A closed packed structure of uniform spheres has the edge length of $$534$$ pm. Calculate the radius of sphere in pm if it exists in simple cubic lattice.
In a simple cubic lattice, we have atoms only at the corners of the lattice.
If we compare the edge length (a) to radius (r), we get ratio of $$2:1$$.
The two atoms joins each other along the edge of a lattice.
So, the answer is $$\dfrac { a }{ 2 } =r$$
Hence, the radius of sphere is $$267$$ pm.
A solid made up of ions of A and B which possesses edge length of unit cell equal to $$0.564$$ nm has four formula units. Among the two ions, the smaller one occupies the interstitial void and the larger ions occupy the space lattice with ccp type of arrangement. One molecule of solid has mass $$ 9.712\times10^{-23}$$ g. The ionic radius (in $$\mathring A$$) for $$B^{-}$$ ion assuming anion - anion contact, is :
A crystal has SCC structure and its lattice constant is 3.5$$\overset{\circ}{A}$$. What is the atomic raidus?
A given metal crystallizes out with a cubic structure having edge length of $$361 $$ pm. If there are four metal atoms in one unit cell, what is the radius of one atom?
An ionic solid $$A^+B^-$$ crystallizes as a body centred cubic structure. The distance between cation and anion in the lattice is 338 pm. Calculate the edge length of the unit cell.
The edge length of a body centred cubic unit cell is 390 pm. If the radius of the cation is 150 pm, what if the radius of the anion?
The length of the unit cell edge of a body-centred cubic metal crystal is 352 pm. Calculate the radius of an atom of the metal.
What will be the radius of the simple cubic lattice if it has the edge length of $$534$$ pm?
The edge length of the unit of $$LiCl(NaCl-$$ like structure) is $$514pm$$. Assuming that the lithium-ion is small enough that the chloride ions is small enough that the chloride ions are in contact, the ionic radius of chloride ion (in pm) is $$\left( \cfrac { 1 }{ \sqrt { 2 } } =0.7 \right) $$