
The density of copper metal is 8.95 g$\text{c}{{\text{m}}^{-3}}$, if the radius of copper atom is 127.8 pm is the copper r unit cell a simple cubic, a body centered cubic or a face centered cubic structure?( At mass of Cu = 63.54 g/mole and ${{\text{N}}_{a}}$ = 6.02 × ${{10}^{23}}$ $\text{mol}{{\text{e}}^{-1}}$)
Answer
447.3k+ views
Hint: We need to first find the densities of copper metal taking the different values of Z and r according to simple cube, bcc and fcc. For simple cube Z = 1 and a = 2r , for bcc Z = 2 and a = $\dfrac{\text{4r}}{\sqrt{3}}$ and for fcc Z = 4 and a = $\dfrac{\text{4r}}{\sqrt{2}}$ and then the by applying the formula for density = $\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
, we can find their densities and after that we can easily know whether the copper unit cell is simple cube, bcc or fcc.
Complete step by step answer:
To know about the unit cell of copper, we have to find the densities of copper metal in the simple cube, bcc, and fcc.
So, for simple cubic, we know that Z=1 and a=2r
r= 127.8 pm (given)
then,
\[\begin{align}
& \Rightarrow a=2\times 127.8\\
& \Rightarrow \text{ }=\text{ }255.6pm \\
& \Rightarrow \text{ }=2.55\times {{10}^{-12}}m\text{ }(1m\text{ }={{10}^{12}}pm) \\
& \Rightarrow \text{ }=2.55\times {{10}^{-8}}cm\text{ }(1m=100cm) \\
\end{align}\]
As we know that, density =$\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
Atomic mass of Cu=63.54 $\text{gmol}{{\text{e}}^{-1}}$ (given)
${{\text{N}}_{a}}$= $6.02\times ~{{10}^{23}}\text{mol}{{\text{e}}^{-1}}$ (given), then;
density = $\dfrac{1\times 63.5}{{{(2.55)}^{3}}\times {{10}^{-24}}\times 6.023\times {{10}^{23}}}$
=$\dfrac{1\times 63.54\times 10}{16.58\times 6.023}$
=$\dfrac{635.4}{99.86}$
=6.36 g$\text{c}{{\text{m}}^{-3}}$
So, the density of a simple cube is not the same as that of the density of copper metal given in the statement. So, the copper (Cu) unit cell is not a simple cube.
For body centered, we know that Z=2 and a=$\dfrac{\text{4r}}{\sqrt{3}}$
r= 127.8 pm (given)
Then,
\[\begin{align}
& \Rightarrow a=\dfrac{4\times 127.8}{1.732}\\
& \Rightarrow \text{ } = \text{ }295.15pm\\
& \Rightarrow \text{ } =295.15\times {{10}^{-12}}m\text{ }(1m\text{ }={{10}^{12}}pm) \\
& \Rightarrow \text{ } = 2.95\times {{10}^{-8}}cm\text{ }(1m = 100cm) \\
\end{align}\]
As we know that, density =$\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
Atomic mass of Cu = 63.54 $\text{gmol}{{\text{e}}^{-1}}$ (given)
${{\text{N}}_{a}}$ = $6.02\times ~{{10}^{23}}\text{mol}{{\text{e}}^{-1}}$(given), then
Density = $\dfrac{2\times 63.5}{{{(2.95)}^{3}}\times {{10}^{-24}}\times 6.023\times {{10}^{23}}}$
=$\dfrac{2\times 63.54\times 10}{25.67\times 6.023}$
=$\dfrac{1270.8}{154.61}$
=8.21 $g \text{c}{{\text{m}}^{-3}}$
So, the density of body centered cube (bcc) is nearly equal to that of the density of copper metal given in the statement but not the same. So, the copper (Cu) unit cell is not a body centered cube.
For face centered, we know that Z=4 and a=$\dfrac{\text{4r}}{\sqrt{2}}$
r= 127.8 pm (given)
Then,
\[\begin{align}
& \Rightarrow a=\dfrac{4\times 127.8}{1.414}\\
& \Rightarrow \text{ }=\text{ 361}\text{.52}pm \\
& \Rightarrow \text{ }=361.52\times {{10}^{-12}}m\text{ }(1m\text{ }={{10}^{12}}pm) \\
& \Rightarrow \text{ }=3.61\times {{10}^{-8}}cm\text{ }(1m=100cm) \\
\end{align}\]
As we know that, density =$\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
Atomic mass of Cu=63.54 $\text{gmol}{{\text{e}}^{-1}}$ (given)
${{\text{N}}_{a}}$=$6.02\times ~{{10}^{23}}\text{mol}{{\text{e}}^{-1}}$(given), then
Density = $\dfrac{4\times 63.5}{{{(3.61)}^{3}}\times {{10}^{-24}}\times 6.023\times {{10}^{23}}}$
=$\dfrac{4\times 63.54\times 10}{47.04\times 6.023}$
=$\dfrac{2540}{282}$
=8.96 g$\text{c}{{\text{m}}^{-3}}$
The density of face centered is same as that of the density of copper metal given in the statement.
Hence, the copper (Cu) unit cell is face centered cubic(fcc).
Note: Don’t get confused in the a and r, a is the edge length (i.e. edge means the line segment where the two edges meet and the cube have 12 edges and the length of one edge to another is called as the edge length) and r is the radius of the cube. If we know, edge length and density, we can calculate radius from the density formula i.e. $\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$ and vice versa.
, we can find their densities and after that we can easily know whether the copper unit cell is simple cube, bcc or fcc.
Complete step by step answer:
To know about the unit cell of copper, we have to find the densities of copper metal in the simple cube, bcc, and fcc.
So, for simple cubic, we know that Z=1 and a=2r
r= 127.8 pm (given)
then,
\[\begin{align}
& \Rightarrow a=2\times 127.8\\
& \Rightarrow \text{ }=\text{ }255.6pm \\
& \Rightarrow \text{ }=2.55\times {{10}^{-12}}m\text{ }(1m\text{ }={{10}^{12}}pm) \\
& \Rightarrow \text{ }=2.55\times {{10}^{-8}}cm\text{ }(1m=100cm) \\
\end{align}\]
As we know that, density =$\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
Atomic mass of Cu=63.54 $\text{gmol}{{\text{e}}^{-1}}$ (given)
${{\text{N}}_{a}}$= $6.02\times ~{{10}^{23}}\text{mol}{{\text{e}}^{-1}}$ (given), then;
density = $\dfrac{1\times 63.5}{{{(2.55)}^{3}}\times {{10}^{-24}}\times 6.023\times {{10}^{23}}}$
=$\dfrac{1\times 63.54\times 10}{16.58\times 6.023}$
=$\dfrac{635.4}{99.86}$
=6.36 g$\text{c}{{\text{m}}^{-3}}$
So, the density of a simple cube is not the same as that of the density of copper metal given in the statement. So, the copper (Cu) unit cell is not a simple cube.
For body centered, we know that Z=2 and a=$\dfrac{\text{4r}}{\sqrt{3}}$
r= 127.8 pm (given)
Then,
\[\begin{align}
& \Rightarrow a=\dfrac{4\times 127.8}{1.732}\\
& \Rightarrow \text{ } = \text{ }295.15pm\\
& \Rightarrow \text{ } =295.15\times {{10}^{-12}}m\text{ }(1m\text{ }={{10}^{12}}pm) \\
& \Rightarrow \text{ } = 2.95\times {{10}^{-8}}cm\text{ }(1m = 100cm) \\
\end{align}\]
As we know that, density =$\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
Atomic mass of Cu = 63.54 $\text{gmol}{{\text{e}}^{-1}}$ (given)
${{\text{N}}_{a}}$ = $6.02\times ~{{10}^{23}}\text{mol}{{\text{e}}^{-1}}$(given), then
Density = $\dfrac{2\times 63.5}{{{(2.95)}^{3}}\times {{10}^{-24}}\times 6.023\times {{10}^{23}}}$
=$\dfrac{2\times 63.54\times 10}{25.67\times 6.023}$
=$\dfrac{1270.8}{154.61}$
=8.21 $g \text{c}{{\text{m}}^{-3}}$
So, the density of body centered cube (bcc) is nearly equal to that of the density of copper metal given in the statement but not the same. So, the copper (Cu) unit cell is not a body centered cube.
For face centered, we know that Z=4 and a=$\dfrac{\text{4r}}{\sqrt{2}}$
r= 127.8 pm (given)
Then,
\[\begin{align}
& \Rightarrow a=\dfrac{4\times 127.8}{1.414}\\
& \Rightarrow \text{ }=\text{ 361}\text{.52}pm \\
& \Rightarrow \text{ }=361.52\times {{10}^{-12}}m\text{ }(1m\text{ }={{10}^{12}}pm) \\
& \Rightarrow \text{ }=3.61\times {{10}^{-8}}cm\text{ }(1m=100cm) \\
\end{align}\]
As we know that, density =$\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
Atomic mass of Cu=63.54 $\text{gmol}{{\text{e}}^{-1}}$ (given)
${{\text{N}}_{a}}$=$6.02\times ~{{10}^{23}}\text{mol}{{\text{e}}^{-1}}$(given), then
Density = $\dfrac{4\times 63.5}{{{(3.61)}^{3}}\times {{10}^{-24}}\times 6.023\times {{10}^{23}}}$
=$\dfrac{4\times 63.54\times 10}{47.04\times 6.023}$
=$\dfrac{2540}{282}$
=8.96 g$\text{c}{{\text{m}}^{-3}}$
The density of face centered is same as that of the density of copper metal given in the statement.
Hence, the copper (Cu) unit cell is face centered cubic(fcc).
Note: Don’t get confused in the a and r, a is the edge length (i.e. edge means the line segment where the two edges meet and the cube have 12 edges and the length of one edge to another is called as the edge length) and r is the radius of the cube. If we know, edge length and density, we can calculate radius from the density formula i.e. $\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$ and vice versa.
Recently Updated Pages
Master Class 12 Social Science: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
The gas that burns in oxygen with a green flame is class 12 chemistry CBSE

Most of the Sinhalaspeaking people in Sri Lanka are class 12 social science CBSE

And such too is the grandeur of the dooms We have imagined class 12 english CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What I want should not be confused with total inactivity class 12 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
