Video Transcript
A semiconductor crystal is doped
with aluminum atoms. Which of the following graphs
correctly represents the relation between free electrons 𝑛 and aluminum ions 𝑁 sub
A−?
In this question, we’ve been asked
to identify the relationship between the number of free electrons and dopant ions in
a semiconductor that’s been doped with aluminum atoms.
To begin, it’s important to note
that a neutral aluminum atom has three electrons in its outermost shell. This means that aluminum is a
trivalent dopant atom. And so doping with aluminum atoms
produces a p-type semiconductor. Recall that aluminum atoms that
have been added to the semiconductor lattice can also be referred to as negative
acceptor ions, since they accept free electrons from the lattice to complete their
outermost electron shell.
This symbol represents the number
of acceptor ions in the semiconductor. The A stands for acceptor. And the negative sign refers to the
aluminum atom becoming a negatively charged ion once it accepts a free electron from
the lattice. The 𝑁 simply refers to the
concentration of these ions. Now, as more aluminum atoms accept
free electrons from the lattice, there are fewer free electrons remaining in the
lattice. Remember that the number of free
electrons in the lattice is represented by 𝑛. Thus, as the density of acceptor
ions increases, the density of free electrons decreases. It’s our job to recognize which of
these graphs best shows this relationship.
Notice that for each graph, the
vertical axis represents the density of free electrons, and the horizontal axis
represents the density of acceptor ions. The correct graph will show an
inverse relation between these quantities. In other words, as one quantity
increases, the other must decrease, and vice versa.
Now, looking at option (A), the
graph shows a horizontal line. This implies that adding more
acceptor ions doesn’t change the number of free electrons in the lattice. We know this is not correct, so
let’s eliminate (A). Next, (B) actually does show an
inverse relationship between the quantities represented by the graph’s axes. So (B) looks like a good
option. Moving on to (C), this graph shows
a straight line with a negative slope. Here, the concentration of free
electrons does decrease as the number of aluminum ions increases. So let’s keep this option around as
well. Next, (D) shows a straight line
with a positive slope. This suggests that more aluminum
ions means more free electrons, which we know is incorrect. So we should eliminate (D).
We’re left with two choices
then. To figure out which is correct,
let’s clear some space and refer to a formula that relates the two quantities in
question. Recall that for a p-type
semiconductor, the concentration of free electrons is given by this formula. Now, this term in the numerator is
a constant value that we actually don’t need to be concerned with to answer this
question. So, let’s turn this equation into a
proportionality by setting the constant term in the numerator equal to one, thus
holding its place in the fraction. The relationship then reads that
the amount of free electrons is proportional to one divided by the number of
acceptor ions. This is not a linear
relationship.
Because the quantity represented by
the horizontal axis is in the denominator, the graph of the function cannot be a
straight line. Thus, we should eliminate answer
choice (C). Rather, the graph is hyperbolic,
and looks like this. We can see that this corresponds to
answer choice (B). There is an inverse relationship
between the number of free electrons and the number of aluminum ions in the doped
semiconductor. So, (B) is the correct answer.
