Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:12.4 (a) For the solidification of nickel, calculate the critical radius r* and the activations

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Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:12.4 (a) For the solidification of nickel, calculate the critical radius r* and the activations

free energy AG* if nucleation is homogeneous. Values for the latent heat of fusion and«

surface free energy are —2.53 x 109 J/m3 and 0.255 J/m2, respectively. Use the<

supercooling value found in Table 12.1.

(b) Now calculate the number of atoms found in a nucleus of critical size. Assume a

lattice parameter of 0.360 nm for solid nickel at its melting temperature.<

@

= Table 12.1 Degree of Supercooling AT (Homogeneous Nucleation) for Several Metals<

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VIPSolutions · Accepted Answer

Let's break the problem into parts and solve each step-by-step.

---

## **Part (a): Critical Radius $ r^* $ and Activation Free Energy $ \Delta G^* $**

### **Given Data**
- Latent heat of fusion, $ \Delta H_f = -2.53 \times 10^9 $ J/m³  
- Surface free energy, $ \gamma = .255 $ J/m²
- Degree of supercooling for Nickel, $ \Delta T = 319^\circ C $ (from table)
- Melting temperature of Nickel, $ T_m = 1455^\circ C = 1728 $ K (standard value for Ni)

### **Formulas**

#### **Critical Radius ($ r^* $):**
$$
r^* = \frac{2 \gamma}{|\Delta G_v|}
$$
Where $ \Delta G_v $ is the free energy change per unit volume:
$$
|\Delta G_v| = \frac{\Delta H_f \Delta T}{T_m}
$$

#### **Activation Free Energy ($ \Delta G^* $):**
$$
\Delta G^* = \frac{16 \pi \gamma^3}{3 (\Delta G_v)^2}
$$

---

### **Step 1: Calculate $ |\Delta G_v| $**
$$
|\Delta G_v| = \frac{2.53 \times 10^9 \times 319}{1728}
$$
$$
|\Delta G_v| = \frac{2.53 \times 10^9 \times 319}{1728}
$$
$$
= \frac{807.07 \times 10^9}{1728}
$$
$$
= 4.67 \times 10^8 \text{ J/m}^3
$$

---

### **Step 2: Calculate Critical Radius $ r^* $**
$$
r^* = \frac{2 \times .255}{4.67 \times 10^8}
$$
$$
= \frac{.51}{4.67 \times 10^8}
$$
$$
= 1.09 \times 10^{-9} \text{ m} = 1.09 \text{ nm}
$$

---

### **Step 3: Calculate Activation Free Energy $ \Delta G^* $**
$$
\Delta G^* = \frac{16 \pi (.255)^3}{3 \times (4.67 \times 10^8)^2}
$$
$$
= \frac{16 \times 3.1416 \times .0166}{3 \times 2.18 \times 10^{17}}
$$
$$
= \frac{.834}{6.54 \times 10^{17}}
$$
$$
= 1.28 \times 10^{-18} \text{ J}
$$

---

## **Part (b): Number of Atoms in the Critical Nucleus**

### **Given Data**
- Lattice parameter $ a = .360 $ nm = $ .360 \times 10^{-9} $ m
- Assume fcc structure for nickel ($ 4 $ atoms per unit cell)

### **Step 1: Volume of Critical Nucleus**
$$
V^* = \frac{4}{3} \pi (r^*)^3
$$
$$
V^* = \frac{4}{3} \pi (1.09 \times 10^{-9})^3
$$
$$
= \frac{4}{3} \pi \times 1.30 \times 10^{-27}
$$
$$
= 5.44 \times 10^{-27} \text{ m}^3
$$

### **Step 2: Volume of Unit Cell**
$$
V_{cell} = a^3 = (.36 \times 10^{-9})^3 = 4.66 \times 10^{-29} \text{ m}^3
$$

### **Step 3: Number of Unit Cells in Critical Nucleus**
$$
\text{Number of unit cells} = \frac{V^*}{V_{cell}} = \frac{5.44 \times 10^{-27}}{4.66 \times 10^{-29}} = 116.7
$$

### **Step 4: Number of Atoms**
$$
\text{Number of atoms} = 116.7 \times 4 \approx 467
$$

---

## **Final Answers**

### **(a)**
- **Critical radius, $ r^* $ = 1.09 nm**
- **Activation free energy, $ \Delta G^* = 1.28 \times 10^{-18} $ J**

### **(b)**
- **Number of atoms in critical nucleus ≈ 467**

---

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