In bcc lattice why XRD 100 peak is not observed ?

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SUMMARY

The absence of the XRD (100) peak in body-centered cubic (BCC) lattices is due to destructive interference caused by the arrangement of atoms within the unit cell. The structure factor, calculated using the formula S(hkl)=f{1+exp[-iπ(h+k+l)]}, reveals that S equals zero when h+k+l is an odd integer, leading to the non-observation of certain planes, including (100), (300), (111), and (221). In contrast, reflections such as (200), (110), and (222) are observed due to their even integer sums. The phase cancellation occurs because the intervening atomic planes have equal scattering power, resulting in a net zero contribution for the (100) reflection.

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Can someone tell me that in bcc lattice why xrd 100 peak does not appear?
 
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The planes passing through the base atoms are a/2 away from the planes passing through the planes defined by the atoms on the edge of the cell. And is only a shifted copy of the latter. Hence it will reflect x rays exactly as does one of the edge planes and you get destructive interference.
 
If you calculate the structure factor of the BCC lattice using:
$$S(hkl)=f{1+exp[-i\pi(h+k+l)]}$$, where f= form factor. The value o f S is zero when the exponential has the value -1, which is whenever the argument is: -i*pi* (odd integer).
So we have:
S=0 when h+k+l= odd integer
S=2f when h+k+l= even integer
So therefore the diffraction pattern does not contain these planes: (100), (300), (111), (221)
But it occurs for fx.: (200), (110), (222)

The (100) reflection occurs when reflections from the planes that bound the cubic cell differ in phase by 2*pi. In BCC there is an intervening plane of atoms, which is equal to in scattering power to the other planes. Situated between them it gives a reflection retarded in phase by pi w.r.t. the first plane, thereby cancelling the contribution from that plane. So the cancellation occurs since the planes are identical in composition.
 

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