S
OLIDS
TATE ANDS
URFACEC
HEMISTRY(
LECTURE4)
Dr. Saedah Rwede Al-Mhyawi Assistant professor
O
BJECTIVES By the end of this section you should:
understand the concept of diffraction in crystals
be able to derive and use Bragg’s law
B
RAGG’
SL
AW The Braggs were awarded the Nobel Prize in physics in 1915 for their work in determining crystal structures
beginning with NaCl, ZnS and diamond.
They developed a relationship to explain
why the cleavage faces of crystals appear to reflect X-ray beams at certain angles of
incidence X-ray beam .
X-
RAYS X-rays are formed when electrons are accelerated
through an electric field, and then crashed into a metal target.
The sudden stopping, or breaking, of the electrons gives rise to the x-ray emission.
Diffraction pattern formed on the screen OBSTACLE
INCOMING WAVE
SCREEN
Constructive interference occurs only when
n λ = AB + BC AB=BC
n λ = 2AB Sinθ =AB/d
AB=d sin θ n λ =2dsin θ
D
ERIVINGB
RAGG’
SL
AWλ = 2d
hklsin θ
hklB
RAGG’
SL
AW The relationship describing the angle at which a beam of X-rays of a particular wavelength diffracts from a crystalline surface
WHERE
n: order of reflection λ : wavelength of X-ray
θ : angle of reflection
d: distance between tow layer
I
NDEXING AC
UBICP
ATTERN Bragg’s Law tells us the location of a peak with indices hkl, θhkl, is related to the interplanar spacing, d, as follows:
λ = 2dhkl sin θhkl 1/d = 2 sin θ/ λ 1/d2 = 4 sin2 θ/ λ2
Earlier we saw that for a cubic phase the d-
values can be calculated from the Miller indices (hkl)
1/d2 = (h2 + k2 + l2)/a2
Combining these two equations we get the following relationship
sin2 θ /(h2 + k2 + l2) = λ2 /4 a2
Need to find values of h,k,l for that give a constant when divided by each sin2 θ.
X rays of wavelength 0.154 nm are diffracted from a crystal at an angle of 14.170. Assuming that n = 1, what is the distance (in pm) between layers in the crystal?
nl = 2d sin q n = 1 q = 14.170
l = 0.154 nm = 154 pm
d = nl
2sinq = 1 x 154 pm
2 x sin14.17 = 314.0 pm
In the figure below we show the MEASURED variation of the reflected X-ray intensity as a function of incident angle for a crystalline sample of SILICON CARBIDE. A major peak is
observed at approximately 2q = 36º and for the incident wavelength of 1.9 Å, what inter-plane spacing it corresponds to?
Solution: According to Bragg’s law,
1 . 18 3
sin 2
9 . 1 sin
sin 2
2
q
l l
q d
d
Example
Example
8 10
4 2
10 1
sin 2 sin
2
1010
n
d n n d
n l q q l
n =
1 2 3 4 5 6 7 8
q 7.2º 14.5º 22.0º 30.0º 38.7º 48.6º 61.0º 90º
A beam of X-rays with a wavelength of 1 Å is directed at a sample whose crystal planes have a spacing of 4 Å. How many diffraction peaks can be observed as the sample orientation is rotated with
respect to the beam?
Solution: From the formula below we note that the largest allowed value of n is 8 so that EIGHT different peaks will be observed in experiment
Sinθ
-(n/8)
Consider the case where X-rays of wavelength l are incident on a set of crystal planes of spacing d. What is the LARGEST
possible value of l for which diffraction by these planes can be observed
Solution: The largest value of l when n = 1 and sin(q) = 1. This yields a corresponding MAXIMUM value for l
d 1 2 d
sin 2
l l
q
THE LARGEST ALLOWED VALUE OF l IS TWICE THE INTER-ATOMIC SPACING!
Example
E
XAMPLE A crystal of table salt (NaCl) is radiated with X- rays of wavelength 0.3 nm. The first Bragg peak is observed at an angle (q) of 32.13º. Calculate the atomic spacing between the planes in NaCl.
nl = 2d sin q
d = nl
2sinq = 1 x 0.3 nm
2 x sin32.13 = 0.282 nm
E
XAMPLE A piece of Al metal is irradiated with X-rays of wavelength 0.25 nm (d=0.405 nm). Calculate the angles for the first two peaks.
nλ = 2d sin θ
sin θ = nλ/2d
sin θ = 1 × 0.25 /2 × 0.405 = 0.0054◦
sin θ = 2 × 0.25 /2 × 0.405 = 39.12◦
