[10] Al metal has the fcc crystal structure. An X-ray diffraction experiment is

Question

[10] Al metal has the fcc crystal structure. An X-ray diffraction experiment is carried out on Al powder, and diffraction spots (maxima in scattered X-ray intensity) are found at the following Bragg angles : 19.48° 22.64° 33.00 39.68 41.83° 50.35° 57.05 59.420 a) [6] For each angle, identify the Miller indices (hkl) of the associated lattice planes. (I suggest you make a table with a few columns containing helpful information. For example, including a column with the quantity h'+k'+1' will probably be useful.) From the information in your table, calculate the cubic lattice constant for Al. b) [4] Given that Al has atomic weight of 27.0 grams and a density of 2.70 g/cm*, calculate Avogadro's number from your data.

Explanation / Answer

Using Bragg's law 2dsin =

asumming = 0.154 nm here ( hasn't been provided in the question) and substituting in the above formula for each of and we find d, that is the diffraction spacing parameter/interplane spacing.

For,
= 19.48 d = 0.230 nm
= 33.0 d =0.141 nm
= 41.83 d = 0.115 nm

d = a/(h2 + l2 + k2 )1/2, a is the unit cell dimension and h,k,l are the miller's indices

a = 2 x atomic radius of AI x 1.414 = 2 x 0.143 nm x 1.414 = 0.404 nm

therefore for each d we can find (h2 + l2 + k2 )1/2 now that we know the value of a.

(h2 + l2 + k2 )1/2 d   (h2 + l2 + k2 )   (h, l, k)--approx peaks

1.756 0.230 3.085 (1,1,1)
2.865 0.141 8.209 (2 ,2, 0) (2,0,2)
3.513 0.115 12.34 (2, 2, 2)

For FCC principal diffraction planes are those whose miller's indices are all even or all odd

rho(density) = n.A/Volume.Avogadro's number

Volume = Volume of cube = a3 (a has been calculated above)
A = Atomic mass
n= number of atoms in the structure

therefore avogadro's number = n.A/volume.density = 4 x 27 / (0.404 x 10-7 cm)3 x 2.70

avogadro's number = 6.023 x 1023

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