mixture of several phases.
Dear Volker
Thanks for your reply. The other two phases are gamma (Ni5Zn21) and eta. could you please explain more how to complete the diffraction data for each phase?
in the equation, in my case, lambda is equal to 0.0025 nm and L is equal to 250 mm. For R (ring's radius??) should I put the radius which is measured by software with unit (1/nm) directly? or I should convert it to nm?
Hi Volker,
Thanks for your reply.
I have some question regarding your answers.
1) in your lecture note, where are 2.54 cm and 1024 pixel came from? my picture is 2048pixle* 2048 pixel based on digitalmicrograph software.
2) Should I inter 0.04 (1/nm) in the "scale" section for my image based on your attached file and then measure the ring radius by software?
3) what is "n" in the formula: 1/d=(n. (pixel size))/lambda*L?
my problem is that when I calculate "d" based on "r" values, it is too big by several order of magnitude!!
For example: based on software, r=59 (1/nm)= 0.017 nm, so d= (0.0025nm *250 mm)/0.017 will be 367,647,058 angstrom which is HUGE. it should be around 3.6 A. I appreciate if you could guide me.
Hi Dear Volker
Thank you so much for your reply.
I did the calibration based on what you guided me.but still the calculated d value is so big. I will attached what I did here:
after calibration considering these facts:
the pixel size for our TF20 TEM is: 15 micron
so pixel size for my image= 15 micron/ (0.0025 nm * 200mm)= 0.03 (1/nm)
I entered 0.03 in calibration section in image info as shown below and put the scal bar again:
so this time I calculated d again but it is 10^7 times bigger than what it should be, can you guide me please?
R= 13.03/2= 6.525 (1/nm)= 0.15 nm
d=(0.0025*200*10^6)/0.15= 3.33 * 10^6 nm= 3.33 * 10^7 angstrom
Thanks Volker, I should ask the TEM technician to see if he can do the calibration with Au standard.
wish me luck ;) Thanks for your time and consideration :)
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