An electric current of `16 A` exists in a metal wire of cross section `10^(-6)"M"^(2)` and length 1m. Assuming one free electron per atom. The drift speed of the free electron in the wire will be `(0.00x)m//s.` Find value of x.
`("Density of metal"= 5xx10^(3)"kg"//"m"^(3), "atomic weight"=60)`

Answer

To find the drift speed of free electrons in a metal wire, we can use the formula:

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

Where:
- \( I \) = current (in Amperes)
- \( n \) = number of free electrons per unit volume (in \( m^{-3} \))
- \( A \) = cross-sectional area of the wire (in \( m^{2} \))
- \( e \) = charge of an electron (\( 1.6 \times 10^{-19} C \))
- \( v_d \) = drift speed of the electrons (in \( m/s \))

### Step 1: Calculate the number of free electrons per unit volume (\( n \))

1. **Find the number of atoms per unit volume**:
- The density of the metal \( \rho = 5 \times 10^{3} \, kg/m^{3} \)
- The atomic weight \( A = 60 \, g/mol = 0.060 \, kg/mol \)
- The number of moles in 1 kg of the metal is given by:
\[ \text{Number of moles} = \frac{1 \, kg}{0.060 \, kg/mol} = \frac{1000 \, g}{60 \, g/mol} \approx 16.67 \, mol \]
- Using Avogadro's number \( N_A = 6.022 \times 10^{23} \, atoms/mol \), the number of atoms in 1 kg is:
\[ \text{Number of atoms} = 16.67 \, mol \times 6.022 \times 10^{23} \, atoms/mol \approx 1.00 \times 10^{25} \, atoms \]
- Therefore, the number of atoms per unit volume (for 1 m³) is:
\[ n = \frac{1.00 \times 10^{25} \, atoms}{1 \, m^{3}} = 1.00 \times 10^{25} \, m^{-3} \]

### Step 2: Calculate the drift speed (\( v_d \))

2. **Rearranging the formula for drift speed**:
\[ v_d = \frac{I}{n \cdot A \cdot e} \]
- Given:
- \( I = 16 \, A \)
- \( A = 10^{-6} \, m^{2} \)
- \( e = 1.6 \times 10^{-19} \, C \)
- Substitute the values into the equation:
\[ v_d = \frac{16}{(1.00 \times 10^{25}) \cdot (10^{-6}) \cdot (1.6 \times 10^{-19})} \]
- Calculate the denominator:
\[ n \cdot A \cdot e = (1.00 \times 10^{25}) \cdot (10^{-6}) \cdot (1.6 \times 10^{-19}) = 1.6 \times 10^{0} = 1.6 \]
- Now calculate \( v_d \):
\[ v_d = \frac{16}{1.6} = 10 \, m/s \]

### Step 3: Formatting the answer

3. **Express the drift speed in the required format**:
- The problem states that \( v_d = 0.00x \, m/s \).
- Since \( v_d = 0.010 \, m/s \), we can see that \( x = 10 \).

### Final Answer:
The value of \( x \) is \( 2 \).

---