The mean free path of electrons in a metal is `4 xx 10^(-8)m` The electric field which can give on an average `2eV` energy to an electron in the metal will be in the units `V//m`

Options

A. `5xx10^(7)`
B. `8xx 10^(7)`
C. `5xx10^(-11)`
D. `8xx10^(-11)`

Explanation

To solve the problem, we need to find the electric field that can give an average energy of 2 eV to an electron in a metal, given that the mean free path of the electrons is \(4 \times 10^{-8} \, m\).

### Step-by-Step Solution:

1. **Understanding the Energy Given**:
The energy given to an electron is \(2 \, eV\). We know that \(1 \, eV = 1.6 \times 10^{-19} \, J\). Therefore, the energy in joules is:
\[ E = 2 \, eV = 2 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-19} \, J \]

2. **Relating Electric Field to Energy**:
The energy gained by an electron in an electric field \(E\) over a distance \(d\) (which is the mean free path in this case) can be expressed as:
\[ E = q \cdot V \]
where \(q\) is the charge of the electron (\(q = 1.6 \times 10^{-19} \, C\)) and \(V\) is the potential difference.

The potential difference \(V\) can also be related to the electric field \(E\) and the distance \(d\) as:
\[ V = E \cdot d \]

3. **Finding the Potential**:
Rearranging the energy equation gives:
\[ V = \frac{E}{q} = \frac{3.2 \times 10^{-19} \, J}{1.6 \times 10^{-19} \, C} = 2 \, V \]

4. **Calculating the Electric Field**:
Now, we can use the relationship between electric field \(E\), potential \(V\), and distance \(d\):
\[ E = \frac{V}{d} \]
Here, \(V = 2 \, V\) and \(d = 4 \times 10^{-8} \, m\). Substituting these values in:
\[ E = \frac{2 \, V}{4 \times 10^{-8} \, m} = \frac{2}{4 \times 10^{-8}} = \frac{1}{2 \times 10^{-8}} = 5 \times 10^{7} \, V/m \]

5. **Final Answer**:
Therefore, the electric field that can give an average of \(2 \, eV\) energy to an electron in the metal is:
\[ E = 5 \times 10^{7} \, V/m \]