Question

An element having atomic mass 60 has face-centered cubic unit cells.The edge length of the unit cell is 400pm.Find the density of the element.

Answer 1

Answer:

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Answer 2

Given:

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

Atomic mass of the element = 60

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

To find:

Density of the element = ?

Formula to be used:

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

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

Calculation:

Unit cell mass = $Number of atoms in the unit cell\times  Mass of each atom$

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

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

Number of atom in fcc unit cell = 4

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

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

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

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

$\frac{4\times60}{6.023\times10^{23}}$$\times$$\frac{1}{64\times10^{-24}}$

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

Conclusion:

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

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