Animated Solution for Physics - Semiconductors: If the ratio of the concentration of electrons to that of holes in a semiconductor is $\frac{7}{5}$ and the ratio of currents is $\frac{7}{4}$, then what is the ratio of their drift velocities?

When you connect a simple copper wire to a battery, the story is quite straightforward: a sea of free electrons begins to drift, creating an electric current. But when you replace that copper wire with a piece of semiconductor material, the plot thickens. Semiconductors are fascinating because they don't just have one type of charge carrier; they have two!

Imagine you are standing inside a silicon crystal. When an electric field is applied, you would see negatively charged electrons rushing in one direction (opposite to the field), and positively charged holes (which are essentially missing electrons) migrating in the exact opposite direction. Because they have opposite charges and move in opposite directions, their individual currents actually add up to give the total current!

To solve any problem involving the flow of charge carriers, we need a reliable mathematical tool. The fundamental equation connecting the macroscopic world of electric current to the microscopic world of moving charges is:

Let's break down this beautiful equation: $I$ is the electric current. $n$ is the charge carrier concentration (number of carriers per unit volume). $e$ is the elementary charge (the charge of a single electron or hole). $A$ is the cross-sectional area of the material. $v_d$* is the drift velocity (the average speed at which the carriers are pushed by the electric field).

In our specific problem, we are given the ratio of the electron concentration to the hole concentration ($n_e/n_h=7/5$) and the ratio of their respective currents ($I_e/I_h=7/4$). We need to find the ratio of their drift velocities.

Since both the electrons and the holes are flowing through the exact same piece of semiconductor, the cross-sectional area $A$ is identical for both. Furthermore, the magnitude of the charge $e$ is the same for an electron and a hole.

Let's write down the current equations for both carriers and divide them:

This is where the magic happens. The constants $e$ and $A$ appear in both the numerator and the denominator, so they cancel out perfectly! We are left with a much cleaner, more elegant relationship:

Now, we simply substitute the values given in the problem:

To isolate the ratio of the drift velocities, we just need to perform a quick cross-multiplication. We multiply both sides by $5/7$:

The $7$ in the numerator cancels with the $7$ in the denominator, leaving us with our final answer:

Why are Electrons Faster?

We found that the ratio is $5/4$, which means $(v_d{)}_e>(v_d{)}_h$. The electrons are drifting faster than the holes! But why?

In a semiconductor, electrons travel through the conduction band, where they are relatively free and have a lower "effective mass." Holes, on the other hand, travel through the valence band by jumping from one atomic bond to another. This process is inherently more sluggish, giving holes a higher effective mass and lower mobility. Therefore, for the exact same applied electric field, electrons will always attain a higher drift velocity than holes. It's a beautiful confirmation of quantum mechanics hidden inside a simple algebra problem!