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 \( 2 \, \text{eV} \) to an electron in a metal with a mean free path \( \lambda = 4 \times 10^{-8} \, \text{m} \).

### Step-by-Step Solution:

1. **Understand the relationship between energy, electric field, and mean free path:**
The average energy gained by an electron in an electric field can be expressed as:
\[ \text{Energy} = Q \cdot E \cdot \lambda \]
where:
- \( Q \) is the charge of the electron,
- \( E \) is the electric field,
- \( \lambda \) is the mean free path.

2. **Convert the energy from eV to Joules:**
The energy given is \( 2 \, \text{eV} \). We need to convert this to Joules using the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ 2 \, \text{eV} = 2 \times 1.6 \times 10^{-19} \, \text{J} = 3.2 \times 10^{-19} \, \text{J} \]

3. **Substitute the known values into the equation:**
The charge of an electron \( Q \) is approximately \( 1.6 \times 10^{-19} \, \text{C} \). The mean free path \( \lambda \) is given as \( 4 \times 10^{-8} \, \text{m} \). Now we can substitute these values into the equation:
\[ 3.2 \times 10^{-19} \, \text{J} = (1.6 \times 10^{-19} \, \text{C}) \cdot E \cdot (4 \times 10^{-8} \, \text{m}) \]

4. **Rearranging the equation to solve for \( E \):**
To isolate \( E \), we rearrange the equation:
\[ E = \frac{3.2 \times 10^{-19} \, \text{J}}{(1.6 \times 10^{-19} \, \text{C}) \cdot (4 \times 10^{-8} \, \text{m})} \]

5. **Calculating the value of \( E \):**
First, calculate the denominator:
\[ (1.6 \times 10^{-19}) \cdot (4 \times 10^{-8}) = 6.4 \times 10^{-27} \]
Now substitute this back into the equation for \( E \):
\[ E = \frac{3.2 \times 10^{-19}}{6.4 \times 10^{-27}} = 5 \times 10^{7} \, \text{V/m} \]

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