Answer
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Hint: The cubic unit cell is the smallest repeating unit where all the angles are 90 degrees and all lengths are equal with each axis (x, y, z). The unit cell is of three types: simple cubic cell, body-centered cubic cell and Face-centered cubic. In a cubic unit cell, 8 atoms are present in each corner of the cell and the atom is shared by the 8 neighboring cells. In this question we need to find the number of atoms by which we can find the nature of the cubic unit cell.
Complete step by step answer:
It is given that the molar mass of the element is 27 $gmo{l^{ - 1}}$, the edge length of the cell is $4.05 \times {10^{ - 8}}$cm. The density is 2.7 $gc{m^{ - 3}}$.
27 $gmo{l^{ - 1}}$= 0.027 Kg
$4.05 \times {10^{ - 8}}$cm = $4.05 \times {10^{ - 10}}$m.
2.7 $gc{m^{ - 3}}$= $2.7 \times {10^{ - 3}}Kg{m^{ - 3}}$
The formula relating the edge length, density, molar mass is shown below.
$d = \dfrac{{Z \times M}}{{{a^3} \times {N_A}}}$
Where,
d is the density
Z is the number of atoms in the unit cell.
M is the molar mass
a is the edge length
${N_A}$ is the Avagadro’s number
Here, we have to calculate the number of atoms present in the cell, so the formula is rearranged as shown below.
$Z = \dfrac{{d \times {a^3} \times {N_A}}}{M}$
To calculate the number of atoms, substitute the values in the above equation.
$\Rightarrow Z = \dfrac{{(2.7 \times {{10}^3}){{(4.05 \times {{10}^{ - 10}})}^3}(6.022 \times {{10}^{23}})}}{{0.027}}$
$\Rightarrow Z = 4$
The number of atoms present in the cubic unit cell is 4.
As the number of atoms is 4, the given cubic cell is a face-centered cubic cell (FCC) or cubic close packed structure (CCP).
Note:
Make sure to convert the molar mass given in gram per mole into kilogram per mole, the value of edge length given in centimeters into meters, and the value of density given in gram cubic centimeter into kilogram cubic meter.
Complete step by step answer:
It is given that the molar mass of the element is 27 $gmo{l^{ - 1}}$, the edge length of the cell is $4.05 \times {10^{ - 8}}$cm. The density is 2.7 $gc{m^{ - 3}}$.
27 $gmo{l^{ - 1}}$= 0.027 Kg
$4.05 \times {10^{ - 8}}$cm = $4.05 \times {10^{ - 10}}$m.
2.7 $gc{m^{ - 3}}$= $2.7 \times {10^{ - 3}}Kg{m^{ - 3}}$
The formula relating the edge length, density, molar mass is shown below.
$d = \dfrac{{Z \times M}}{{{a^3} \times {N_A}}}$
Where,
d is the density
Z is the number of atoms in the unit cell.
M is the molar mass
a is the edge length
${N_A}$ is the Avagadro’s number
Here, we have to calculate the number of atoms present in the cell, so the formula is rearranged as shown below.
$Z = \dfrac{{d \times {a^3} \times {N_A}}}{M}$
To calculate the number of atoms, substitute the values in the above equation.
$\Rightarrow Z = \dfrac{{(2.7 \times {{10}^3}){{(4.05 \times {{10}^{ - 10}})}^3}(6.022 \times {{10}^{23}})}}{{0.027}}$
$\Rightarrow Z = 4$
The number of atoms present in the cubic unit cell is 4.
As the number of atoms is 4, the given cubic cell is a face-centered cubic cell (FCC) or cubic close packed structure (CCP).
Note:
Make sure to convert the molar mass given in gram per mole into kilogram per mole, the value of edge length given in centimeters into meters, and the value of density given in gram cubic centimeter into kilogram cubic meter.
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