# Question Video: Representing the Density of Free Electrons in a Donor Ion-Doped Semiconductor Physics • 9th Grade

A doped semiconductor that contains donor 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 donor ions in the semiconductor is represented by 𝑁_(D) ^(+). 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] 𝑛 = (𝑁_(D) ^(+)) − 𝑝 [B] 𝑛 = 𝑝 − (𝑁_(D) ^(+)) [C] 𝑛 = 𝑝 + (𝑁_(D) ^(+)) [D] 𝑛 = 𝑝/(𝑁_(D) ^(+)) [E] 𝑛 = (𝑁_(D) ^(+))/𝑝

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### Video Transcript

A doped semiconductor that contains donor 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 donor ions in the semiconductor is represented by 𝑁 subscript D superscript plus. 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 free electrons equals the density of donor ions minus the density of vacancies. (B) The density of free electrons equals the density of vacancies minus the density of donor ions. (C) The density of free electrons equals the density of vacancies plus the density of donor ions. (D) The density of free electrons equals the density of vacancies divided by the density of donor ions. (E) The density of free electrons equals the density of donor ions divided by the density of vacancies.

This question is asking about a doped semiconductor. However, let’s begin by thinking about a pure undoped sample. Recall that for a pure semiconductor, an electron becomes free by receiving enough thermal energy to overcome its atomic bond. And so, it leaves behind a vacancy or electron hole in its place. And by the way, throughout this question, we’re assuming that there is some thermal energy in the sample. If we were working at absolute zero, there would be no thermal energy, no free electrons, and no vacancies, which isn’t very interesting. So, we’re assuming the temperature and therefore the quantities 𝑛 and 𝑝 to be greater than zero.

So, for a pure semiconductor, we know that for each free electron, there’s a vacancy somewhere in the atomic lattice where that electron once was bound. Therefore, for a pure sample, we know that the density of free electrons must be equal to the density of vacancies or 𝑛 equals 𝑝. Now, let’s talk about our doped sample.

Positive donor ions have been added to the atomic lattice, so we know that the concentration of donor ions is greater than zero. Recall that they’re called positive donor ions because each one came to the lattice with one more electron than the surrounding atoms that made up the original undoped sample. And this extra electron gets donated to the lattice as a free electron, leaving behind a positively charged ion. Therefore, doping with these donor ions has increased the concentration of free electrons in the sample without creating any new vacancies in the sample.

So, 𝑛 does not equal 𝑝 for the doped semiconductor. And it’s our job to figure out which of these formulas does correctly model our doped sample. Answer choice (A) states that the density of free electrons equals the density of donor ions minus the density of vacancies. The subtraction of 𝑝 implies that the vacancies, which were all left behind by now freed electrons, somehow count against or take away from the concentration of free electrons. We know this isn’t true. Therefore, (A) is incorrect.

Answer choice (B) says that the density of free electrons equals the density of vacancies minus the density of donor ions, or that the density of donor ions somehow takes away from or decreases the density of free electrons. This isn’t true, either. We know that the donor ions donate free electrons, which increases not decreases the density of free electrons. (B) is also incorrect.

Now, before we look at (C), let’s skip ahead to answer choices (D) and (E). (D) supposes that the density of free electrons equals the density of vacancies divided by the density of donor ions. And (E) suggests that the density of free electrons equals the density of donor ions divided by the density of vacancies. Let’s think about these. Again, recall that for an undoped sample, 𝑛 must equal 𝑝. So, suppose that we have a doped sample and that we can somehow make it more and more pure as if we’re steadily removing the donor ions and the extra electrons they brought with them. What should happen is that as the concentration of donor ions goes to zero or as the sample becomes more pure, 𝑛, the concentration of free electrons, goes to the value 𝑝, the concentration of vacancies. But both of these formulas suggest otherwise.

Looking at (D), as the concentration of donor ions goes towards zero or as the denominator goes towards zero, this entire quantity and therefore the concentration of free electrons would increase without limit to ∞. And answer choice (E) says that as the concentration of donor ions goes to zero or as the numerator goes to zero, this entire quantity and therefore the concentration of free electrons would also go to zero. But we know that as a sample gets more and more pure, the concentration of free electrons just becomes the concentration of vacancies, which is neither ∞ nor zero. Therefore, (D) and (E) are incorrect as well.

Finally, let’s look at (C), which says that the density of free electrons equals the density of vacancies plus the density of donor ions. This means that both the vacancies and donor ions make a positive contribution to the concentration of free electrons. We know this is true. When we dope with donor ions, the density of free electrons equals the amount of free electrons before doping plus the amount of new free electrons added by the donor ions. Therefore, in our doped semiconductor, the density of free electrons equals the density of vacancies plus the density of donor ions.