Question

the density of zinc is 7.14 g/cm^3 and has an atomic radius of 1.332A. Calculate the height of its unit cell. the atomic weight is 65.37 grams/mole.

Answer 1

Answer : The height of the unit cell is, $4.947\AA$

Explanation : Given,

Density of zinc = $7.14g/cm^3$

Atomic radius = $1.332\AA$

Atomic weight of zinc = 65.37 g/mole

As we know that, zinc has haxagonal close packed crystal structure. The number of atoms in unit cell of HCP is, 6.

Formula used for density :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      .............(1)

where,

$\rho$ = density  of zinc

Z = number of atom in unit cell  = 6 atoms/unit cell (for HCP)

M = atomic mass

$(N_{A})$ = Avogadro's number  = $6.022\times 10^{23}atoms/mole$

a = edge length of unit cell

$a^3$ = volume of unit cell

Now put all the values in above formula (1), we get

$7.14g/cm^3=\frac{(6\text{ atoms per unit cell})\times (65.37g/mole)}{(6.022\times 10^{23}atoms/mole)\times a^{3}}$

$V=a^{3}=9.122\times 10^{-23}cm^3$

$V=9.122\times 10^{-29}m^3$

Now we have to calculate the height of the unit cell.

Formula used :

$V=6\sqrt {3}r^2h$

where,

V = volume of unit cell

r = atomic radius = $1.332\AA=1.332\times 10^{-10}m$

conversion used : $(1\AA=10^{-10}m)$

h = height of the unit cell

Now put all the given values in this formula, we get:

$9.122\times 10^{-29}m^3=6\sqrt {3}\times (1.332\times 10^{-10}m)^2h$

$h=4.947\times 10^{-10}m=4.947\AA$

Therefore, the height of the unit cell is, $4.947\AA$