Concept explainers
Interpretation:
The structure factors and the form of diffraction pattern under the given conditions has to be determined.
Concept Introduction:
The structure factor is given by
Fhkl= ∑ffj ei ϕhkl(j)
Where,
ϕhkl(j)= 2π(hxj+kyj+lzj)= Phase difference
fj= Scattering factor
fjei ϕhkl(j)= phase factor
Explanation of Solution
The location of the atoms in a bcc unit cell can be depicted as follows
Figure 1
Calculation of structure factor for bcc unit cell:
The atoms at the corner will give a weight of 18 and the atom at the body center will give a weight of 12. Consider all the atoms are identical.
Table for calculated values of phase difference:
AtomWeightxyzϕhkl11/8000021/80102πk31/81102π(h+k)41/81002πh51/81012π(h+l)61/81112π(h+k+l)71/80112π(k+l)81/80012πl91/21/21/21/22π(12h+12k+12l)
Calculation of phase factor:
For corner atoms:
fj ei ϕhkl(j)= fA(ei .0 +ei .2πk +ei .2π(h+k)+ei .2πh + ei .2π(h+l)+ ei .2π(h+k+l)+ ei .2π(k+l)+ei .2πl)
The phase factor for the corner atoms is +1 and as they have a weight of 18, the total contribution to the structure factor is fA.
For the atom at body center,
fj ei ϕhkl(j)= fB[ei .2π(12h+12k+12l)]= fB[ei π(h+k+l)]
Hence, the structure factor is given by
Fhkl= fA+ fB ei π(h+k+l)=fA+fB[(−1)(h+k+l)]
If (h+k+l)= even, Fhkl= fA+fB
If (h+k+l)= odd, Fhkl=fA
Diffraction pattern:
When,
(a) fA=f, fB=0: fB=12fAFhkl= fA+ fB ei π(h+k+l)=fA+fB[(−1)(h+k+l)]=f
This implies that there are no systematic absences.
(b) fB=12fA: Fhkl= fA+ fB ei π(h+k+l)=fA+12fA[(−1)(h+k+l)]=fA[1+12(−1)(h+k+l)]
When h+k+l is odd, Fhkl=fA(1−12)=12fA
When h+k+l is even, Fhkl=fA(1+12)=32fA
That is, there is an alternation of intensity as (I α F2hkl) according to whether h+k+l is odd or even.
(c) fA=fB=f: Fhkl= fA+ fB ei π(h+k+l)=f+f[(−1)(h+k+l)]=f[1+ (−1)(h+k+l)]
If h+k+l is odd, Fhkl= 0. Thus, all h+k+l odd lines are missing.
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Chapter 15 Solutions
PHYSICAL CHEMISTRY. VOL.1+2 (LL)(11TH)
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