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PS204 Tutorial 3

1.

The Bragg angle corresponding to a reflection for which (h2 + k2 + l2) = 8 is found to be 14.35 degrees. Determine the lattice parameter of the crystal. X-rays of wavelength 0.071 nm are used. If there are two other reflections with smaller Bragg angles, what is the crystal structure?

h2 + k2 + l2 = 8 is possible only for (hkl) = (220). Using nλ = 2d sinθ and taking n = 1 and θ = 14.35 degrees, we get d220 = 0.143.10-9 m Now using

d hkl =

a h + k 2 + l2 2

We find that the lattice parameter a = 0.405.10-9 m Reflections for smaller angles θ means for larger d, and larger d means smaller h2 + k2 + l2. Possible (hkl) smaller than (220) are (200) - all even - and (111) - all odd. This means that the structure must be FCC (check notes)

2.

Sylvine (KCI) crystallizes in the form of a simple cubic structure. Determine the interatomic spacing d and the angle at which an x-ray of wavelength 0.11787 nm is reflected in third order. The density of KCI is 1990 kg.m-3 and its molecular weight is 74.6.

Density of 1990 kg.m-3 means 1990 kg occupies 1m3. So, one molecule, of mass 74.6 amu, occupies (74.6.1.66.10-27/1990) m3 or 0.06.10-27 m3. For simple cubic structure, unit cell contains equivalent of one basis unit (here one KCl molecule). Volume of unit cell is a3 so a3 = 0.06.10-27 m3 → a = 3.9.10-10 m Using nλ = 2d sinθ and taking n = 3, given λ = 0.11787.10-9 m, and (assuming we are dealing with (100) reflections) taking d = 3.9.10-10 m gives θ = 27 degrees.

3.

In a Laue photograph of an fcc crystal whose lattice parameter is 0.45 nm, determine the minimum distance from the centre of the diffraction pattern at which reflections can occur from the planes of maximum spacing, if the potential across the x-ray tube is 50 kV and the crystal to film distance is 5.0 cm.

Picture shows geometry for Laue process. We need to know the angle that meets the Bragg condition for diffraction by the planes of maximum separation for an FCC crystal

We are not given the wavelength λ of the x-rays but we are given the x-ray tube voltage as V = 50 kV. We assume that the minimum λ is for x-ray energy of 50 keV, and using result (from notes) that minimum λ = 1240/V in nm → λ = 0.0248 nm. For FCC crystal, (hkl) all even or all odd. a h + k 2 + l2

From

d hkl =

Using

a = 0.45 nm we get d111 = 0.45/√3 = 0.26 nm

2

maximum separation is for min (hkl), i.e. for (hkl) = (111).

In Bragg equation nλ = 2d sinθ we have n = 1, λ = 0.0248 nm, and d111 = 0.26 nm, and so we get the value of sinθ = 0.0477 → θ = 2.7 degrees. x

θ 5 cm

x/5 = tanθ = tan(2.7) = 0.047 → x = 0.236 cm

Here, x shows the distance above the centre spot at which the (111) reflections appear on the film

A beam of thermal neutrons emitted from the opening of a reactor is diffracted by the (111) planes of a nickel crystal at an angle of 28.5 degrees. Calculate the effective temperature of the neutrons. Ni has an fcc lattice and its lattice parameter is 0.352 nm. The (111) planes separation is given by d111 = a/√(h2 + k2 + l2) where a is lattice constant. Using a = 0.352 nm gives the (111) planes separation as 0.2 nm, and using θ = 28.5O in the standard equation λ = 2d(111) sinθ gives the neutron wavelength λ = 0.191 nm. To get the requested temperature use kBT = ½ mv2 and λ = h/mv where m and v are neutron mass and velocity. These give T = mv2/2kB = m2v2/2mkB and m2v2 = h2/λ2 So, T = h2/2mkBλ2 = (6.63.10-34)2/(2x1.66.10-27x1.38.10-23x(0.191.10-9)2) T = 4.39.10-67/0.167.10-68 = 263 K Equivalent temperature of neutrons is 263 K

A simple cubic crystal is illuminated with x-rays of wavelength 0.09 nm at a glancing angle. The crystal is rotated and the angles at which Bragg reflection occurs are measured. Which set of crystal planes will give the smallest angle for first order reflection? If this angle is 8.9 degrees determine the spacing between these crystal planes, At what angle will first order reflection be obtained from the (110) planes? From equation nλ = 2d sinθ we will get smallest angle θ for largest separation d. a ; largest d is for smallest value of h2 + k2 + l2, h + k 2 + l2 i.e. for (100) planes. For these planes, separation d = lattice parameter a.

Largest d is found from

d hkl =

2

So, using λ = 2d sinθ with λ = 0.09 nm and θ = 8.9 degrees, we get d100 = a = 0.29 nm The separation of the (110) planes is given by a/√2 = 0.205 nm. Using λ = 2d sinθ again we find that for these planes sinθ = 0.219 and so θ110 = 12.7 degrees

The first three lines from the powder diffraction pattern of a cubic crystal have the following S values: 24.95, 29.11, and 40.9 mm. The camera radius is 57.3 mm. Molybdenum Kα radiation of wavelength 0.071 nm is used. Determine the structure and lattice parameter of the material.

The value of S/radius gives the angle 2θ in radians. The radius is given as 57.3 mm, so we can calculate the angles θi for the first three lines in the powder diffraction

sin θ =

We use the general formula

λ2

2

4a

2

(h

2

+k2 +l2)

and we look for some integer value for h2 + k2 + l2. From notes: We multiply the values of sin2θ by some constant value to give nearly integer values for all the h2 + k2 + l2 values. Integer values are then assigned. Radius is 57.3 mm S (mm) 24.9 29.11 40.9

2θ (rad) 0.434 0.508 0.714

θ (rad) 0.217 0.254 0.357

sinθ 0.215 0.251 0.349

sin2θ 0.046 0.063 0.122

multiply by ??

How do we choose the multiplier? First we try just the inverse of the first value of sin2θ as we know that will give us 1 for the first line: S (mm)

2θ (rad)

θ (rad)

sinθ

sin2θ

24.9 29.11 40.9

0.434 0.508 0.714

0.217 0.254 0.357

0.215 0.251 0.349

0.046 0.063 0.122

multiply by 1/0.046 i.e. by 21.74 1 1.34 .2.65

Now we must get a set of approximate integers instead of 1, 1.34, and 2.65. Multiply all values by 3 and we get the numbers in the second last column below.

S (mm)

2θ (rad)

θ (rad)

sinθ

sin2θ

24.9 29.11 40.9

0.434 0.508 0.714

0.217 0.254 0.357

0.215 0.251 0.349

0.046 0.063 0.122

Multiply by 3

multiply by 1/0.046 i.e. by 21.74 1 1.34 .2.65

3 4.02 7.95

h2 + k2 + l2 3 4 8

So, h2 + k2 + l2 takes the values 3, 4, 8 for the first three lines, and we see that these correspond to the (111), (200) and (220) reflections. These correspond to the pattern expected for an FCC crystal (check notes).

To get the lattice parameter use

sin θ = 2

λ2 4a

2

(h

2

+k +l 2

2

)

Given λ = 0.071 nm and taking the first line in the data 0.046 = ((0.071)2/4a2)x3. From this we get a2 = 0.082 and so a = 0.287 nm...