Question

The copper with an atomic radius of 0.1278 nm has an FCC unit cell structure. The wavelength under the X-ray diffraction is 0.1542 nm. (15%) (a) Determine the interplanar spacing dnktfor the crystallographic plane (111) (b) Calculate the diffraction angle 20 for the crystallographic plane (111)

Answer 1

Answer:

a) d = 0.2086 nm

b) $2\theta = 43.36 degree$]

Explanation:

given data:

atomic radius = 0.1278 nm

\lamda = 0.1542 nm

a) interplanar spacing (d)is given as $=\frac{a}{\sqrt {h^{2} +k^{2}+l^{2}}}$

   

a is the size of unit cell

a\sqrt{2} = 4*r

$a = \frac{4*0.1278}{\sqrt{2}}$

a = 0.3614 nm

miller indices for crystallofgraphic plane [111] is reciprocal i.e.

[h,k,t] = [1,1,1,]

$d = \frac{0.3614}{\sqrt {1^{2} +1^{2}+1^{2}}}$

d = 0 .2086 nm

b) diffraction angle

by bragg's law and for first order reflection (n=1)

$2d*sin\theta = 1.\lamda$

$2*0.2086*sin\theta = 0.1542$

$\theta = 21.68degree$

$2\theta = 43.36 degree$

Answer 2

A) The interplanar spacing d_nkt for the crystallographic plane is; 0.2087 nm

B) The diffraction angle for the crystallographic plane is; 43.36°

Crystal structures

A) We are given;

• Atomic radius; r = 0.1278 nm

• wavelength; λ = 0.1542 nm

Formula for interplanar spacing is;

d_nkt = a/√(n² + k² + t²)

For FCC structures, n, k and t are 1,1,1.

Also, a = 2r√2

a = 2*0.1278√2

a = 0.3615 nm

Thus;

d_nkt = 0.3615/√(1² + 1² + 1²)

d_nkt = 0.2087 nm

B) Formula for the diffraction angle is gotten from;

d_nkt = nλ/(2 sin θ)

where  θ is half of the diffraction angle

n is the order of reflection. This is first order and so n = 1

Thus;

0.2087 * 2 sin θ = 0.1542

sin θ = 0.1542/(0.2087 * 2)

sin θ = 0.3694

θ = sin⁻¹ 0.3694

θ = 21.68°

Since that is half of the diffraction angle, then we can say that;

Diffraction angle = 2 * 21.68° = 43.36°

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