Question

The density of a fcc element(atomic mass 60.2) is 6.25 g/cm3. Calculate the edge length of the unit cell *

Answer 1

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Answer 2

Given:

The atomic mass of the element = 60.2 g/mol

The density of the element = 6.25 g/cm³

To find:

The edge length of the unit cell

Solution:

We know that,

For an fcc element, the no. of atoms per unit cell = 4

To find the density of a unit cell, we will use the following formula:

$\boxed{\boxed{\bold{\rho = \frac{Z\times M}{a^3\times N_A} }}}$

where

$\rho$ → density (g/cm³)

Z → no. of atoms per unit cell

M → atomic mass (g/mol)

a → the edge length of the unit cell (cm)

$N_A$ → Avogadro's number = 6.023 × 10²³ atoms per mole

Now, rearranging the above formula and then substituting the given values in it we get,

$a^3 = \frac{Z\:\times\: M}{\rho\:\times \:N_A}$

$\implies a^3 = \frac{4\:\times\:60.2}{6.25\:\times\:6.023 \: \times\: 10^2^3 }$

$\implies a^3 = \frac{240.8}{37.643 \: \times\: 10^2^3 }$

$\implies a^3 = 6.4\times 10^-^2^3$

$\implies a^3 = 64\times 10^-^2^4$

taking cube root on both sides

$\implies a = \sqrt[3]{64\times 10^-^2^4}$

$\implies a = 4 \times 10^-^8\: cm$

$\implies a = 400 \times 10^-^1^0\:cm$

$\implies \bold{a = 400\:pm}$

Thus, the edge length of the unit cell is 400 pm.

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