Question

The energy band diagram of a p-type semiconductor bar of length $L$ under equilibrium condition (i.e ., the Fermi energy $E_F$ is constant) is shown in the figure.The valence 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. $\frac{\Delta }{2qL}$
• B. $\frac{3\Delta }{2qL}$
• C. $\frac{\Delta }{qL}$
• D. $\frac{2\Delta }{qL}$

Answer 1

The correct option is C $\frac{\Delta }{qL}$

Ans : (d)

The built in electric field is due to non- uniform doping (the semiconductor is under equilibrium) and Electric field is basically the negative gradient of potential Energy .These two concepts helps to get the answer .
$E=\frac{1}{q}\frac{d{E}_v}{dx}$
$=\frac{1}{q}\frac{\Delta }{L}=\frac{\Delta }{qL}$