Solid State
If a is the edge length, In BCC, the distance between two nearest atoms will be:
Number of atoms per unit cell in B.C.C. is:
In BCC, lattice atoms occupies the corners and body center of a cube.
Each corner has a contribution of $$\dfrac {1}{8}$$ and there are 8 corners in a cube. Effective no.of atoms in a cube = $$\dfrac {1}{8} \times 8 = 1$$
Also, a body centre will have contribution of 1 and each cube will have 1 body centre. So effective number of atoms in a cube= 1. So total number of atoms in BCC = 1+1 = 2.
So the answer is option C.
If a is the edge length, In BCC, the distance between two nearest atoms will be:
CsBr crystallizes in a body-centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being $$6.02\times 10^{23}$$ mol$$^{-1}$$, the density of CsBr is:
The radius of the largest sphere which fits properly at the centre of the edge of body centred cubic unit cell is:(Edge length is represented by '$$a$$')
An element has a body centered cubic (bcc) structure with a cell edge of $$288 pm$$. The atomic radius is
Body centred cubic lattice has a co-ordination number of:
An example of a body cube is:
The intermetallic compound $$LiAg$$ crystallizes in a cubic lattice in which both $$Li$$ and $$Ag$$ atoms have coordination numbers of $$8$$. To what crystal class does the unit cell belong?
Packing fraction in BCC lattice is
The low density of alkali metals is due to
What is the theoretical density of crystalline $$K$$?