Question

Give the formula for the density of a crystal whose length of the edge of the unit cell is known?

Answer 1

Answer:

The density of a Unit Cell  =   $\frac{n\times M}{a^{3} \times N_A}$

Explanation:

A unit cell also has a cubic structure, as we are aware. It has one, two, or four atoms scattered across the lattice.

Density = ( Mass of Unit Cell ) / ( Volume of Unit Cell )

$\implies$ Mass of Unit Cell:-

We multiply the number of atoms, "n," by the average mass, "m," to get the mass of a unit cell.

Mass of Unit Cell = m × n

The Avogadro Number can be used to calculate the mass of one atom

($N_A$).

Atomic Mass = ( Molar Mass ) /  ( Avogadro Number )

                      =  $M/N_A$

$\therefore$  Mass of Unit Cell =  $n\times (M/N_A)$

$\implies$ Volume of unit Cell:-

The unit cell is a cubic structure. Assume that the cube's side length is "a."

Unit Cell Volume = $a^3$

$\implies$ Density of a Unit Cell = $\frac{n\times M}{a^{3} \times N_A}$

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Answer 2

The Density of the cell =  $\frac{nM}{a^{3} Ma}$

Explanation:

the formula for the density of a crystal whose length of the edge of the unit cell is known

Density of a unit cell= $Densitiy of unit cell =\frac{mass of unit cell}{volume of unit cell}$----------------1

Mass of unit cell = Total number of atoms belonging to the unit cell mass of one atom -------- 2

Mass of one atom = $\frac{molar mass}{Avogadros number}$  ---------3

When we substitute the value of eqn 3 in eqn 2 we get

Mass of unit cell = $\frac{nM}{Ma}$

For  a cubic unit cell,edge lengths are equal for all side so a = b = c  and The Volume of the unit cell = a x a x a = a3

Substitute this value in eqn1

Density of the cell =  $\frac{nM}{a^{3} Ma}$

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