Q1.

i) 200-reflection

d₂₀₀ = 5.638/(2² + 0² + 0²)^-2
   = 5.638/2
   = 2.819
l = 2dsinq

sinq / l = 1 / 2d
   = 1 / 5.638
   = 0.177

f_Na⁺ = 8.713

f_Cl^- = 13.16

Sⁿf_ncos{360(hx + ky + lz)} =
  8.713cos{360(2x0 + 0x0 + 0x0)} +
  8.713cos{360(2x0 + 0x0.5 + 0x0.5)} +
  8.713cos{360(2x0.5 + 0x0 + 0x0.5)} +
  8.713cos{360(2x0.5 + 0x0.5 + 0x0)} +
  13.16cos{360(2x0.5 + 0x0 + 0x0)} +
  13.16cos{360(2x0 + 0x0.5 + 0x0)} +
  13.16cos{360(2x0 + 0x0 + 0x0.5)} +
  13.16cos{360(2x0.5 + 0x0.5 + 0x0.5)}

   = 8.713cos0 + 8.713cos0 + 8.713cos 360 + 8.713cos360 + 13.16cos360 + 13.16cos0 + 13.16cos360

   = 8.713 + 8.713 + 8.713 + 8.713 + 13.16 + 13.16 + 13.16 + 13.16

   = 87.492

Sⁿf_nsin{360(hx + ky + lz)} =
  8.713sin{360(2x0 + 0x0 + 0x0)} +
  8.713sin{360(2x0 + 0x0.5 + 0x0.5)} +
  8.713sin{360(2x0.5 + 0x0 + 0x0.5)} +
  8.713sin{360(2x0.5 + 0x0.5 + 0x0)} +
  13.16sin{360(2x0.5 + 0x0 + 0x0)} +
  13.16sin{360(2x0 + 0x0.5 + 0x0)} +
  13.16sin{360(2x0 + 0x0 + 0x0.5)} +
  13.16sin{360(2x0.5 + 0x0.5 + 0x0.5)}

   = 8.713sin0 + 8.713sin0 + 8.713sin360 + 8.713sin360 + 13.16sin360 + 13.16sin0 + 13.16sin0 + 13.16sin360

   = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

   = 0

F₂₀₀² = 87.492² + 0²
   = 7654.85

I₂₀₀ = 7654.85

ii) 100-reflection

d₁₀₀ = 5.638/(1² + 0² + 0²)
   = 5.638/1
   = 5.638

l = 2dsinq
sinq / l = 1 / 2d
   = 1 / 11.276
   = 0.089

f_Na⁺ = 17.426

f_Cl^- = 26.32

Sⁿf_ncos{360(hx + ky + lz)} =
  17.426cos{360(1x0 + 0x0 + 0x0)} +
  17.426cos{360(1x0 + 0x0.5 + 0x0.5)} +
  17.426cos{360(1x0.5 + 0x0 + 0x0.5)} +
  17.426cos{360(1x0.5 + 0x0.5 + 0x0)} +
  26.32cos{360(1x0.5 + 0x0 + 0x0)} +
  26.32cos{360(1x0 + 0x0.5 + 0x0)} +
  26.32cos{360(1x0 + 0x0 + 0x0.5)} +
  26.32cos{360(1x0.5 + 0x0.5 + 0x0.5)}

   = 17.426cos360 + 17.426cos360 + 17.426cos180 + 17.426cos180 + 26.32cos180 + 26.32cos360 + 26.32cos360 + 26.32cos180

   = 17.426 + 17.426 - 17.426 - 17.426 - 26.32 + 26.32 + 26.32 - 26.32

   = 0

Sⁿf_nsin{360(hx + ky + lz)} =
  17.426sin{360(1x0 + 0x0 + 0x0)} +
  17.426sin{360(1x0 + 0x0.5 + 0x0.5)} +
  17.426sin{360(1x0.5 + 0x0 + 0x0.5)} +
  17.426sin{360(1x0.5 + 0x0.5 + 0x0)} +
  26.32sin{360(1x0.5 + 0x0 + 0x0)} +
  26.32sin{360(1x0 + 0x0.5 + 0x0)} +
  26.32sin{360(1x0 + 0x0 + 0x0.5)} +
  26.32sin{360(1x0.5 + 0x0.5 + 0x0.5)}

   = 17.426sin360 + 17.426sin360 + 17.426sin180 + 17.426sin180 + 26.32sin180 + 26.32sin360 + 26.32sin360 + 26.32sin180

   = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0

   = 0

F₁₀₀² = 0² + 0²
   = 0

I₂₀₀ = 0

The intensity of the (100) reflection is zero. This is a result of the centring on the bc face which causes complete destructive interference, giving rise to the systematic absence h+k=2n+1.

Q2.

The multiplicity of the (200) reflection is 2, arising from the equivalence of the (200) and (-200) reflections.

Q3.

P₂₀₀ = (1 + cos²2q) / 2

d₂₀₀ = 2.819

l = 2dsinq
q = sin^-1(l / 2d)
   = sin^-1(0.274)
   = 15.87°
2q = 31.741°

P₂₀₀ = (1 + cos²31.741) / 2
   = (1 + 0.723) / 2
   = 0.862

The polarisation factor is 0.862.

Q4.

L = 1 / sin²qcosq

d₂₀₀ = 2.819

sinq = 0.274
sin²q = 0.0748
cosq = 0.9619

L = 1 / (0.0748)(0.9619)
   = 1 / 0.0719
   = 13.899

The Lorentz factor is 13.899.

Q5.

I₁₁₁ corrected for multiplicity, polarisation and the Lorentz factor is 52347c units.

I₂₀₀ corrected is F₂₀₀ x j₂₀₀ x P₂₀₀ x L₂₀₀ x c units

   = 7654.85 x 2 x 0.862 x 13.899 x c

   = 183339.45c units.

I₁₁₁ / I₂₀₀ = 52347 / 183339

   = 1 / 3.5

Q6.

Detector calibration : if the detector is not calibrated correctly, theta values will be miscalculated.

Monochromator misaligned : resulting in the miscalculation of theta values.

Divergence of diffracted X-ray beam : if a laboratory diffractometer was used and corrective devices, such as Soller slits, were not used to reduce divergence of the diffracted X-ray beam, the resulting peak is broadened and will give rise to possible errors in theta values.

Sample size : eg. powder too fine to give sufficient crystallites at the correct orientation for diffraction.

Sample preparation : in transmission geometry, the powder should completely fill the capilliary tube, without "gaps" which would reduce the theoretical value of the measured intensity or "clumps" which may absorb more X-rays and reduce intensity.

Atmospheric conditions : e.g. loss of water from sample leading to sample shrinkage, or absorption of moisture from the atmoshpere leading to sample expansion. Both these effects change the position of the incident X-ray beam and therefore alter theta.

Sample alignment : if the sample is not centred correctly and precesses through the X-ray beam, certain positions will have cause less sample to be in the beam than others, leading to incorrect intensity values.