Mathematical analysis of cubic system and Bragg’s equation Questions in English

(C) For a $BCC$ unit cell, the relationship between the edge length $a$ and the atomic radius $r$ is given by $4r = \sqrt{3}a$.
Given $a = 352 \ pm$.
Substituting the value: $r = \frac{\sqrt{3} \times 352}{4}$.
$r = \frac{1.732 \times 352}{4} = 152.4 \ pm$.

(A) The density formula for a unit cell is $d = \frac{Z \cdot M}{N_A \cdot a^3}$.
For a $BCC$ structure,the number of atoms per unit cell is $Z = 2$.
Given: $d = 8.55 \ g \ cm^{-3}$,$M = 93 \ g \ mol^{-1}$,$N_A \approx 6.022 \times 10^{23} \ mol^{-1}$.
Rearranging for $a^3$: $a^3 = \frac{Z \cdot M}{d \cdot N_A}$.
Substituting the values: $a^3 = \frac{2 \times 93}{8.55 \times 6.022 \times 10^{23}} \approx 3.61 \times 10^{-23} \ cm^3$.
Therefore,the edge length $a = (3.61 \times 10^{-23})^{1/3} \ cm$.

(A) The formula for density is $d = \frac{Z \cdot M}{N_{A} \cdot a^3}$.
Rearranging to solve for $Z$: $Z = \frac{d \cdot N_{A} \cdot a^3}{M}$.
Given: $d = 7 \ g \ cm^{-3}$, $a = 300 \ pm = 300 \times 10^{-10} \ cm = 3 \times 10^{-8} \ cm$, $M = 52 \ g \ mol^{-1}$, and $N_{A} = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $Z = \frac{7 \times 6.022 \times 10^{23} \times (3 \times 10^{-8})^3}{52}$.
$Z = \frac{7 \times 6.022 \times 10^{23} \times 27 \times 10^{-24}}{52} \approx \frac{1138.158 \times 10^{-1}}{52} \approx \frac{113.8}{52} \approx 2.18 \approx 2$.
Since $Z = 2$, the unit cell is Body centred cubic $(BCC)$.

(A) $\text{Volume of metal} = \frac{\text{Mass}}{\text{Density}} = \frac{38.6 \ g}{19.3 \ g \ cm^{-3}} = 2 \ cm^{3}$
$\text{Number of unit cells} = \frac{\text{Total volume of metal}}{\text{Volume of one unit cell}}$
$\text{Number of unit cells} = \frac{2 \ cm^{3}}{6.18 \times 10^{-23} \ cm^{3}} \approx 3.236 \times 10^{22}$

(A) For a body-centered cubic $(bcc)$ unit cell,the relationship between edge length $(a)$ and radius $(r)$ is given by $\sqrt{3} a = 4r$.
Given $r = 1.86 \times 10^{-8} \text{ cm}$.
$a = \frac{4r}{\sqrt{3}} = \frac{4 \times 1.86 \times 10^{-8}}{1.732} \approx 4.3 \times 10^{-8} \text{ cm}$.

(D) Volume of one unit cell $= (a)^{3} = (100 \ pm)^{3} = (100 \times 10^{-10} \ cm)^{3} = 10^{-24} \ cm^{3}$.
Total volume of $100 \ g$ of the element $= \frac{\text{Mass}}{\text{Density}} = \frac{100 \ g}{10 \ g/cm^{3}} = 10 \ cm^{3}$.
Number of unit cells $= \frac{\text{Total volume}}{\text{Volume of one unit cell}} = \frac{10 \ cm^{3}}{10^{-24} \ cm^{3}} = 1 \times 10^{25}$ unit cells.

(D) For a $bcc$ crystal,the relationship between atomic radius $r$ and edge length $a$ is given by: $r = \frac{\sqrt{3}}{4} a$
Therefore,$a = \frac{4r}{\sqrt{3}}$.
Substituting the given value $r = 1.33 \times 10^{-8} \ cm$:
$a = \frac{4 \times 1.33 \times 10^{-8} \ cm}{1.732} = \frac{5.32 \times 10^{-8} \ cm}{1.732} \approx 3.07 \times 10^{-8} \ cm$.

(D) For a $bcc$ unit cell, the number of atoms per unit cell $n = 2$.
Given density $\rho = 4 \ g \ cm^{-3}$ and edge length $a = 500 \ pm = 500 \times 10^{-10} \ cm = 5 \times 10^{-8} \ cm$.
The volume of the unit cell $a^{3} = (5 \times 10^{-8} \ cm)^{3} = 125 \times 10^{-24} \ cm^{3}$.
The formula for atomic mass $M$ is $M = \frac{\rho \times a^{3} \times N_{A}}{n}$.
Substituting the values: $M = \frac{4 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2} = 2 \times 125 \times 0.6022 = 150.55 \ g \ mol^{-1}$.
Thus, the atomic mass is approximately $150.5 \ g \ mol^{-1}$.

(D) For a $fcc$ unit cell,the relationship between the radius of the atom $(r)$ and the edge length of the unit cell $(a)$ is given by: $r = \frac{a}{2\sqrt{2}}$.
Given,$a = 0.3524 \ nm$.
Substituting the value of $a$ in the formula:
$r = \frac{0.3524 \ nm}{2 \times 1.414} = \frac{0.3524 \ nm}{2.828} = 0.1246 \ nm$.

(B) For an $fcc$ type unit cell, the relationship between edge length $a$ and atomic radius $r$ is given by $r = \frac{a}{2 \sqrt{2}}$.
Rearranging for $a$, we get $a = 2 \sqrt{2} \times r$.
Given $r = 127.6 \ pm$, substituting the value:
$a = 2 \times 1.414 \times 127.6 \ pm = 360.85 \ pm$.
Rounding to the nearest integer, $a \approx 361 \ pm$.

(D) The formula for density is $\rho = \frac{n \times M}{a^{3} \times N_{A}}$.
Given: $a = 4 \ \mathring{A} = 4 \times 10^{-8} \ cm$,$\rho = 2.7 \ g \ cm^{-3}$,$M = 27 \ g \ mol^{-1}$,$N_{A} = 6.022 \times 10^{23} \ mol^{-1}$.
Rearranging for $n$: $n = \frac{\rho \times a^{3} \times N_{A}}{M}$.
Substituting the values: $n = \frac{2.7 \times (4 \times 10^{-8})^{3} \times 6.022 \times 10^{23}}{27}$.
$n = \frac{2.7 \times 64 \times 10^{-24} \times 6.022 \times 10^{23}}{27} = 0.1 \times 64 \times 0.6022 \approx 3.85$.
Since the number of atoms per unit cell must be an integer,$n \approx 4$,which corresponds to a face-centered cubic $(FCC)$ unit cell.

(A) For a body-centered cubic $(BCC)$ unit cell, the relationship between the radius $(r)$ and the edge length $(a)$ is given by $r = \frac{\sqrt{3} a}{4}$.
Given that the edge length $a = 351 \ pm$.
Substituting the value of $a$ into the formula:
$r = \frac{\sqrt{3} \times 351}{4} = \frac{1.732 \times 351}{4} = \frac{607.932}{4} = 151.98 \ pm$.

(B) For a face-centred cubic $(fcc)$ unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $r = \frac{a}{2 \sqrt{2}}$.
Therefore, the edge length $a = 2 \sqrt{2} \times r$.
Given $r = 125 \text{ pm}$, we have $a = 2 \times 1.414 \times 125 \text{ pm} = 353.5 \text{ pm}$.

(D) Given: Edge length $a = 250 \ pm = 2.5 \times 10^{-8} \ cm$.
Atomic mass $M = 90.3 \ g \ mol^{-1}$.
For $fcc$ lattice,the number of atoms per unit cell $n = 4$.
Avogadro's number $N_{A} = 6.022 \times 10^{23} \ mol^{-1}$.
The density formula is $\rho = \frac{n \times M}{a^{3} \times N_{A}}$.
Substituting the values: $\rho = \frac{4 \times 90.3}{(2.5 \times 10^{-8})^{3} \times 6.022 \times 10^{23}}$.
$\rho = \frac{361.2}{15.625 \times 10^{-24} \times 6.022 \times 10^{23}} = \frac{361.2}{9.409} \approx 38.40 \ g \ cm^{-3}$.

(C) Given: Density $(\rho)$ = $2.8 \ g \ cm^{-3}$,Edge length $(a)$ = $4 \times 10^{-8} \ cm$,Number of atoms per unit cell $(n)$ = $4$ (for $fcc$),Avogadro's number $(N_{A})$ = $6.022 \times 10^{23} \ mol^{-1}$.
Formula: $\rho = \frac{n \times M}{a^{3} \times N_{A}}$.
Rearranging for molar mass $(M)$: $M = \frac{\rho \times a^{3} \times N_{A}}{n}$.
Substituting the values: $M = \frac{2.8 \times (4 \times 10^{-8})^{3} \times 6.022 \times 10^{23}}{4}$.
$M = \frac{2.8 \times 64 \times 10^{-24} \times 6.022 \times 10^{23}}{4}$.
$M = 0.7 \times 64 \times 0.6022 = 26.97896 \approx 27 \ g \ mol^{-1}$.

(A) Edge length $a = 16 \times 10^{-8} \ cm$.
Volume of one unit cell $= a^{3} = (16 \times 10^{-8} \ cm)^{3} = 4096 \times 10^{-24} \ cm^{3}$.
Since the unit cell contains $24$ molecules,the volume occupied by $24$ molecules is $4096 \times 10^{-24} \ cm^{3}$.
Volume of $1$ mole of molecules $= (\text{Volume of } 24 \text{ molecules} / 24) \times N_{A}$.
Volume of $1$ mole $= (4096 \times 10^{-24} / 24) \times 6.022 \times 10^{23} \ cm^{3} \ mol^{-1}$.
Volume of $1$ mole $\approx 170.67 \times 10^{-24} \times 6.022 \times 10^{23} \approx 102.77 \ cm^{3} \ mol^{-1}$.

(A) For a $bcc$ unit cell,the relationship between the radius $r$ and the edge length $a$ is given by $r = \frac{\sqrt{3} a}{4}$.
Rearranging the formula to solve for $a$,we get $a = \frac{4 r}{\sqrt{3}}$.
Substituting the given value $r = 1.86 \times 10^{-8} \ cm$:
$a = \frac{4 \times 1.86 \times 10^{-8} \ cm}{1.732} = 4.29 \times 10^{-8} \ cm$.

(B) Given edge length $a = 300 \ pm = 300 \times 10^{-10} \ cm = 3 \times 10^{-8} \ cm$.
For a $bcc$ unit cell, the relationship between radius $r$ and edge length $a$ is given by $r = \frac{\sqrt{3} a}{4}$.
Substituting the values: $r = \frac{1.732 \times 3 \times 10^{-8} \ cm}{4}$.
$r = \frac{5.196 \times 10^{-8} \ cm}{4} = 1.299 \times 10^{-8} \ cm$.

(B) $1$. Edge length $a = 3 \mathring{A} = 3 \times 10^{-8} \ cm$.
$2$. Volume of the unit cell $V = a^{3} = (3 \times 10^{-8} \ cm)^{3} = 27 \times 10^{-24} \ cm^{3}$.
$3$. Mass of one unit cell = $\text{Volume} \times \text{Density} = 27 \times 10^{-24} \ cm^{3} \times 8 \ g/cm^{3} = 216 \times 10^{-24} \ g$.
$4$. Number of unit cells in $100 \ g = \frac{100 \ g}{216 \times 10^{-24} \ g/\text{unit cell}} = 0.4629 \times 10^{24} = 4.63 \times 10^{23}$ unit cells.

(A) For an $fcc$ (face-centered cubic) unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $a = 2 \sqrt{2} r$ or $r = \frac{a}{2 \sqrt{2}}$.
Given edge length $a = 620 \ pm$.
Substituting the value: $r = \frac{620}{2 \times 1.414} = \frac{620}{2.828} \approx 219.2 \ pm$.
Thus, the radius of the $Xe$ atom is $219.2 \ pm$.

(B) For $fcc$ structure,the number of atoms per unit cell $Z = 4$.
Density $d = \frac{Z \cdot M}{a^3 \cdot N_A}$.
Rearranging for edge length $a$: $a^3 = \frac{Z \cdot M}{d \cdot N_A}$.
Substituting the values: $a^3 = \frac{4 \times 98.99}{3.4 \times 6.022 \times 10^{23}}$.
$a^3 = 193.389 \times 10^{-24} \text{ cm}^3$.
$a = \sqrt[3]{193.389} \times 10^{-8} \text{ cm} = 5.783 \times 10^{-8} \text{ cm}$.
Since $1 \text{ cm} = 10^8 \mathring A$,$a = 5.783 \mathring A$.

$(A)$ For a $bcc$ unit cell, the number of atoms per unit cell $(n)$ is $2$.
The edge length $(a)$ is $288 \ pm = 288 \times 10^{-10} \ cm = 2.88 \times 10^{-8} \ cm$.
The density formula is $\rho = \frac{n \times M}{a^{3} \times N_{A}}$.
Rearranging for molar mass $(M)$: $M = \frac{\rho \times a^{3} \times N_{A}}{n}$.
Substituting the values: $M = \frac{7.2 \times (2.88 \times 10^{-8})^{3} \times 6.022 \times 10^{23}}{2}$.
$M = \frac{7.2 \times 23.8878 \times 10^{-24} \times 6.022 \times 10^{23}}{2} \approx \frac{103.56}{2} \approx 51.78 \ g \ mol^{-1}$.
Thus, the correct option is $A$.

(B) For a face-centered cubic $(fcc)$ unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $a = 2 \sqrt{2} r$ or $r = \frac{a}{2 \sqrt{2}}$.
Given edge length $a = 316.5 \text{ pm}$.
Substituting the value: $r = \frac{316.5 \text{ pm}}{2 \times 1.414} = \frac{316.5 \text{ pm}}{2.828} \approx 111.91 \text{ pm}$.
Therefore, the correct option is $B$.

(B) For a $bcc$ lattice, the number of atoms per unit cell, $n = 2$.
Given: edge length $a = 500 \ pm = 5 \times 10^{-8} \ cm$, density $\rho = 7.5 \ g \ cm^{-3}$, mass $m = 300 \ g$.
Volume of unit cell $a^{3} = (5 \times 10^{-8} \ cm)^{3} = 125 \times 10^{-24} \ cm^{3}$.
Using the density formula $\rho = \frac{nM}{a^{3} N_{A}}$, we find the molar mass $M$:
$M = \frac{\rho \times a^{3} \times N_{A}}{n} = \frac{7.5 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2} \approx 282.3 \ g \ mol^{-1}$.
Number of atoms in $300 \ g$ of metal = $\frac{\text{mass}}{\text{molar mass}} \times N_{A} = \frac{300}{282.3} \times 6.022 \times 10^{23} \approx 6.4 \times 10^{23} \ \text{atoms}$.

(A) The volume of the given sample is calculated as: $\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{36 \ g}{7.2 \ g \ cm^{-3}} = 5 \ cm^{3}$.
The number of unit cells in the sample is given by the ratio of the total volume of the sample to the volume of one unit cell:
$\text{Number of unit cells} = \frac{\text{Total Volume}}{\text{Volume of one unit cell}} = \frac{5 \ cm^{3}}{24.99 \times 10^{-24} \ cm^{3}}$.
$\text{Number of unit cells} \approx 0.20008 \times 10^{24} = 2.0 \times 10^{23}$.

(B) For a $bcc$ (body-centered cubic) structure,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by:
$r = \frac{\sqrt{3}}{4} a$
Given,$a = 4.29 \times 10^{-8} \ cm$.
Substituting the value of $a$:
$r = \frac{\sqrt{3}}{4} \times 4.29 \times 10^{-8} \ cm$
$r = 0.433 \times 4.29 \times 10^{-8} \ cm$
$r \approx 1.857 \times 10^{-8} \ cm$
Rounding to two decimal places,we get $r \approx 1.86 \times 10^{-8} \ cm$.

(B) The density formula for a unit cell is given by $\rho = \frac{nM}{a^3 N_{A}}$.
Substituting the given values: $0.96 = \frac{n \times 23}{48}$.
Rearranging for $n$: $n = \frac{0.96 \times 48}{23}$.
$n = \frac{46.08}{23} \approx 2.003$.
Since $n$ represents the number of atoms per unit cell,the value is $2$.

(D) The formula for density $(\rho)$ of a unit cell is given by: $\rho = \frac{M \times n}{a^3 \times N_A}$
Where $M$ is the molar mass,$n$ is the number of atoms per unit cell,$a^3$ is the volume of the unit cell,and $N_A$ is the Avogadro constant.
Given: $\rho = 20 \ g \ cm^{-3}$,$M = 190 \ g \ mol^{-1}$,and $a^3 \cdot N_A = 38 \ cm^3 \ mol^{-1}$.
Substituting the values into the formula: $20 = \frac{190 \times n}{38}$
$n = \frac{20 \times 38}{190}$
$n = \frac{760}{190} = 4$
Therefore,the number of atoms present in the unit cell is $4$.

(D) The volume of the cube corresponding to one $CsCl$ ion pair is given as $V = 7.014 \times 10^{-23} \ cm^{3}$.
According to the problem,the smallest $Cs^{+}-Cs^{+}$ inter-nuclear distance $(a)$ is equal to the side length of this cube.
Therefore,$a^{3} = 7.014 \times 10^{-23} \ cm^{3}$.
$a = (7.014 \times 10^{-23})^{1/3} \ cm$.
$a = 4.124 \times 10^{-8} \ cm$.
Since $1 \ \mathring{A} = 10^{-8} \ cm$,we have $a = 4.124 \ \mathring{A}$.
Rounding to the nearest value,the distance is approximately $4.1 \ \mathring{A}$.
Thus,the correct option is $D$.

(B) For a simple cubic unit cell,the relation between radius '$r$' and edge length '$a$' is $r = \frac{a}{2}$.
For a body-centered cubic $(bcc)$ unit cell,the relation is $r = \frac{a \sqrt{3}}{4}$.
For a face-centered cubic $(fcc)$ unit cell,the relation is $r = \frac{a}{2 \sqrt{2}}$.
Thus,the ratio of the radii for simple cubic : $bcc$ : $fcc$ is $\frac{1}{2} \ a : \frac{\sqrt{3}}{4} \ a : \frac{1}{2 \sqrt{2}} \ a$.

(D) For a face-centred cubic $(FCC)$ crystal, the number of atoms per unit cell is $Z = 4$.
Edge length $a = 300 \ pm = 300 \times 10^{-10} \ cm = 3 \times 10^{-8} \ cm$.
Density $d = 10 \ g \ cm^{-3}$.
Using the density formula: $d = \frac{Z \times M}{N_A \times a^3}$, where $M$ is the molar mass.
$M = \frac{d \times N_A \times a^3}{Z} = \frac{10 \times 6.022 \times 10^{23} \times (3 \times 10^{-8})^3}{4}$.
$M = \frac{10 \times 6.022 \times 10^{23} \times 27 \times 10^{-24}}{4} = \frac{162.594}{4} = 40.6485 \ g \ mol^{-1}$.
Now, the number of atoms in $4.5 \ g$ is calculated as:
$\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{4.5}{40.6485} \times 6.022 \times 10^{23} \approx 6.66 \times 10^{22}$ atoms.

(C) Given,metallic radius $r = \sqrt{3} \ \mathring{A} = \sqrt{3} \times 10^{-10} \ m$.
For a body-centred cubic $(BCC)$ lattice,the relationship between radius $r$ and edge length $a$ is $r = \frac{\sqrt{3} a}{4}$.
Rearranging for $a$: $a = \frac{4r}{\sqrt{3}} = \frac{4 \times \sqrt{3} \times 10^{-10} \ m}{\sqrt{3}} = 4 \times 10^{-10} \ m$.
The volume of the unit cell is $V = a^3 = (4 \times 10^{-10} \ m)^3 = 64 \times 10^{-30} \ m^3 = 6.4 \times 10^{-29} \ m^3$.

(A) Given, metal crystallises in $bcc$ lattice, therefore $Z = 2$.
Edge length $a = 300 \ pm = 300 \times 10^{-10} \ cm$.
Density $d = 6.15 \ g \ cm^{-3}$.
Using the formula for density: $d = \frac{Z \times M}{a^3 \times N_A}$.
Rearranging for molar mass $M$: $M = \frac{d \times a^3 \times N_A}{Z}$.
Substituting the values: $M = \frac{6.15 \times (300 \times 10^{-10})^3 \times 6.022 \times 10^{23}}{2}$.
$M = \frac{6.15 \times 27 \times 10^{-24} \times 6.022 \times 10^{23}}{2} = \frac{99.965}{2} \approx 49.98 \ g \ mol^{-1}$.
Thus, the molar mass is approximately $50 \ g \ mol^{-1}$.

(B) For an $FCC$ unit cell,the relationship between the edge length $a$ and the metallic radius $r$ is given by $4r = \sqrt{2}a$.
Given $r = \sqrt{2} \ \mathring{A}$.
Substituting the value of $r$: $a = \frac{4r}{\sqrt{2}} = \frac{4 \times \sqrt{2}}{\sqrt{2}} = 4 \ \mathring{A}$.
The volume of the unit cell is $V = a^{3} = (4 \ \mathring{A})^{3} = 64 \ \mathring{A}^{3}$.
Since $1 \ \mathring{A} = 10^{-10} \ m$,then $1 \ \mathring{A}^{3} = 10^{-30} \ m^{3}$.
Therefore,$V = 64 \times 10^{-30} \ m^{3} = 6.4 \times 10^{-29} \ m^{3}$.

(C) For a $BCC$ crystal,the number of atoms per unit cell,$Z = 2$.
The density formula is $d = \frac{Z \times M}{a^3 \times N_A}$.
Given: $d = 10 \ g \ cm^{-3}$,$a = 200 \ pm = 200 \times 10^{-10} \ cm = 2 \times 10^{-8} \ cm$,$N_A = 6 \times 10^{23} \ mol^{-1}$,$Z = 2$.
Calculating molar mass $(M)$:
$M = \frac{d \times a^3 \times N_A}{Z} = \frac{10 \times (2 \times 10^{-8})^3 \times 6 \times 10^{23}}{2} = \frac{10 \times 8 \times 10^{-24} \times 6 \times 10^{23}}{2} = 24 \ g \ mol^{-1}$.
Number of moles in $2.4 \ g$ = $\frac{2.4 \ g}{24 \ g \ mol^{-1}} = 0.1 \ mol$.
Number of atoms = $\text{moles} \times N_A = 0.1 \times 6 \times 10^{23} = 6 \times 10^{22} \ atoms$.

(C) The formula for the body diagonal of a cube is given by $\sqrt{3} \times a$, where $a$ is the edge length of the cube.
Given, $a = 300 \ pm$.
Body diagonal $= \sqrt{3} \times 300 \ pm$.
Using $\sqrt{3} \approx 1.732$, we get:
Body diagonal $= 1.732 \times 300 \ pm = 519.6 \ pm$.

(D) For a $B.C.C.$ (Body-Centered Cubic) lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $r = \frac{\sqrt{3}}{4} a$.
Given the edge length $a = 4.29 \ \mathring{A}$.
Substituting the value of $a$ into the formula:
$r = \frac{\sqrt{3}}{4} \times 4.29$
$r = 0.433 \times 4.29$
$r \approx 1.857 \ \mathring{A}$.

(C) For a $bcc$ lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3}a$,which implies $a = \frac{4r}{\sqrt{3}}$.
Given $r = 2.598 \ \mathring{A} = 2.598 \times 10^{-8} \ cm$.
Substituting the value of $r$: $a = \frac{4 \times 2.598 \times 10^{-8}}{\sqrt{3}} = \frac{4 \times 2.598 \times 10^{-8}}{1.732} = 4 \times 1.5 \times 10^{-8} = 6 \times 10^{-8} \ cm$.
The volume of the unit cell is $V = a^3 = (6 \times 10^{-8} \ cm)^3 = 216 \times 10^{-24} \ cm^3 = 2.16 \times 10^{-22} \ cm^3$.

(C) In a simple cubic lattice, the atoms are present at the corners of the cube.
For a simple cubic unit cell, the edge length $a$ is related to the radius of the atom $r$ by the formula $a = 2r$.
Given that the radius of the metal atom is $r = x \ pm$.
Therefore, the edge length $a = 2 \times x = 2x \ pm$.
The volume of the unit cell is given by $V = a^3$.
Substituting the value of $a$, we get $V = (2x)^3 = 8x^3 \ pm^3$.

(C) According to Bragg's law: $n \lambda = 2 d \sin \theta$.
Given: $n = 1$,$\lambda = 1.54 \mathring{A}$,and $2 \theta = 60^{\circ}$,so $\theta = 30^{\circ}$.
Substituting the values into the formula: $1 \times 1.54 \mathring{A} = 2 \times d \times \sin 30^{\circ}$.
Since $\sin 30^{\circ} = 0.5$,we have: $1.54 \mathring{A} = 2 \times d \times 0.5$.
$1.54 \mathring{A} = d$.
Converting $\mathring{A}$ to $cm$ $(1 \mathring{A} = 10^{-8} \ cm)$:
$d = 1.54 \times 10^{-8} \ cm$.

(C) The formula for density $(d)$ of a unit cell is given by: $d = \frac{Z \times M}{a^3 \times N_A}$
Where $Z$ is the number of atoms per unit cell,$M$ is the molar mass $(23 \ g \ mol^{-1})$,$a$ is the edge length $(5 \ \mathring{A} = 5 \times 10^{-8} \ cm)$,and $N_A$ is Avogadro's number $(6.022 \times 10^{23} \ mol^{-1})$.
Rearranging for $Z$: $Z = \frac{d \times a^3 \times N_A}{M}$
Substituting the values: $Z = \frac{0.613 \times (5 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{23}$
$Z = \frac{0.613 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{23}$
$Z = \frac{0.613 \times 125 \times 0.6022}{23} \approx \frac{46.14}{23} \approx 2$
Thus,the effective number of atoms per unit cell is $2$.

(B) For an $fcc$ lattice,the edge length $a = 4 \ \mathring{A} = 4 \times 10^{-8} \ cm$.
$1$. The closest distance between atoms $(x)$ in an $fcc$ lattice is given by $x = \frac{a}{\sqrt{2}}$.
$x = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ \mathring{A} \approx 2.828 \ \mathring{A}$.
$2$. The density $(y)$ is given by the formula $d = \frac{Z \times M}{N_A \times a^3}$.
For $fcc$,$Z = 4$. Given $M = 197 \ g \ mol^{-1}$,$N_A = 6 \times 10^{23} \ mol^{-1}$,and $a = 4 \times 10^{-8} \ cm$.
$y = \frac{4 \times 197}{6 \times 10^{23} \times (4 \times 10^{-8})^3} = \frac{788}{6 \times 10^{23} \times 64 \times 10^{-24}} = \frac{788}{38.4} \approx 20.52 \ g \ cm^{-3}$.
Thus,$x = 2\sqrt{2}$ and $y = 20.52$.

(C) For a simple cubic lattice,the number of atoms per unit cell $(Z)$ is $1$.
The formula for density $(d)$ is $d = \frac{Z \times M}{N_A \times a^3}$,where $M$ is the atomic weight,$N_A$ is Avogadro's number,and $a$ is the edge length.
Substituting the values: $7.2 = \frac{1 \times 250}{6.02 \times 10^{23} \times a^3}$.
$a^3 = \frac{250}{7.2 \times 6.02 \times 10^{23}} \approx 5.767 \times 10^{-23} \ cm^3$.
$a = (57.67 \times 10^{-24})^{1/3} \approx 3.86 \times 10^{-8} \ cm = 3.86 \ \mathring{A}$.
For a simple cubic lattice,the radius $(r)$ is related to the edge length $(a)$ by $r = \frac{a}{2}$.
$r = \frac{3.86}{2} = 1.93 \ \mathring{A}$.

(A) $1$. For a $BCC$ unit cell, the number of atoms per unit cell $(Z)$ is $2$.
$2$. The edge length $(a)$ is $288 \ pm = 288 \times 10^{-10} \ cm$.
$3$. The volume of the unit cell $(V)$ is $a^3 = (288 \times 10^{-10} \ cm)^3 \approx 2.39 \times 10^{-23} \ cm^3$.
$4$. The density $(\rho)$ is given by $\rho = \frac{Z \times M}{N_A \times a^3}$.
$5$. Rearranging to find the molar mass $(M)$: $M = \frac{\rho \times N_A \times a^3}{Z} = \frac{7.2 \times 6.022 \times 10^{23} \times 2.39 \times 10^{-23}}{2} \approx 51.8 \ g \ mol^{-1}$.
$6$. The number of moles in $208 \ g$ is $n = \frac{\text{mass}}{M} = \frac{208}{51.8} \approx 4.015 \ mol$.
$7$. The number of atoms is $n \times N_A = 4.015 \times 6.022 \times 10^{23} \approx 24.18 \times 10^{23} \approx 24.2 \times 10^{23}$.

(B) For a simple cubic lattice, the relationship between the edge length $(a)$ and the radius of the atom $(r)$ is given by $a = 2r$.
The volume of the unit cell $(V)$ is given by $V = a^3$.
Given $V = 6.4 \times 10^7 \ pm^3$.
Therefore, $a^3 = 6.4 \times 10^7 \ pm^3$.
Taking the cube root of both sides, $a = (64 \times 10^6)^{1/3} \ pm = 400 \ pm$.
Since $a = 2r$, we have $r = a / 2 = 400 / 2 = 200 \ pm$.

(A) The density formula for a cubic crystal is $\rho = \frac{Z \times M}{N_A \times a^3}$.
Given values are $\rho = 7.6 \ g \ cm^{-3}$,$M = 56 \ g \ mol^{-1}$,$N_A = 6.022 \times 10^{23} \ mol^{-1}$,and $a = 290 \ pm = 290 \times 10^{-10} \ cm$.
Rearranging the formula for $Z$: $Z = \frac{\rho \times N_A \times a^3}{M}$.
Substituting the values: $Z = \frac{7.6 \times 6.022 \times 10^{23} \times (290 \times 10^{-10})^3}{56}$.
$Z = \frac{7.6 \times 6.022 \times 10^{23} \times 24389 \times 10^{-30}}{56}$.
$Z = \frac{1115780.8 \times 10^{-7}}{56} \approx \frac{0.111578}{0.056} \approx 1.99$.
Thus,the value of $Z$ is approximately $2$.

(C) For a $CCP$ structure,the number of atoms per unit cell $Z = 4$.
The density formula is $d = \frac{ZM}{N_A a^3}$.
Rearranging for $a^3$: $a^3 = \frac{ZM}{N_A d} = \frac{4 \times 107.9}{(6.022 \times 10^{23}) \times 10.5}$.
$a^3 = \frac{431.6}{63.231 \times 10^{23}} \approx 6.825 \times 10^{-23} \ cm^3$.
Converting $cm^3$ to $\mathring{A}^3$: $1 \ cm^3 = (10^8 \ \mathring{A})^3 = 10^{24} \ \mathring{A}^3$.
$a^3 = 6.825 \times 10^{-23} \times 10^{24} \ \mathring{A}^3 = 68.25 \ \mathring{A}^3$.
Therefore,$a = \sqrt[3]{68.25} \ \mathring{A}$.

(B) Given: Density $(d) = 8.92 \ g \ cm^{-3}$,Edge length $(a) = 3.61 \times 10^{-8} \ cm$,Avogadro's number $(N_A) = 6.022 \times 10^{23} \ mol^{-1}$.
For an $fcc$ lattice,the number of atoms per unit cell $(Z) = 4$.
The formula for density is $d = \frac{Z \times M}{a^3 \times N_A}$,where $M$ is the atomic weight.
Rearranging for $M$: $M = \frac{d \times a^3 \times N_A}{Z}$.
Substituting the values: $M = \frac{8.92 \times (3.61 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{4}$.
$M = \frac{8.92 \times 47.0458 \times 10^{-24} \times 6.022 \times 10^{23}}{4}$.
$M = \frac{252.712}{4} = 63.178 \ u$.

(A) The number of atoms in $210 \ g$ of $A$ is $\frac{210}{M_A} \times N_A$ and in $594 \ g$ of $B$ is $\frac{594}{M_B} \times N_A$,where $M_A$ and $M_B$ are molar masses.
Given: $\frac{210}{M_A} = \frac{594}{M_B} \implies \frac{M_B}{M_A} = \frac{594}{210} = 2.828$.
Density formula: $d = \frac{Z \times M}{N_A \times a^3}$.
For $A$ ($fcc$,$Z=4$): $7 = \frac{4 \times M_A}{N_A \times a^3}$.
For $B$ ($bcc$,$Z=2$): $d_B = \frac{2 \times M_B}{N_A \times a^3}$.
Taking the ratio: $\frac{d_B}{7} = \frac{2 \times M_B}{4 \times M_A} = \frac{1}{2} \times \frac{M_B}{M_A}$.
Substituting the ratio: $\frac{d_B}{7} = \frac{1}{2} \times \frac{594}{210} = \frac{1}{2} \times 2.828 = 1.414$.
$d_B = 7 \times 1.414 = 9.9 \ g \ cm^{-3}$.

(C) The effective number of atoms in an $fcc$ unit cell is $Z = 4$.
The formula for density is $d = \frac{Z \times M}{N_{A} \times a^3}$.
Given: $d = 11.21 \ g \ cm^{-3}$,$a = 4 \ \mathring{A} = 4 \times 10^{-8} \ cm$,$N_{A} = 6.023 \times 10^{23} \ mol^{-1}$.
Rearranging for molar mass $(M)$: $M = \frac{d \times N_{A} \times a^3}{Z}$.
$M = \frac{11.21 \times 6.023 \times 10^{23} \times (4 \times 10^{-8})^3}{4}$.
$M = \frac{11.21 \times 6.023 \times 10^{23} \times 64 \times 10^{-24}}{4}$.
$M = 11.21 \times 6.023 \times 16 \times 0.1 = 108.0 \ g \ mol^{-1}$.