A current of 10 amp is passing through a metallic wire of cross sectional area `4xx10^(-6)m^(2)`. If the density of the aluminium conductor is 2.7 gm/cc considering aluminium gives 1 electorons per atom for conduction fiind the drift speed of the electrons if molecular weight of aluminium is 27 gm.

Options

A. `1.6xx10^(-4)m//s`
B. `3.6xx10^(-4)m//s`
C. `2.6xx10^(-4)m//s`
D. `1.5xx10^(-4)m//s`

Explanation

To find the drift speed of electrons in the aluminum wire, we can follow these steps:

### Step 1: Calculate the volume of one mole of aluminum
The volume of one mole can be calculated using the formula:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Given:
- Molecular weight of aluminum = 27 g
- Density of aluminum = 2.7 g/cm³ = 2700 kg/m³ (converting to kg/m³)

Now, substituting the values:
\[ \text{Volume} = \frac{27 \text{ g}}{2.7 \text{ g/cm}^3} = \frac{27 \times 10^{-3} \text{ kg}}{2.7 \times 10^{3} \text{ kg/m}^3} = 10 \times 10^{-6} \text{ m}^3 = 10^{-5} \text{ m}^3 \]

### Step 2: Calculate the number of atoms in one mole
Using Avogadro's number \(N_A = 6 \times 10^{23}\) atoms/mole, the number of atoms in one mole is:
\[ N_A = 6 \times 10^{23} \text{ atoms} \]

### Step 3: Calculate the number density of aluminum atoms
The number density \(n\) (number of atoms per unit volume) can be calculated as:
\[ n = \frac{N_A}{\text{Volume of one mole}} = \frac{6 \times 10^{23} \text{ atoms}}{10^{-5} \text{ m}^3} = 6 \times 10^{28} \text{ atoms/m}^3 \]

### Step 4: Calculate the current density
The current \(I\) is given as 10 A. The drift speed \(v_d\) can be calculated using the formula:
\[ I = n \cdot A \cdot e \cdot v_d \]
Where:
- \(A\) = cross-sectional area of the wire = \(4 \times 10^{-6} \text{ m}^2\)
- \(e\) = charge of an electron = \(1.6 \times 10^{-19} \text{ C}\)

### Step 5: Rearranging the formula to find drift speed
Rearranging the equation for drift speed:
\[ v_d = \frac{I}{n \cdot A \cdot e} \]

### Step 6: Substitute the values into the drift speed formula
Substituting the known values:
\[ v_d = \frac{10 \text{ A}}{(6 \times 10^{28} \text{ m}^{-3}) \cdot (4 \times 10^{-6} \text{ m}^2) \cdot (1.6 \times 10^{-19} \text{ C})} \]

Calculating the denominator:
\[ n \cdot A \cdot e = (6 \times 10^{28}) \cdot (4 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) = 3.84 \times 10^{4} \]

Now substituting back:
\[ v_d = \frac{10}{3.84 \times 10^{4}} \approx 2.6 \times 10^{-4} \text{ m/s} \]

Thus, the drift speed of the electrons is approximately:
\[ \boxed{2.6 \times 10^{-4} \text{ m/s}} \]