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

Answer

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

### Step-by-Step Solution:

1. **Understanding the Energy Relation**:
The average energy (E_avg) gained by an electron in an electric field can be expressed as:
\[ E_{\text{avg}} = q \cdot E \cdot \lambda \]
where:
- \(E_{\text{avg}}\) is the average energy (in joules),
- \(q\) is the charge of the electron (approximately \(1.6 \times 10^{-19} \, \text{C}\)),
- \(E\) is the electric field (in V/m),
- \(\lambda\) is the mean free path (in meters).

2. **Convert Energy from eV to Joules**:
Given that the average energy is \(2 \, \text{eV}\), we need to convert this energy into joules:
\[ E_{\text{avg}} = 2 \, \text{eV} = 2 \times 1.6 \times 10^{-19} \, \text{J} = 3.2 \times 10^{-19} \, \text{J} \]

3. **Substituting Known Values**:
Now we can substitute the known values into the energy relation:
\[ 3.2 \times 10^{-19} = (1.6 \times 10^{-19}) \cdot E \cdot (4 \times 10^{-8}) \]

4. **Rearranging the Equation**:
Rearranging the equation to solve for \(E\):
\[ E = \frac{3.2 \times 10^{-19}}{(1.6 \times 10^{-19}) \cdot (4 \times 10^{-8})} \]

5. **Calculating the Electric Field**:
Now we can calculate \(E\):
\[ E = \frac{3.2 \times 10^{-19}}{6.4 \times 10^{-27}} = 5 \times 10^{7} \, \text{V/m} \]

### Final Answer:
The electric field required is:
\[ E = 5 \times 10^7 \, \text{V/m} \]