The width of depletion region in p-n junction diode is `500 nm` and an intrinsic electric field of `6xx 10^(5) Vm^(-1)` is also found to exist in it. What is the kinetic energy which a conduction electron must have in order to diffuse from the n-side to p-side?

Options

A. 0.03 eV
B. 0.030 eV
C. 0.45 eV
D. 0.60 eV

Explanation

To solve the problem, we need to find the kinetic energy that a conduction electron must have to diffuse from the n-side to the p-side of a p-n junction diode. We will use the given width of the depletion region and the intrinsic electric field to calculate the potential barrier, and then use that to find the kinetic energy.

### Step-by-Step Solution:

1. **Convert the width of the depletion region to meters**:
\[ \text{Width of depletion region} = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \]

2. **Identify the intrinsic electric field**:
\[ \text{Electric field} (E) = 6 \times 10^{5} \, \text{Vm}^{-1} \]

3. **Calculate the potential barrier (Vb)**:
The potential barrier can be calculated using the formula:
\[ V_b = E \times \text{Width} \]
Substituting the values:
\[ V_b = (6 \times 10^{5} \, \text{Vm}^{-1}) \times (500 \times 10^{-9} \, \text{m}) \]
\[ V_b = 6 \times 500 \times 10^{-4} \, \text{V} = 3 \times 10^{-4} \, \text{V} = 0.3 \, \text{V} \]

4. **Calculate the kinetic energy (KE)**:
The kinetic energy of the conduction electron can be calculated using the formula:
\[ KE = e \times V_b \]
Where \( e \) is the charge of an electron, approximately \( 1.6 \times 10^{-19} \, \text{C} \). However, since we are looking for the kinetic energy in electronvolts (eV), we can directly use the potential barrier in volts:
\[ KE = V_b = 0.3 \, \text{eV} \]

5. **Final Answer**:
The kinetic energy which a conduction electron must have in order to diffuse from the n-side to the p-side is:
\[ KE = 0.3 \, \text{eV} \]