### Video Transcript

An electron with an energy of 5.00 electron volts impacts on a barrier of width 0.30 nanometers. Find the probability that the electron will tunnel through the barrier if the barrier height is 7.00 electron volts.

Let’s call the energy of the electron, 5.00 electron volts, 𝐸 and we’ll call the barrier height of 7.00 electron volts 𝑉. Along with these energies, we’re told that the barrier width is 0.30 nanometers, a value we’ll call 𝑤. We want to solve for the probability that the electron will tunnel through the barrier, what we’ll call capital 𝑃.

Let’s begin our solution by drawing a diagram. In this example, we have an electron with energy 𝑒 encountering a barrier. Classically, 100 times out of 100 the electron would bounce off the barrier and not be able to pass through. But at this small scale with quantum mechanical effects at play, there is a nonzero chance that the electron will tunnel through the barrier, impossible from the classical energy point of view but possible in the quantum world. We want to solve for the probability of that happening. That is, if we fired a million electrons with the same energy 𝑒 at the barrier, how many would tunnel through?

With a barrier of this shape, there’s a mathematical relationship that tells us what that probability is. It says that the probability of an electron tunneling through a potential barrier is equal to 16 times the energy of the electron 𝐸 multiplied by the difference between the potential barrier, 𝑉, and 𝐸 divided by the energy of the potential barrier squared multiplied by 𝑒 to the negative 𝑘𝑤, where 𝑤 is the width of the barrier, and 𝑘 equals two 𝜋 over Planck’s constant, ℎ, times the square root of two times the object’s mass, that’s trying to tunnel through, multiplied by the potential difference between the barrier height and the energy of the approaching electron.

When we apply this relationship to our scenario, we see that for all the variables in this expression, we’ve been given values, with the exception of 𝑘. So let’s solve for 𝑘 now. As we do, we’ll assume that Planck’s constant, ℎ, is exactly 6.626 times 10 to the negative 34th joule seconds, and 𝑚, the mass of the approaching electron, is exactly 9.1 times 10 to the negative 31st kilograms.

When we plug in for the values of the variables in the expression for 𝑘, we will add in one extra factor, which is a conversion factor between electron volts and joules. And the purpose of this factor is to make the units across our entire expression consistent. Entering this expression on our calculator, we find that 𝑘, which is a wavenumber, is approximately 7.242 times 10 to the ninth inverse meters.

Knowing the value for 𝑘, we’re now ready to solve for the probability of tunneling 𝑃. When we plug in for the various values in the expression for 𝑃, we make sure to express the barrier width in units of meters to be consistent with the units in the rest of our expression. When we enter these values on our calculator, we find that the probability of an electron tunneling through the barrier is 0.042 or 4.2 percent.

So if we fired 1000 electrons with the energy 𝐸 at the barrier, approximately 42 would make it across.