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

Explanation

To find the electric field that can give an average energy of 2 eV to an electron in a metal with a mean free path of \(4 \times 10^{-8} \, \text{m}\), we can follow these steps:

### Step 1: Understand the relationship between energy, electric field, and mean free path
The work done on an electron by the electric field is equal to the energy gained by the electron. This can be expressed as:
\[ \text{Work Done} = \text{Charge} \times \text{Electric Field} \times \text{Mean Free Path} \]
In this case, the average energy gained by the electron is given as \(2 \, \text{eV}\).

### Step 2: Convert energy from electron volts to joules
We need to convert the energy from electron volts to joules for our calculations. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, \(2 \, \text{eV}\) can be converted to joules:
\[ 2 \, \text{eV} = 2 \times 1.6 \times 10^{-19} \, \text{J} = 3.2 \times 10^{-19} \, \text{J} \]

### Step 3: Write down the equation for work done
From the relationship established in Step 1, we can write:
\[ 3.2 \times 10^{-19} \, \text{J} = e \times E \times (4 \times 10^{-8} \, \text{m}) \]
where \(e\) is the charge of an electron, approximately \(1.6 \times 10^{-19} \, \text{C}\), and \(E\) is the electric field in \(V/m\).

### Step 4: Substitute the values into the equation
Substituting the values into the equation gives:
\[ 3.2 \times 10^{-19} = (1.6 \times 10^{-19}) \times E \times (4 \times 10^{-8}) \]

### Step 5: Solve for the electric field \(E\)
Rearranging the equation to solve for \(E\):
\[ E = \frac{3.2 \times 10^{-19}}{(1.6 \times 10^{-19}) \times (4 \times 10^{-8})} \]

### Step 6: Calculate the value of \(E\)
Calculating the denominator:
\[ (1.6 \times 10^{-19}) \times (4 \times 10^{-8}) = 6.4 \times 10^{-27} \]
Now substituting 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 required to give an average energy of \(2 \, \text{eV}\) to an electron in the metal is:
\[ E = 5 \times 10^{7} \, \text{V/m} \]