Question

A potential barrier of \(0.6 \mathrm{~V}\) exists across a P-N junction. If the depletion region is \(1 \mu \mathrm{m}\) wide, what is the intensity of electric field in the region? (A) \(4 \times 10^{5} \mathrm{Vm}^{-1}\) (B) \(5 \times 10^{5} \mathrm{Vm}^{-1}\) (C) \(6 \times 10^{5} \mathrm{Vm}^{-1}\) (D) \(2 \times 10^{5} \mathrm{Vm}^{-1}\)

Short answer

(C) \(6 \times 10^{5} Vm^{-1}\)

Step-by-step solution

Identify Known Values

The given values are the potential barrier \(V = 0.6V\) and the width of the depletion region \(d = 1µm = 1 \times 10^{-6} m\) (converted from micrometers to meters for consistent units).

Identify the Unknown

The unknown value is intensity of the electric field \(E\) in the depletion region.

Apply the Electric Field Equation

The relationship between the electric field, potential difference and distance is given by the formula \(E = V / d\).

Carry out the Calculation

Simply substitute the given values into the equation, \[ E = \frac{V}{d} = \frac{0.6 V}{1 \times 10^{-6} m} = 6 \times 10^{5} Vm^{-1} \]

Comparing the Answer to the Options

From the calculated value, the intensity of the electric field in the depletion region is found to be \(6 \times 10^{5} Vm^{-1}\). This answer matches with option (C). Therefore, the correct answer is (C) \(6 \times 10^{5} Vm^{-1}\).