Video Transcript
The diagram shows a lattice of silicon atoms in a semiconductor. The left side of the lattice has been doped with donor atoms. This is called the n-side. The right side of the lattice has been doped with acceptor atoms. This side of the lattice is called the p-side. The regions on either side of the dividing line are of equal size, and the ion
concentration is the same on both sides. The semiconductor is at thermal equilibrium. Toward which side of the lattice will free electrons tend to move by diffusion? (A) The p-side, (B) the n-side, (C) both sides.
The question asks us to work out which side of the junction the free electrons will
tend to move toward by diffusion. Here, we have a doped semiconductor for which the n-side is on the left and the
p-side is on the right. The n-side has been doped with phosphorus, which has five electrons in its outer
shell. This means that each phosphorus atom has one free electron. The p-side is doped with boron, which has three electrons in the outer shell. Therefore, each boron atom provides one vacancy or carrier of effective positive
charge.
The question states that the two regions have the same size and have the same
concentration of ions, that is, the density of acceptor or donor atoms. This means that there will be the same number of free electrons on the n-side and
vacancies on the p-side. Since negatively charged electrons repel each other, then free electrons at the
boundary between the n-side and the p-side will have a greater net repulsive force
on them from the n-side than from the p-side. That is, there’s an overall net force on these free electrons from the n-side to the
p-side. These electrons will therefore tend to move toward the p-side. This means that the correct answer is option (A), the p-side.