ASSIGNMENT III

Problem I

Calculation of F₂₀₀ for (200) reflection

Calculation of atomic scattering factor for Na⁺ and Cl^-

The atomic scattering factors for Na⁺ and Cl^- ions was calculated using Bragg's equation to compute Sin q/l. The value obtained for (200) reflection was 0.1773. This value is exactly at the mid of 0.15 and 0.2. Therefore after consulting the International Crystallographic table f_Na⁺ and f_Cl^- was found to be 8.7125 and 13.16.

Comment: A small quible: 0.1773 is not exactly at the mid of 0.15 and 0.2. However I agree that it is a reasonable approximation.
4 Marks

Calculating the Cosine part of F₂₀₀

---

å n	fn cos{360 (hx + ky + lz)}	=		8.7125 cos {360 (2´0 + 0´0 + 0´0)}
			+	8.7125 cos {360 (2´0 + 0´1/2 + 0´1/2 )}
			+	8.7125 cos {360 (2´1/2 + 0´0 + 0´1/2 )}
			+	8.7125 cos {360 (2´1/2 + 0´1/2 + 0´0)}
			+	13.16 cos {360 (2´1/2 + 0´0 + 0´0)}
			+	13.16 cos {360 (2´0 + 0´1/2 + 0´0)}
			+	13.16cos {360 (2´0 + 0´0 + 0´1/2 )}
			+	13.16 cos {360 (2´1/2 + 0´1/2 + 0´1/2 )}
		=		8.7125 cos 0 + 8.7125 cos 0 + 8.7125 cos 360 + 8.7125 cos 360
			+	13.16 cos 360+ 13.16 cos 0 + 13.16 cos 0 + 13.16 cos 360
		=		8.7125 + 8.7125 + 8.7125 + 8.7125 + 13.16 + 13.16 + 13.16 + 13.16
		=		87.49

Calculating the Sine part of F₂₀₀

---

å n	fn sin {2p (hx + ky + lz)}	=		8.7125 sin {360 (2´0 + 0´0 + 0´0)}
			+	8.7125 sin {360 (2´0 + 0´1/2 + 0´1/2 )}
			+	8.7125 sin {360 (2´1/2 + 0´0 + 0´1/2 )}
			+	8.7125 sin {360 (2´1/2 + 0´1/2 + 0´0)}
			+	13.16 sin {360 (2´1/2 + 0´0 + 0´0)}
			+	13.16 sin {360 (2´0 + 0´1/2 + 0´0)}
			+	13.16 sin {360 (2´0 + 0´0 + 0´1/2 )}
			+	13.16 sin {360 (2´1/2 + 0´1/2 + 0´1/2 )}
		=		8.98 sin 0 + 8.98 sin 0 + 8.98 sin 360 + 8.98 sin 360
			+	13.16 sin 360 + 13.16 sin 0 + 13.97 sin 360 + 13.97 sin 360
		=		0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
		=		0

Combining the Cosine and Sine parts of F₂₀₀gives F₂₀₀ = 87.49 + i.0 and the intensity is

I200 = F2002 = 7654.5units

Comment: Correct.
5 Marks

Calculation of F₁₀₀ for (100) reflection

Calculation of atomic scattering factor for Na⁺ and Cl^-

---

For the (100) reflection the Sin q/l was half of that for the (200) reflection i.e. 0.08865. By looking into the International Crystallographic table and upon substituting the values in the str ucture factor equation, both the sine and cosine part vanishes. Thus the intensity of the (100) is zero.

Comment: Correct numerical answer but I would have also liked some explanation behind the zero intensity (see model answer).
5 Marks

Problem II

Comment: Exactly right.
3 Marks

Problem III

using the formula,

2q	= 31.7 °
cos (2q)	= 0.8508
cos2 (2q)	= 0.723
1 + cos2 (2q)	= 1.723
P200 = (1 + cos2 (2q ))/2	= 0.861

Thus polarization factor reduces the Intensity of 200 reflection to 86%.

Comment: Correct.
4 Marks

Problem IV

Using the value for theta already obtained in the polarisation calculation:

										q	= 15.85°
										sin q	= 0.2731
										sin2q	= 0.07459
										cos q	= 0.962
										sin2q cos q	= 0.07175
										L = 1/sin2q cos q	= 13.94

Comment: Correct.
3 Marks

Problem V

Substituting the values obtained for different factors from the previous problems in the above equation, gives the value 0.095. Thus the intensity from (111) plane is one tenth of (200) plane. This is in consistent with the observed results.

Comment: Exactly right.
3 Marks

Problem VI

1. Poorly ground sample
0.5 Mark
2. Filter may not work properly
0 Mark
3. Thick powder samples
0 Mark
4. absorption by the capillary
0.5 Mark

Comment: See model answer for reasons. Details above a little too brief.

Comment: A good solid assignment; in particular your numerical answers were extremely well done, but a little more thought was needed on the last question.