Question Video: Representing the Density of Vacancies in an Acceptor Ion-Doped Semiconductor Physics • 9th Grade

A doped semiconductor that contains acceptor ions and is at thermal equilibrium is modeled using three variables. The density of free electrons in the semiconductor is represented by 𝑛. The density of acceptor ions in the semiconductor is represented by 𝑁 subscript 𝐴 superscript minus. The density of vacancies in the semiconductor is represented by 𝑝. Which of the following formulas correctly represents the relationship between these variables in the semiconductor? [A] 𝑝 = 𝑁 subscript 𝐴 superscript minus + 𝑛 [B] 𝑝 = 𝑛 − 𝑁 subscript 𝐴 superscript minus [C] 𝑝 = 𝑁 subscript 𝐴 superscript minus − 𝑛 [D] 𝑝 = 𝑛/𝑁 subscript 𝐴 superscript minus [E] 𝑝 = 𝑁 subscript 𝐴 superscript minus/𝑛

05:41

Video Transcript

A doped semiconductor that contains acceptor ions and is at thermal equilibrium is modeled using three variables. The density of free electrons in the semiconductor is represented by 𝑛. The density of acceptor ions in the semiconductor is represented by 𝑁 subscript 𝐴 superscript minus. The density of vacancies in the semiconductor is represented by 𝑝. Which of the following formulas correctly represents the relationship between these variables in the semiconductor? (A) The density of vacancies equals the density of acceptor ions plus the density of free electrons. (B) The density of vacancies equals the density of free electrons minus the density of acceptor ions. (C) The density of vacancies equals the density of acceptor ions minus the density of free electrons. (D) The density of vacancies equals the density of free electrons divided by the density of acceptor ions. (E) The density of vacancies equals the density of acceptor ions divided by the density of free electrons.

This question is asking about a doped semiconductor. But we’re gonna begin by first thinking about a pure undoped sample. Recall that for an undoped sample, free electrons and vacancies are created in pairs whenever an electron receives enough thermal energy to break its atomic bond, become freed, and leave behind a vacancy or electron hole in its place. Recall that in order to have thermal energy, the temperature of the semiconductor must be above absolute zero. If we had a pure sample at absolute zero, there would be no vacancies and no free electrons. So, throughout the course of this question, we’re going to assume the temperature and, therefore, the quantities 𝑝 and 𝑛 to be greater than zero.

So, for a pure semiconductor, we know that for each vacancy in an electron shell, there’s a free electron somewhere in the sample that used to occupy a space in that shell. Therefore, in a pure semiconductor, the density of vacancies must be equal to the density of free electrons, or 𝑝 equals 𝑛.

Keeping this in mind, let’s move on to think about our doped sample, which contains acceptor ions. So we know that the concentration of acceptor ions is greater than zero. Recall that acceptor ions are first introduced to the atomic lattice with one fewer outermost electron than the surrounding atoms that made up the original undoped sample. So there’s room in its outermost shell for one more electron. And that room for one more possible electron constitutes one vacancy. Therefore, doping with these acceptor ions has increased the concentration of vacancies in the semiconductor without creating any new free electrons. So, once the sample is doped, we can no longer say that 𝑝 equals 𝑛.

So now we need to figure out which of these formulas actually does correctly model our doped semiconductor. Let’s start down here, where answer choice (E) says that the density of vacancies equals the density of acceptor ions divided by the density of free electrons. And answer choice (D) is pretty similar, saying that the density of vacancies equals the density of free electrons divided by the density of acceptor ions. To determine whether either of these makes sense, let’s consider the process of taking a doped sample and somehow making it more and more pure, as if we’re removing the acceptor ions along with the vacancies they came with. If we were to do this, we would be reducing the density of acceptor ions down closer and closer to zero.

Let’s see how these two formulas would react to this process. First, looking at (E), if the density of acceptor ions goes to zero, that means the numerator would go to zero, causing this entire quantity and, therefore, the density of vacancies to go to zero as well. Now, looking at (D), if the density of acceptor ions goes to zero, the denominator would be going to zero, causing this entire quantity and, therefore, the density of vacancies to increase without limit to ∞. But because for a pure semiconductor 𝑝 equals 𝑛, we know that as a doped sample becomes more and more pure as the density of acceptor ions goes to zero, the density of vacancies should just become the density of free electrons, which as we established earlier is not zero and it’s definitely not ∞. So we know (E) and (D) are incorrect.

Option (C) says that the density of vacancies equals the density of acceptor ions minus the density of free electrons. The subtraction of 𝑛 implies that the concentration of free electrons somehow counts against the concentration of vacancies. This isn’t true. Each free electron had to have created a vacancy. And therefore, the density of free electrons contributes to, not takes away from, the density of vacancies. Answer choice (C) is wrong as well.

Answer choice (B) says that the density of vacancies equals the density of free electrons minus the density of acceptor ions. This implies that adding acceptor ions would decrease the density of vacancies. But we’ve already established that adding acceptor ions does the opposite. It increases the density of vacancies. Therefore, (B) is incorrect as well.

And we’re left with answer choice (A), which says that the density of vacancies equals the density of acceptor ions plus the density of free electrons. In a doped sample, the total amount of vacancies must equal the amount of vacancies before it was doped plus the amount of vacancies contributed by the acceptor ions. This agrees with the formula. And therefore, in our doped sample, the density of vacancies equals the density of acceptor ions plus the density of free electrons.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.