There are `8.4xx10^(22)` free electrons per `cm^(3)` in copper. The current in the wire is `0.21A` (e=`1.6xx10^(-19)C`). Then the drifts velocity of electrns in a copper wire of `1mm^(2)` cross section, will be:-

Options

A. `2.12xx10^(-5)m//s`
B. `0.78xx10^(-5)m//s`
C. `1.56xx10^(-5)m//s`
D. None of these

Answer

To find the drift velocity of electrons in a copper wire, we can use the formula:

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

Where:
- \( I \) is the current (in amperes)
- \( n \) is the number of free electrons per unit volume (in \( m^{-3} \))
- \( A \) is the cross-sectional area of the wire (in \( m^{2} \))
- \( e \) is the charge of an electron (in coulombs)
- \( v_d \) is the drift velocity (in \( m/s \))

### Step 1: Convert the number of free electrons from \( cm^{-3} \) to \( m^{-3} \)

Given:
\[ n = 8.4 \times 10^{22} \, cm^{-3} \]

To convert to \( m^{-3} \):
\[ n = 8.4 \times 10^{22} \, cm^{-3} \times 10^{6} \, m^{-3} = 8.4 \times 10^{28} \, m^{-3} \]

### Step 2: Convert the cross-sectional area from \( mm^{2} \) to \( m^{2} \)

Given:
\[ A = 1 \, mm^{2} \]

To convert to \( m^{2} \):
\[ A = 1 \, mm^{2} = 1 \times 10^{-6} \, m^{2} \]

### Step 3: Use the formula to find the drift velocity

Given:
- Current, \( I = 0.21 \, A \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, C \)

Using the formula:
\[ I = n \cdot A \cdot e \cdot v_d \]

Rearranging for \( v_d \):
\[ v_d = \frac{I}{n \cdot A \cdot e} \]

Substituting the values:
\[ v_d = \frac{0.21}{(8.4 \times 10^{28}) \cdot (1 \times 10^{-6}) \cdot (1.6 \times 10^{-19})} \]

### Step 4: Calculate \( v_d \)

Calculating the denominator:
\[ n \cdot A \cdot e = (8.4 \times 10^{28}) \cdot (1 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) \]
\[ = 8.4 \times 1.6 \times 10^{28 - 6 - 19} \]
\[ = 13.44 \times 10^{3} \]
\[ = 1.344 \times 10^{4} \]

Now substituting back into the drift velocity formula:
\[ v_d = \frac{0.21}{1.344 \times 10^{4}} \]
\[ = 1.56 \times 10^{-5} \, m/s \]

### Final Answer:
The drift velocity of electrons in the copper wire is:
\[ v_d = 1.56 \times 10^{-5} \, m/s \]