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=(Z* M)/(N_(A)* 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* 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})* (65.37g/mole)}{(6.022* 10^(23)atoms/mole)* a^(3)}`

`V=a^(3)=9.122* 10^(-23)cm^3`

`V=9.122* 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* 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* 10^(-29)m^3=6\sqrt {3}* (1.332* 10^(-10)m)^2h`

`h=4.947* 10^(-10)m=4.947\AA`

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