The drift velocity of electrons in silver wire with cross-sectional area `3.14 × 10^(-6) m^2` carrying a current of 20 A is. Given atomic weight of Ag = 108, density of silver = `10.5 xx 10^3 kg//m^3`

Options

A. `2.798 xx 10^(-4)` m/sec
B. `67.98 xx10^(-4)` m/sec
C. `0.67 xx 10^(-4)` m/sec
D. `6.798 xx 10^(-9)` m/sec

Answer

To find the drift velocity of electrons in a silver wire, we can use the formula relating current (I), number of charge carriers per unit volume (n), charge of an electron (e), cross-sectional area (A), and drift velocity (v_d):

\[ I = n \cdot A \cdot e \cdot v_d \]

### Step-by-step Solution:

**Step 1: Calculate the mass of silver in 1 m³.**
Given the density of silver (ρ) is \(10.5 \times 10^3 \, \text{kg/m}^3\), the mass of silver in 1 m³ is:

\[ \text{Mass} = \text{Density} \times \text{Volume} = 10.5 \times 10^3 \, \text{kg/m}^3 \times 1 \, \text{m}^3 = 10.5 \times 10^3 \, \text{kg} \]

**Hint:** Density is defined as mass per unit volume.

---

**Step 2: Calculate the number of moles of silver in 1 m³.**
The atomic weight of silver (Ag) is 108 g/mol. Convert this to kg:

\[ \text{Molar mass} = 108 \, \text{g/mol} = 0.108 \, \text{kg/mol} \]

Now, calculate the number of moles (n_moles) in 1 m³:

\[ n_{\text{moles}} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{10.5 \times 10^3 \, \text{kg}}{0.108 \, \text{kg/mol}} \]

Calculating this gives:

\[ n_{\text{moles}} \approx 97222.22 \, \text{mol} \]

**Hint:** The number of moles can be found by dividing the mass by the molar mass.

---

**Step 3: Calculate the number of atoms in 1 m³.**
Using Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{atoms/mol}\)), we find the total number of atoms (n) in 1 m³:

\[ n = n_{\text{moles}} \times N_A = 97222.22 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \]

Calculating this gives:

\[ n \approx 5.85 \times 10^{28} \, \text{atoms/m}^3 \]

**Hint:** To find the number of atoms, multiply the number of moles by Avogadro's number.

---

**Step 4: Use the formula to find the drift velocity (v_d).**
Now we can rearrange the formula for current to solve for drift velocity:

\[ v_d = \frac{I}{n \cdot A \cdot e} \]

Where:
- \(I = 20 \, \text{A}\)
- \(A = 3.14 \times 10^{-6} \, \text{m}^2\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\)

Substituting in the values:

\[ v_d = \frac{20}{(5.85 \times 10^{28}) \cdot (3.14 \times 10^{-6}) \cdot (1.6 \times 10^{-19})} \]

Calculating the denominator:

\[ n \cdot A \cdot e \approx (5.85 \times 10^{28}) \cdot (3.14 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) \approx 2.93 \times 10^{4} \]

Now substituting back to find \(v_d\):

\[ v_d \approx \frac{20}{2.93 \times 10^{4}} \approx 6.81 \times 10^{-9} \, \text{m/s} \]

**Final Answer:**
The drift velocity of electrons in the silver wire is approximately \(6.81 \times 10^{-9} \, \text{m/s}\).

---