Question

If the potential barrier across a p–n junction is 0.6 V. Then the electric field intensity, in the depletion region having the width of 6×10^–6 m, will be ______× 10⁵ N/C.

Answer 1

Correct Answer: 1

To find the electric field intensity in the depletion region of a p–n junction, we use the formula for electric field intensity \(E\):

\(E = \frac{V}{d}\)

where \(V\) is the potential barrier and \(d\) is the width of the depletion region.

Given:

• Potential barrier, \(V = 0.6 \text{ V}\)
• Width of depletion region, \(d = 6 \times 10^{-6} \text{ m}\)

Substitute the given values into the formula:

\(E = \frac{0.6}{6 \times 10^{-6}}\)

Calculate the electric field:

\(E = 0.1 \times 10^{6}=1.0 \times 10^5 \text{ N/C}\)

The electric field intensity is \(1.0 \times 10^5 \text{ N/C}\), which is within the expected range of 1,1.