An orthorrhombic material containing $Mn$ and $P$ has the following lattice parameters $a=5\mathrm{.}916b$

$=5\mathrm{.}260=3\mathrm{.}173$ and belongs to a spacial group with the symbols bnm$.$ Inside the

conventional cell there are $8$ atoms, $4Mn$ and $4P,$ situated at $($with respect to the origin of the cell$)$

$+-(u,v,\frac{1}{4})$ y $+-(\frac{1}{2}-u,v+\frac{1}{2},\frac{1}{4})$ where $u(Mn)=0\mathrm{.}20,v(Mn)=0\mathrm{.}005,u(P)=0\mathrm{.}57$ y $v(P)=0\mathrm{.}19.a$

Draw the conventional cell with all its atoms b$)$ Find the kind of latttice, basis and formula of the

material as well as its density $($atomic weights for Mn y P are, respectively, $55\mathrm{.}94$ y $30\mathrm{.}97).$ c$)$ Find

the point group from which the space group comes. d$)$ Plot the symmetry elements contained in the

indexes of the space group in $2$D projections of the cell.