Answer:
The answer is "0.147 nm and 99.63 mol %"
Explanation:
In point (a):
[tex]\to nk1 = 062[/tex]
[tex]\to \text{Bragg angle}[/tex] [tex]\theta =37.21^{\circ}[/tex]
[tex]\to \text{diffraction angle}[/tex] [tex]2 \theta = 74.42^{\circ}[/tex]
[tex]\to \lambda = 0.1790 nm[/tex]
find:
d(062)=?
formula:
[tex]\to nx = 2d \sin \theta[/tex]
[tex]\to d(062) = \frac{1 \times 0.1790^{\circ}}{2 \times \sin 37.21^{\circ}}\\[/tex]
[tex]= \frac{0.1790^{\circ}}{2 \times 0.604738126}\\\\= \frac{0.29599589}{2}\\\\= 0.147 \\[/tex]
In point (b):
[tex]\to Mg_2SiO_4\longleftrightarrow Fe_2SiO_4[/tex]
[tex]d= 0.14774 \ \ \ \ \ olivine = 0.147 \ \ \ \ \ 0.15153[/tex]
formula:
[tex]\to d=\frac{a}{\sqrt{n^2+k^2+i^2}}\\[/tex]
that's why the composition value equal to 99.63 %