Question

Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3.165 A^o.

(a) What is the atomic radius of tungsten in this structure?

(b) Calculate the density of tungsten.

Short answer

a.The radius of body centered cubic structure of tungsten is \(1.370\;{A^o}\)

b.The density of tungsten is \(19.25\;{\rm{g}}/{\rm{mL}}\)

Step-by-step solution

Define the coordination number

a.The relationship between edge length and radius for body centred cubic structure is as follows:

\(a = \dfrac{4}{{\sqrt 3 }}r\). Here, \(a\) is edge length and \(r\) is radius

b.To calculate the density of tungsten, use the following formula

density \( = \dfrac{{Z \times M}}{{{N_o} \times {a^3}}}\). Here, \(Z\) is number of atoms present in the body centred cubic that is two. \({\rm{M}}\) is molar mass of metal; molar mass of tungsten is \(183.84\;{\rm{g}}/{\rm{mol}}\),\({N_o}\)is Avogadro's number that is \(6.023 \times {10^{23}}\)

Calculate the atomic radius of body centered cubic structure of tungsten. 

To calculate the radius of body-centered cubic structure of tungsten use the formula \(a = \dfrac{4}{{\sqrt 3 }}r\).

Here, \(a\) is the edge length and \(r\) is radius.

Calculate the radius of body-centered cubic structure of tungsten as follows:

\(a = \dfrac{4}{{\sqrt 3 }}r\)

\(r = \dfrac{{a\sqrt 3 }}{4}\)

\( = \dfrac{{3.165{A^o}\sqrt 3 }}{4}\)

\( = 1.370\;{A^o}\)

Calculate the density of tungsten.

To calculate density of tungsten, substitute edge length, Avogadro's number, number of atoms present in the body centred cubic and its molar mass is as follows:

\(\begin{aligned}{\rm{Density}} &= \dfrac{{Z \times M}}{{{N_o} \times {a^3}}}\\ &= \dfrac{{2 \times 183.84\;{\rm{g}}}}{{6.023 \times {{10}^{23}} \times {{\left( {3.165\;{A^o} \times \dfrac{{{{10}^{ - 8}}\;{\rm{cm}}}}{{1\;{A^o}}}} \right)}^3}}}\\ &= 19.25\;{\rm{g}}/{\rm{mL}}\end{aligned}\)

Hence, density of tungsten is \(19.25\;{\rm{g}}/{\rm{mL}}\).