Question

Nickel has a face-centered cubic structure and has a density of 8.90 g/cm3. what is its atomic radius?

Answer 1

Atomic radius,
`r=1.24* 10^(-8)\ cm`

Step-by-step explanation:

It is given that,

Density of nickel, d = 8.9 g/cm³

The density of a unit cell is given by :

`d=(Z* A)/(N_0* a^3)`

Where

Z = no of atoms per unit cell

A = molar mass of an element in g/mol

N₀ = Avogadro's number

a = edge length

The edge length of FCC crystal is,
`a=\sqrt8r`.............(1)

r = atomic radius

Nickel has a FCC structure and for FCC, Z = 4

`a^3=(Z* A)/(N_0* d)`

For nickel A = 58.69 g/mol

`a^3=(4* 58.69)/(6.022* 10^(23)* 8.9)`

`a^3=4.38* 10^(-23)\ cm`

`a=3.52* 10^(-8)\ cm`

Atomic radius is,
`r=(a)/(\sqrt8)`

`r=(3.52* 10^(-8))/(\sqrt8)`

`r=1.24* 10^(-8)\ cm`

Hence, this is the required solution.

Answer 2

density = (mass/volume)

Mass = # atoms (in unit cell) x (mol / 6.022 x 10^23 atoms) x (58.69 g/mol)
The last number is the atomic mass of nickel

The number of atoms = 8(1/8) + 6(1/2) = 4

The volume (of the entire unit cell) = side^3 = (r x 8^(1/2))^3

6.84 g/cm3 = [4 atoms x (mol / 6.022 x 10^23 atoms) x (58.69 g/mol)] / [r x 8^(1/2)]^3

r = 1.36 x 10^-8 cm