Is Bi2Te3 Rhombohedral or Hexagonal, and How Do You Analyze Its XRD Data?

[mahedi]

need urgent information

Dear All,

Can anyone inform me the following things on urgent basis ?

- The crystal structure of Bi2Te3 is Rhombohedral or Hexagonal ?
- How can I calculate the grain size from the XRD data/curve for Bi2Te3 ?
- how to calculate the lattice parameter of Bi2Te3 (Rhombohedral/Hexagonal crystal structure) from the XRD data/curve ?

Providing such information would be very much helpful to me.

Sincerely
Mahedi

 

[drizzle]

not so sure about your first question, Bi2Te3 can posses both structures..
to calculate the grain size [g]:
g=0.9λ/Bcosθ
where,
0.9 is the shape factor
λ is the wavelength of the Cu K_α1 radiation [λ =1.5406×10^-10 m]
B is the band width at half maximum [BWHM], and here’s how to get it:
in the diffraction pattern and at an exact Bragg angle, find the maximum intensity value, divide it by 2, you would have 2 points with the same value on both sids of the peak [illustrated below]:
http://img12.imageshack.us/img12/5416/31327198.png

B= 2θ₂-2θ₁ [B should be measured in Radians]

To find the lattice parameters, you use two equations
1] Bragg’s law:

n λ= 2d_(hkl)sinθ
set n=1

2] the function that describe the inter-planar distance [d] as a function of the unit cell parameters and Miller indices, h, k and l. i.e. in the case of the hexagonal system:

1/d²_(hkl)=4/3[h²+hk+k²]/a² + l²/c²

hope I’m helping, good luck.

 

[mahedi]

Dear Sir,
Thanks for your kind information regarding my inquiry. Sorry for late reply as I was in vacation.