Question

The energy band diagram of a p-type semiconductor bar of length \( L \) under equilibrium condition (i.e., the Fermi energy level \( E_F \) is constant) is shown in the figure. The valance band \( E_V \) is sloped since doping is non-uniform along the bar. The difference between the energy levels of the valence band at the two edges of the bar is \( \Delta \). 

If the charge of an electron is \( q \), then the magnitude of the electric field developed inside this semiconductor bar is 
 

• A. \( \frac{\Delta}{qL} \)
• B. \( \frac{2\Delta}{qL} \)
• C. \( \frac{\Delta}{2qL} \)
• D. \( \frac{3\Delta}{2qL} \)

Hint:

The electric field in a semiconductor is related to the gradient of the energy bands. For a linear change in energy over distance, the field is given by \( E = \frac{\Delta}{L} \).

Answer 1

The Correct Option is A

The electric field is related to the gradient of the energy band in the semiconductor. Since the energy difference across the length \( L \) of the bar is \( \Delta \), the electric field can be calculated as: \[ E = \frac{\Delta}{L} \] Now, considering the charge of the electron is \( q \), the magnitude of the electric field developed inside the semiconductor bar is: \[ E = \frac{\Delta}{qL} \] Thus, the correct answer is option (A).
Final Answer: (A) \( \frac{\Delta}{qL} \)