Chapter 11.13, Problem 72AAP

To determine

The density of zinc telluride $\left(\text{ZnTe}\right)$ in grams per cubic centimeter.

Answer to Problem 72AAP

The density of zinc telluride $\left(\text{ZnTe}\right)$ in grams per cubic centimeter is $\text{4}\text{.09}\text{g}/{\text{cm}}^3$.

Explanation of Solution

Write the expression to calculate lattice constant of zinc telluride structure $\left(a\right)$.

 $a=\frac{4}{\sqrt{3}}\left(r_{{\text{Zn}}^{\text{2+}}}+R_{{\text{Te}}^{2-}}\right)$                                                                                                ...... (I)

Here, ionic radius of ${\text{Zn}}^{\text{2+}}$ ion is $r_{{\text{Zn}}^{\text{2+}}}$ and ionic radius of ${\text{Te}}^{2-}$ ion is $R_{{\text{Te}}^{2-}}$.

Write the expression to calculate mass of unit cell of zinc telluride $\left(m\right)$.

 $m=\frac{n_{{\text{Zn}}^{\text{2+}}}{M}_{{\text{Zn}}^{\text{2+}}}+n_{{\text{Te}}^{2-}}{M}_{{\text{Te}}^{2-}}}{N_A}$                                                                                    ...... (II)

Here, number of atoms and molar mass of ${\text{Zn}}^{\text{2+}}$ ion are $n_{{\text{Zn}}^{\text{2+}}}$ and $M_{{\text{Zn}}^{\text{2+}}}$, number of atoms and molar mass of ${\text{Te}}^{2-}$ ion are $n_{{\text{Te}}^{2-}}$ and $M_{{\text{Te}}^{2-}}$ and Avogadro's number is $N_A$.

Write the expression to calculate density of zinc telluride in grams per cubic centimeter $\left(\rho \right)$.

 $\begin{array}{c}\rho =\frac{m}{V}\\ =\frac{m}{a^3}\end{array}$                                                                                                                  ...... (III)

Here, volume of unit cell is $V$.

Conclusion:

Substitute $0.083\text{nm}$ for $r_{{\text{Zn}}^{\text{2+}}}$ and $0.211\text{nm}$ for $R_{{\text{Te}}^{2-}}$ in Equation (I).

 $\begin{array}{c}a=\frac{4}{\sqrt{3}}\left(0.083\text{nm}+0.211\text{nm}\right)\\ =0.679\text{nm}\\ =6.79\times {10}^{-8}\text{cm}\end{array}$

There will be 4 zinc ions and 4 tellurium ions present in a single unit cell of zinc telluride.

Substitute $4$ for $n_{{\text{Zn}}^{\text{2+}}}$, $65.38\text{g}/\text{mol}$ for $M_{{\text{Zn}}^{\text{2+}}}$, $4$ for $n_{{\text{Te}}^{2-}}$, $127.6\text{g}/\text{mol}$ for $M_{{\text{Te}}^{2-}}$ and $6.023\times {10}^{23}\text{ions}/\text{mol}$ for $N_A$ in Equation (II).

 $\begin{array}{c}m=\frac{\left(4\right)\left(65.38\text{g}/\text{mol}\right)+\left(4\right)\left(127.6\text{g}/\text{mol}\right)}{6.023\times {10}^{23}\text{ions}/\text{mol}}\\ =1.28\times {10}^{-21}\text{g}\end{array}$

Substitute $1.28\times {10}^{-21}\text{g}$ for $m$ and $6.79\times {10}^{-8}\text{cm}$ for $a$ in Equation (III).

 $\begin{array}{c}\rho =\frac{1.28\times {10}^{-21}\text{g}}{{\left(6.79\times {10}^{-8}\text{cm}\right)}^3}\\ =\text{4}\text{.09}\text{g}/{\text{cm}}^3\end{array}$

Thus, the density of zinc telluride $\left(\text{ZnTe}\right)$ in grams per cubic centimeter is $\text{4}\text{.09}\text{g}/{\text{cm}}^3$.