Question

A metallic element has a density of 7.15 g cm-3
, a lattice constant of 2.880 Å and
an atomic weight of 51.9961 u. Predict the lattice crystal structure for this element

Answer 1

volume of the lattice = $a^3$ = $(2.880\times10^{-8})^3 cm^3 =$ $23.9*10^{-24} cm^3$

let the number of atoms per unit cell be $x$

then mass of one unit cell $= density\times volume =$ $7.15\times23.9\times10^{-24}=17*10^{-23} g$

number of unit cell = $\dfrac{mass \ of \ one \ unit \ cell}{mass \ of \ one \ atom }$ $=\dfrac{17*10^{-23}}{1.66*10^{-24}*52}=2$

From this we can say that the element has body centered cubic crystal