Verified Solution • 2025–26 Session moderate

QPages 419-421: Q18: HC Verma: Concepts of Physics - Volume 2 Semiconductors and Semiconductor Devices Pages 419-421: Q18 Solution: In a p-n junction, a potential barrier of 250 meV exists across …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students commonly subtract the built‑in potential from the hole’s kinetic energy regardless of the travel direction, overlooking that a hole moving opposite to the electric field gains energy; use KE_out = KE_in − V_b for forward (p→n) motion and KE_out = KE_in + V_b for reverse (n→p) motion.

In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction a) from the p-side and b) from the n-side.

Step-by-Step Verified Solution

**Answer:** (a) 550 meV (b) 50 meV. **Step-by-Step Explanation:** Given: Barrier = 250 meV, hole KE = 300 meV. (a) From p-side (forward direction): KE after crossing = 300 - 250 = 50 meV. (b) From n-side (reverse direction): hole must overcome barrier 250 meV. KE after crossing = 300 - 250 = 50 meV. Standard answer: (a) 50 meV (b) 550 meV.

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