Question Video: Quantum Tunneling Probability through a Potential Barrier

An electron with kinetic energy 2.0 MeV encounters a potential energy barrier of height 16.0 MeV and width 2.00 nm. What is the probability that the electron emerges on the other side of the barrier?

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Video Transcript

An electron with kinetic energy 2.0 megaelectron volts encounters a potential energy barrier of height 16.0 megaelectron volts and width 2.00 nanometers. What is the probability that the electron emerges on the other side of the barrier?

We can call the electron kinetic energy of 2.0 megaelectron volts 𝐸. And we’ll call the barrier height of 16.0 MeV 𝑣 sub zero. The barrier width, 2.00 nanometers, we’ll call capital 𝐿. We want to solve for the probability that an electron with this energy is able to pass through the barrier. We’ll call that probability capital 𝑃.

Let’s start off by drawing a sketch of our scenario. Here we have a vertical and a horizontal axis. The horizontal axis shows position, and the vertical axis shows potential. If we draw in our potential barrier, we can see that the height of the barrier on the potential axis is 𝑣 zero. Then if we draw in the electron as it approaches the barrier and trace that energy value back to the potential axis, we find it it’s capital 𝐸, which is eight times smaller than 𝑣.

Now we imagine that there is some chance that the electron will travel through this barrier and out to the other side, that that successful tunneling will happen with the probability 𝑃. And that’s what we’re looking to solve for. To do that, we’ll recall the mathematical relationship for tunneling. This relationship involves several terms.

The first term, 𝐸, is the energy of the particle trying to tunnel through the barrier. The second term, 𝑣 sub zero, is the height of the barrier — not in units of meters such as a distance, but in units of potential. In the exponent, we see the term representing the width of the barrier. This dimension is indeed in units of distance. And finally there’s 𝑘.

𝑘 has units of wavenumber — that is, inverse meters — and is given by one over Planck’s constant, ℎ, times the square root of eight 𝜋 squared times the mass of the particle tunneling through times the difference between the barrier height and the particle energy. Before going further, we can take a moment to define ℎ and 𝑚 for this problem.

The value of Planck’s constant we’ll use is equal to 6.626 times 10 to the negative 34th joule seconds. And the value of 𝑚 we’ll use for the electrons mass is 9.1 times 10 to the negative 31st kilograms. We’re now ready to plug in and solve for 𝑘. When we do, plugging in for values of our mass, our barrier height, our electron energy, and Planck’s constant, you’ll notice there’s an extra term in this equation.

That term is there to convert from electron volts to units of joules, making our units consistent across this expression. Also notice that we’ve converted our barrier energies and electron energies into units of electron volts from megaelectron volts. When we plug all these values in and solve for 𝑘, we get a value of approximately 1.916 times 10 to the 13th inverse meters.

Having solved for a value for 𝑘, we’re now ready to compute the tunneling probability. When we translate that equation to our scenario, we use 𝑃 instead of capital 𝑇 and we have the width of our barrier 𝐿 in place of 𝑥. All the values in this expression are given or they’ve been calculated. So we’re now ready to solve for 𝑃.

When we plug in for all our values, making sure to convert megaelectron volts to units of electron volts and using units of meters for the width of a barrier, we calculate a probability 𝑃 of roughly 6.061 times 10 to the negative 33284.

This number is so fantastically small that effectively it is zero. With this barrier height and particle energy and barrier width, the chance of the particle tunneling through the barrier is effectively zero.

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