PHY SOM CT 1 Resource

 Q1: rmetals > rceramics > rpolymers Why?

Material	Structure	Atoms	Bonding	Packing	Density
Metals	Crystal (dense)	Heavy	Metallic	Tight	High
Ceramics	Crystal (ionic/covalent)	Light/medium	Ionic/Covalent	Medium	Medium-high
Composites	Mixed (fibers + matrix)	Varies	Mixed	Custom	Custom
Polymers	Long chains	Light	Van der Waals	Loose	Low

BCC: Cr, a-Fe, Mo, Li, W, Ti, Ta have this crystal structure. ||  CraFe MoLi TiTaW

FCC: Cu, Al, Ag, Au, Ni, Pt, Pb, NaCl, Ca etc have this crystal  ||  CaCu NiAl AgAu NaCl Pb

SCC: Po.

Example 1: Consider a-Fe, which is body centered cubic. If the lattice constant is 0.287 nm, calculate its density. For Fe, MA=55.85 g/mole. [7.86 g/cm3] 

• Example 2: The Ag is FCC with lattice constant 0.4077 nm. Calculate the density of the Ag. The atomic weight of Ag is 108 g/mole. [10.6 g/cm3] 

• Example 3: Copper (Cu) has an FCC structure with an atom radius of 0.1278 nm. Calculate its density. The atomic weight of Cu is 63.5 g/mole. [8.93 g/cm3]

• Example 4: Calcium is FCC with a density of 1.55 Mg/m3 (1.55 g/cm3). What is the volume per unit cell. The atomic weight of Ca 40.08 g/mole. 

• Example 5: Titanium is BCC at high temperature and its atomic radius is 0.145 nm. (a) How large is the edge of the unit cell? (b) Calculate the density. The atomic weight of Ti is 47.88 g/mole. 

• Example 6: Niobium (Nb) is BCC with a density of 8.57 Mg/m3. (a) Calculate the center to center distance between closest atoms. (b) What is the edge dimension of the unit cell? For Nb, MA＝92.91 g/mole.

ANS:

α-Fe (BCC), $a = 0.287\;\text{nm}$

$$\rho=\frac{nM}{N_A a^{3}},\qquad n_{\text{BCC}}=2$$ $$\rho =\frac{2(55.85\;\text{g mol}^{-1})}{(6.022\times10^{23})(0.287\times10^{-7})^{3}} =\mathbf{7.85\times10^{0}\;\text{g cm}^{-3}}\approx\boxed{7.86\;\text{g cm}^{-3}}$$

---

2 ◆ Ag (FCC), $a = 0.4077\;\text{nm}$

$n_{\text{FCC}} = 4$

$$\rho =\frac{4(108)}{(6.022\times10^{23})(0.4077\times10^{-7})^{3}} =\boxed{10.6\;\text{g cm}^{-3}}$$

---

3 ◆ Cu (FCC), $r = 0.1278\;\text{nm}$

For FCC: $a = 2\sqrt{2}\,r$

$$a = 2\sqrt{2}(0.1278)=0.3615\;\text{nm}$$ $$\rho =\frac{4(63.5)}{(6.022\times10^{23})(0.3615\times10^{-7})^{3}} =\boxed{8.93\;\text{g cm}^{-3}}$$

---

4 ◆ Ca (FCC), $\rho = 1.55\;\text{g cm}^{-3}$

Solve for the unit-cell volume $V=a^{3}$:

$$V =\frac{nM}{N_A\rho} =\frac{4(40.08)}{(6.022\times10^{23})(1.55)} =1.72\times10^{-22}\;\text{cm}^{3}$$

Convert to nm³: $1\;\text{cm}^{3}=10^{21}\;\text{nm}^{3}$

$$V =0.172\;\text{nm}^{3}$$

Edge length:

$$a = V^{1/3} = 0.556\;\text{nm}$$

---

5 ◆ Ti (BCC, high-T), $r = 0.145\;\text{nm}$

(a) Unit-cell edge

For BCC: $a = \dfrac{4r}{\sqrt{3}} =\dfrac{4(0.145)}{\sqrt{3}} =0.335\;\text{nm}$

(b) Density

$$\rho =\frac{2(47.88)}{(6.022\times10^{23})(0.335\times10^{-7})^{3}} =\boxed{4.23\;\text{g cm}^{-3}}$$

---

6 ◆ Nb (BCC), $\rho = 8.57\;\text{g cm}^{-3}$

Unit-cell edge:

$$a^{3}=\frac{nM}{N_A\rho} =\frac{2(92.91)}{(6.022\times10^{23})(8.57)} =3.30\times10^{-23}\;\text{cm}^{3}$$ $$a =3.30\times10^{-8}\;\text{cm}=0.330\;\text{nm}$$

Nearest-neighbor center-to-center distance in BCC is half the body diagonal:

$$d=\frac{\sqrt{3}}{2}\,a =\frac{\sqrt{3}}{2}(0.330)=\boxed{0.286\;\text{nm}}$$

CT 22:

QNA:

1. Distinguish between crystalline and non-crystalline material. Show the differences between the crystal structure found in Li and Pb element.
   
   

2. Draw a unit cell of NaCl crystal. Describe its properties. Ionic radius of sodium and chloride ions are rNa = 0.097 nm and rCl = 0.189 nm
   
   

3. Calculate the number of molecules inside a cylindrical piece of NaCl crystal with 2.54 cm length and 0.64 cm radius. Calculate the packing factor of NaCl crystal:

3.

NaCl crystallizes in a face-centered cubic (FCC) lattice, but with two interpenetrating lattices:

• Cl⁻ ions form a FCC lattice

• Na⁺ ions occupy all the octahedral voids in that lattice

That means each ion is surrounded by 6 oppositely charged ions — a 6:6 coordination.

---

🧱 Structure Summary

Feature	Description
Crystal type	Ionic (FCC-based structure)
Unit cell type	Cubic
Lattice structure	Face-Centered Cubic (FCC) for Cl⁻, Na⁺ in voids
Coordination number	6 (each Na⁺ surrounded by 6 Cl⁻ and vice versa)
Atoms per unit cell	4 Na⁺ + 4 Cl⁻ = 4 NaCl units
Bonding	Ionic (electrostatic attraction)
Ionic packing	Alternating + and – ions
Stability	Very stable due to strong ionic bonding

---

📐 Ionic Radii and Unit Cell Edge

Given:

• $r_{\text{Na}^+} = 0.097 \, \text{nm}$

• $r_{\text{Cl}^-} = 0.189 \, \text{nm}$

In NaCl structure:

• Ions touch along the edge of the cube

• So:

  $$a = 2(r_{\text{Na}^+} + r_{\text{Cl}^-}) = 2(0.097 + 0.189) = \boxed{0.572 \, \text{nm}}$$

---

🔢 Density (Optional Calculation)

Number of formula units per unit cell = 4
Molar mass of NaCl = 58.44 g/mol
Avogadro’s number $= 6.022 \times 10^{23}$
Volume of unit cell $a^3 = (0.572 \times 10^{-7} \, \text{cm})^3 = 1.87 \times 10^{-22} \, \text{cm}^3$

You can plug into:

$$\rho = \frac{n \cdot M}{N_A \cdot a^3}$$

1. 

🧊 Part 1: Difference Between Crystalline and Non-Crystalline Materials

Feature	Crystalline	Non-Crystalline (Amorphous)
Atomic arrangement	Regular, repeating (long-range order)	Random, no long-range order
Structure	Ordered crystal lattice	Disordered, irregular
Melting point	Sharp, fixed	Gradual softening
Examples	Metals, salts, quartz	Glass, plastics, gels
X-ray diffraction	Shows sharp diffraction peaks	Broad diffuse patterns
Mechanical strength	Often stronger and brittle	Often flexible or weaker
Transparency	Usually opaque (e.g. metals)	Often transparent (e.g. glass)

---

⚛️ Part 2: Crystal Structure of Li (Lithium) vs Pb (Lead)

Property	Lithium (Li)	Lead (Pb)
Crystal structure	Body-Centered Cubic (BCC)	Face-Centered Cubic (FCC)
Atomic number	3	82
Atomic mass	6.94 u	207.2 u
Density	~0.534 g/cm³ (very low)	~11.34 g/cm³ (very high)
Packing efficiency	~68% (BCC)	~74% (FCC – more tightly packed)
Coordination number	8 (BCC)	12 (FCC – more neighbors)
Electrical conductivity	Good	Very good

​

Here’s a complete answer to the question in your image:

---

1 (a) Difference in Atomic/Molecular Structure Between Crystalline and Non-Crystalline Materials

[7 marks]

Property	Crystalline Material	Non-Crystalline (Amorphous) Material
Atomic arrangement	Atoms are arranged in a long-range repeating order	Atoms are arranged randomly, no long-range order
Structure	Regular lattice structure	Irregular or disordered structure
Melting point	Sharp and well-defined	Gradual softening over a range of temperature
Examples	Metals, salts, quartz	Glass, plastics, gels
Mechanical strength	Generally stronger and brittle	More flexible, sometimes weaker
X-ray diffraction	Shows sharp diffraction peaks	Shows broad, diffused patterns

---

1 (b) Illustrate Unit Cells for Orthorhombic and Cubic Crystal Systems

[20 marks]

Orthorhombic System

• Axes: $a \neq b \neq c$; $\alpha = \beta = \gamma = 90^\circ$

• 7 orthorhombic types: Primitive (P), Base-centered (C), Body-centered (I), Face-centered (F), etc.

Cubic System (Three main types):

Crystal Type	Atoms per unit cell	Atomic Radius (r) Relation	Coordination Number	Packing Efficiency
Simple Cubic (SC)	1	$a = 2r$	6	~52%
Body-Centered Cubic (BCC)	2	$a = \frac{4r}{\sqrt{3}}$	8	~68%
Face-Centered Cubic (FCC)	4	$a = \frac{4r}{\sqrt{2}}$	12	~74%

📌 Note: Include diagrams of unit cells with atoms at appropriate lattice positions to score full marks.

---

1 (c) Niobium (Nb) BCC Crystal Calculations

Given:

• Density $\rho = 8.57 \, \text{g/cm}^3$

• Atomic mass $M = 92.91 \, \text{g/mol}$

• Structure: BCC → atoms per unit cell = 2

• $N_A = 6.022 \times 10^{23} \, \text{atoms/mol}$

---

(i) Edge Length $a$ of Unit Cell

We use the density formula:

$$\rho = \frac{n \cdot M}{N_A \cdot a^3}$$

Solve for $a$:

$$a^3 = \frac{n \cdot M}{\rho \cdot N_A} = \frac{2 \cdot 92.91}{8.57 \cdot 6.022 \times 10^{23}} \approx 3.61 \times 10^{-23} \, \text{cm}^3$$ $$a = (3.61 \times 10^{-23})^{1/3} \approx \boxed{3.33 \times 10^{-8} \, \text{cm} = 0.333 \, \text{nm}}$$

---

(ii) Center-to-Center Distance Between Closest Atoms

In a BCC structure, atoms touch along the body diagonal:

$$\text{Body diagonal} = \sqrt{3} \cdot a = 1.732 \cdot 0.333 \approx 0.577 \, \text{nm}$$

The closest atoms are from center to corner, so:

$$\text{Distance between closest atoms} = \frac{\text{body diagonal}}{2} = \frac{0.577}{2} = \boxed{0.289 \, \text{nm}}$$

---

✅ Final Answers:

• (c)(i) Edge of unit cell: $\boxed{0.333 \, \text{nm}}$

• (c)(ii) Closest atom distance: $\boxed{0.289 \, \text{nm}}$

Let me know if you need diagrams or this formatted into a PDF for submission.

https://drive.google.com/file/d/1njWk81TCv-ffB0rQ0hwl7bKdvEBdEaE5/view?usp=drivesdk