A potential barrier of 0.5V exists across a pn junction diode. If the width of depletion layer is `10^(-6) `m, then the strength of the electric field at the junction is

Options

A. `2 xx 10^5 Vm^(-1)` from n to p side
B. `2 xx 10^(-7) Vm^(-1) ` from p to n side
C. `5 xx 10^5 Vm^(-1) ` from n to p side
D. `5 xx 10^(-7) Vm^(-1) ` from p to n side

Answer

To solve the problem of finding the strength of the electric field at the junction of a pn junction diode, we can follow these steps:

### Step 1: Understand the relationship between potential, electric field, and distance
The electric field (E) across a potential barrier (V) can be calculated using the formula:
\[ E = \frac{V}{d} \]
where:
- \( E \) is the electric field strength (in volts per meter, V/m),
- \( V \) is the potential barrier (in volts, V),
- \( d \) is the width of the depletion layer (in meters, m).

### Step 2: Identify the given values
From the problem, we have:
- Potential barrier \( V = 0.5 \, \text{V} \)
- Width of the depletion layer \( d = 10^{-6} \, \text{m} \)

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the formula:
\[ E = \frac{0.5 \, \text{V}}{10^{-6} \, \text{m}} \]

### Step 4: Perform the calculation
Calculating the electric field:
\[ E = \frac{0.5}{10^{-6}} = 0.5 \times 10^{6} = 5 \times 10^{5} \, \text{V/m} \]

### Step 5: State the direction of the electric field
The electric field in a pn junction diode points from the n-region to the p-region due to the nature of charge carriers (electrons and holes). Thus, the electric field strength is:
\[ E = 5 \times 10^{5} \, \text{V/m} \, \text{(from n to p region)} \]

### Final Answer
The strength of the electric field at the junction is \( 5 \times 10^{5} \, \text{V/m} \).

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