SIMPLE CUBIC .                   FCC                                  BCC

UNIT CELL side A                                                         2r                                             4r/sqrt2                               4r/sqrt3

FACE DIAGONAL                                                         SQRT 2 (2r)                            4r                                       sqrt (2/3)4r

body diagonal                                                       sqrt3 (2r)                                           sqrt (3/2)4r                      4r

NO OF ATOMS                                                              1                                                 4                                              2

cell VOLUME                                                                 8r ^3                                          (32r^3)/sqrt 2     64r^3/3sq3

here r = atomic radius. 

no of atoms in hcp = 6

co ordination number of  HCP , FCC = 12            

co ordination number of bcc                =  8

atomic radius of hcp  r = a/2

atomic radius of simple cubic r = a/2

BCC atomic radius = asqrt(3/4)

fcc   atomic radius = a sqrt2/4

IN FCC , HCP  LATTICES  EMPTY SPACE  = 26%

IN BCC  LATTICES EMPTY SPACE  = 32 %

TOTAL NO OF ATOMS PER UNIT CELL IS GIVEN BY 

                                                     N = Nt + Nf / 2 +Nc/8

 where Nt  = np of interior atoms   , Nf  = no of atoms on faces  , Nc  = no of atoms on corners  .

i hope these wiLL help u !!!!!!!!!!!! I WISH U ALL THE BEST !!!!!!