Chapter 15, Problem 15B.3P

Interpretation Introduction

Interpretation:

The unit cell of silver, the unit cell dimension and mass density of silver has to be determined.

Concept Introduction:

Bragg’s law:

  $n\lambda =2d_{hkl}\sin \theta$

Where,

$\lambda =$ Wavelength of incident light

$d_{hkl}=$ Interplanar distance

$2\theta =$ Glancing angle

$n=$ Order of diffraction

  Density $=\frac{Z\times m}{V\times N_A}$

Where,

$Z=$ Formula unit

$V=$ volume of one unit cell

$N_A=$ Avogadro’s no.

$m=$ Mass of a substance per mole

Answer to Problem 15B.3P

The unit cell of silver, the unit cell dimension and mass density of silver has been found out to be FCC, $408.5518pm$ and $10.5081g/c{m}^3$ respectively.

Explanation of Solution

Given data:

  $\begin{array}{l}\theta ={19.076}^{\circ },{22.171}^{\circ },{32.256}^{\circ }\\ \\ \lambda =154.18pm\end{array}$

Expected reflections:

  $\begin{array}{ccc}\text{Bravis}\text{lattice}& \text{Reflections}\text{present}\text{for}& \text{Reflections}\text{absent}\text{for}\\ \text{Primitive}\text{simple}\text{cubic}& \begin{array}{l}\text{All}\\ \text{Ex}\text{.}\left(\text{100}\right)\text{,}\left(\text{110}\right)\text{,}\left(\text{111}\right)\text{etc}\text{.}\end{array}& \text{None}\\ \text{BCC}& \begin{array}{l}\text{h+k+l=even}\\ \text{Ex}\text{.}\left(\text{110}\right)\text{,}\left(\text{200}\right)\text{,}\left(\text{211}\right)\text{etc}\text{.}\end{array}& \text{h+k+l=odd}\\ \text{FCC}& \begin{array}{l}\text{h,k}\text{and}\text{l=}\text{unmixed}\left(\text{all}\text{even}\text{or}\text{odd}\right)\\ \text{Ex}\text{.}\left(\text{111}\right)\text{,}\left(\text{200}\right)\left(\text{220}\right)\text{etc}\text{.}\end{array}& \text{h,k}\text{and}\text{l=}\text{mixed}\end{array}$

According to Bragg’s relationship,

  $\begin{array}{l}n\lambda =2d_{hkl}\sin \theta \\ \\ d_{hkl}=\frac{n\lambda }{2\sin \theta }=\frac{154.18pm}{2\times \sin \left({19.076}^{\circ }\right)}=235.8775pm\left[bytak\mathrm{in}gn=1\right]\\ \\ d_{hkl}=\frac{n\lambda }{2\sin \theta }=\frac{154.18pm}{2\times \sin \left({22.171}^{\circ }\right)}=204.2811pm\\ \\ d_{hkl}=\frac{n\lambda }{2\sin \theta }=\frac{154.18pm}{2\times \sin \left({32.256}^{\circ }\right)}=144.4436pm\end{array}$

This calculation is according to given data in the question.

Expected $d_{hkl}$ for simple cubic:

  $\begin{array}{l}n\lambda =2d_{hkl}\sin \theta \\ \\ d_{100}=\frac{a}{\sqrt{1^2+0^2+0^2}}=a=235.8775pm\\ \\ d_{110}=\frac{a}{\sqrt{1^2+1^2+0^2}}=\frac{a}{\sqrt{2}}=166.7905pm\\ \\ d_{111}=\frac{a}{\sqrt{1^2+1^2+1^2}}=\frac{a}{\sqrt{3}}=136.1839pm\end{array}$

But it is not matching with the calculated value.

Expected $d_{hkl}$ for BCC:

  $\begin{array}{l}n\lambda =2d_{hkl}\sin \theta \\ \\ d_{110}=\frac{a}{\sqrt{1^2+1^2+0^2}}=\frac{a}{\sqrt{2}}=235.8775pm\\ \\ d_{200}=\frac{a}{\sqrt{2^2+0^2+0^2}}=\frac{a}{2}=\frac{\sqrt{2}a}{2}=\frac{\sqrt{2}\times 235.8775pm}{2}=166.7891pm\\ \\ d_{211}=\frac{a}{\sqrt{2^2+1^2+1^2}}=\frac{a}{\sqrt{6}}=\frac{\sqrt{2}a}{\sqrt{6}}=136.1839pm\end{array}$

But it is not matching with the calculated value.

Expected $d_{hkl}$ for FCC:

  $\begin{array}{l}n\lambda =2d_{hkl}\sin \theta \\ \\ d_{111}=\frac{a}{\sqrt{1^2+1^2+1^2}}=\frac{a}{\sqrt{3}}=235.8775pm\\ \\ d_{200}=\frac{a}{\sqrt{2^2+0^2+0^2}}=\frac{a}{2}=\frac{\sqrt{3}a}{2}=\frac{\sqrt{3}\times 235.8775pm}{2}=204.2759pm\\ \\ d_{220}=\frac{a}{\sqrt{2^2+2^2+1^2}}=\frac{a}{3}=\frac{\sqrt{3}a}{3}=136.1839pm\end{array}$

It is matching with the calculated value.

Hence, the cubic unit cell of silver is FCC and the cell dimension is $\left(\sqrt{3}\times 235.8775pm\right)$$408.5518pm$.

Now, molar mass of silver is $107.9g/mol$ and $Z=4$.

  $Density=\frac{4\times 107.9g/mol}{{\left(408.5518\times {10}^{-10}\right)}^3\times 6.023\times {10}^{23}}=10.5081g/c{m}^3.$

Therefore, the mass density of silver is found out to be $10.5081g/c{m}^3$.