Question

When you have a p-n junction and you connect it to a DC voltage, this will lead
$A.$ The electrons near to the gap between the p-n junction from the n type material to move to the p side.
$B.$ The lack of electrons in the n type side will lead to holes.
$C.$ The holes near to the gap between the p-n junction from the p type material to move to the n side.
$D.$ Both $A$ and $C$.

Answer 1

Hint: We will use the concept of a p-n junction diode and its characteristics under DC voltage. A p-n junction diode is a simple semiconductor device which passes current in one direction only. The current which is unidirectional is called DC current and the voltage corresponding to it is called DC voltage.

Complete step by step answer:
If we connect a p-n junction to a DC voltage then we will get the following phenomena:-
$(i)$ When we join a p type semiconductor material to n type semiconductor together then a very large density gradient is produced between both the sides of the p-n junction.
$(ii)$ Because of this large density gradient the electrons from the donor impurities start to migrate across the p-n junction to fill up the holes created in the p type semiconductor which produce negative ions.
$(iii)$ The movement of electrons is from n type to p type across the p-n junction. Due to this movement the positive charged donor ions are left behind to the negative side of the p-n junction.
$(iv)$ Due to this movement the charge density in the p type semiconductor is filled with negatively charged acceptor ions and the charge density in the n type semiconductor becomes positive.
On the basis of our above analysis we can conclude that both $A$ and $C$ are correct.

Hence, option $(C)$ is correct.

Note:
In solving the problems of a p-n junction nature of biasing should also be considered. Characteristics of a p-n junction diode also change according to DC or AC source. It should be noted that holes move from p side to n side and electrons from n side to p side across the p-n junction.