Question

An element having adormic mass 60 has face centred
cubic unit cells. The edge length of the unit cell is
400 pm. Find out density of the element.​

Answer 1

Explanation:

Edge length of unit cell = 400pm = 400\times 10^{-10}400×10

−10

cm

Atomic mass of the element = 60

Volume of unit cell = (400\times 10^{-10})^3 = 64\times 10^{-24}cm^3(400×10

−10

)

3

=64×10

−24

cm

3

To find:

Density of the element = ?

Formula to be used:

Mass of 1 atom = \frac{Atomic mass}{Avogadro number}Massof1atom=

Avogadronumber

Atomicmass

Density = \frac{Mass}{Volume}Density=

Volume

Mass

Calculation:

Unit cell mass =

Number of atoms in the unit cell = \frac{Unit cell mass}{Mass of each atom}

Massofeachatom

Unitcellmass

Number of atom in fcc unit cell = 8\times\frac{1}{8}+6\times\frac{1}{2}8×

8

1

+6×

2

1

Number of atom in fcc unit cell = 4

Mass of one atom = \frac{Atomic mass}{Avogadro number}

Avogadronumber

Atomicmass

Mass of one atom = \frac{60}{6.023\times10^{23}}

6.023×10

23

60

Mass of four atom = \frac{4\times60}{6.023\times10^{23}}

6.023×10

23

4×60

Density of unit cell = \frac{Mass}{Volume}

Volume

Mass

\frac{4\times60}{6.023\times10^{23}}

6.023×10

23

4×60

\times× \frac{1}{64\times10^{-24}}

64×10

−24

1

Density of unit cell = 6.2gcm^{-3}6.2gcm

−3

Conclusion:

The density of the element is found to be 6.2gcm^{-3}6.2gcm

−3

Answer 2

Explanation:

A fcc element (atomic mass = 60 ) has a cell edge of 400 pm