An electron with initial kinetic energy 32.0 electron volts encounters a square potential barrier with height 41.0 electron volts and width 0.250 nanometers. Find the probability, as a percentage, that the electron will tunnel through this barrier.
The electron initial kinetic energy is 32.0 electron volts. We’ll call 𝐸. The barrier with height 41.0 electron volts has a value we’ll call 𝑉 sub zero. The barrier is 0.250 nanometers wide. That value we’ll call 𝐿. We’re looking for a probability that an electron with this energy will tunnel through a barrier like this. We’ll call this probability capital 𝑃.
We can start out by sketching this situation. If we have a vertical and a horizontal axis with position on the horizontal and electric potential on the vertical, then if we draw in our barrier, we know the height of that barrier has a potential 𝑉 sub zero. Then if we show our electron approaching the barrier, that electron comes in with an energy 𝐸 less than 𝑉 zero. Given that the barrier has a width of 𝐿, we want to know what’s the chance or the probability that the electron when it hits the barrier, will be able to tunnel through to the other side. This is a little like being able to walk through a wall. Something that classically is impossible. But in the world of quantum mechanics, there’s a nonzero chance it can happen.
To figure out this probability, let’s recall that the mathematical equation describing tunneling probability has this form, where 𝐸 is the electron energy, 𝑉 zero is the barrier height, 𝑥 is the barrier width, and 𝑘 as units of wavenumbers and is equal to one over ℎ times the square root of eight 𝜋 squared times the mass of the particle trying to tunnel multiplied by the difference between the barrier height and the particle energy.
Let’s begin solving for the tunneling probability by solving for 𝑘. Before we enter in values to solve for 𝑘, let’s define ℎ and 𝑚. Planck’s constant ℎ, we’ll treat as exactly 6.626 times 10 to the negative 34th joule seconds. Electron mass 𝑚, we’ll treat as exactly 9.1 times 10 to the negative 31st kilograms. We can now plug in for all the values to solve for 𝑘. The values we see to plug in for 𝑘 are mostly expected. We have the mass. We have Planck’s constant. And we have our energy difference between the barrier and the particle energy. The one unexpected value is this last one under the square root sign. This is there to convert our energy, in units initially of electron volts, into a result in units of joules. We put this conversion factor there to make it agree with the units in the rest of our expression. When we calculate out 𝑘, we find it’s approximately equal to 1.5362 times 10 to the 10th inverse meters.
Now the 𝑘 is known, we can use that value in our expression for tunneling probability. In our version of the tunneling probability equation, we have 𝑃 rather than capital 𝑇. And we’re using 𝐿 for the width of our barrier. We can now plug in for the values in this equation. We’ve all our values plugged in. Notice that we use units of meters in our expression for 𝐿, converting it from nanometers. And speaking of units, we see that the units of meters in the exponent cancels out. So the exponent is unitless which is good. And also that the units of electron volts cancel out. So our final answer will be unitless. This is a positive sign.
When we calculate the value for 𝑃, we find a decimal of approximately 0.00130 or, as a percent, 0.130. This is the probability that an incident electron of the barrier will tunnel through.