# Video: Quantum Tunneling Probability through a Potential Barrier

An electron in a long, organic molecule used in a dye laser behaves approximately like a quantum particle in a box of width 4.18 nm. Find the wavelength of the emitted photon when the electron makes a transition from the second excited state to the first excited state.

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### Video Transcript

An electron in a long, organic molecule used in a dye laser behaves approximately like a quantum particle in a box of width 4.18 nanometers. Find the wavelength of the emitted photon when the electron makes a transition from the second excited state to the first excited state.

The statement tells us that the box has a width of 4.18 nanometers, which we’ll call 𝐿. We wanna solve for the wavelength of the photon that’s emitted when the electron transitions from the second to the first excited state, we’ll call that wavelength 𝜆.

Let’s start our solution by drawing a diagram. In this box, which can be modelled as an infinitely deep quantum well, we have an electron which starts out at the second excited state, where 𝑛 equals three. The electron then transitions down to the 𝑛 equals two state losing energy, which is released as a photon, a particle of light, with wavelength 𝜆.

To start solving for that wavelength, we’ll first recall the discrete energy levels of a particle in a box, 𝐸 sub 𝑛. The energy levels in this quantized system, described by the quantum number 𝑛, are equal to 𝑛 squared times ℎ squared over eight 𝑚𝐿 squared, where 𝑚 is the mass of the particle contained in the box, 𝐿 is the width of the box, and ℎ is Planck’s constant which we will assume in this problem is exactly equal to 6.626 times 10 to the negative 34th joule seconds. The box in our example has a width of 𝐿, 4.18 nanometers. The electron in our box exchanges energy as it goes through its transition. That changing in energy, Δ𝐸, is equal to the energy of the electron at 𝑛 equals three minus the energy of the electron at 𝑛 equals two.

Using our particle in a box equation for these two energy levels, we see that Δ𝐸 equals three squared ℎ squared over eight 𝑚𝐿 squared minus two squared ℎ squared over eight 𝑚𝐿 squared, or five ℎ squared over eight 𝑚𝐿 squared. This change in energy that the electron undergoes is equal to the energy of the photon that’s emitted. We can recall that photon energy 𝐸 equals Planck’s constant times the frequency 𝑓, or ℎ times the speed of light 𝑐 over the wavelength 𝜆. So we can write: the photon energy ℎ𝑐 over 𝜆 is equal to Δ𝐸. In this equation, factors of ℏ cancel from each side, and we want to then rearrange to solve for the wavelength 𝜆.

When we do, we see that it’s equal to eight times the mass of the electron times the speed of light times the width of the box squared divided by five times Planck’s constant. We’ll assume that the mass of the electron is exactly 9.1 times 10 to the negative 31st kilograms and that 𝑐 is 3.00 times 10 to the eighth meters per second. We’re now ready to plug in for 𝑚, 𝑐, ℎ, and our given width 𝐿 to solve for 𝜆. When we do, we’re careful to use units of meters in our length value, so that those SI units agree with the units of our other variables.

Calculating 𝜆, we find it’s equal to 11.5 times 10 to the negative six meters, or 11.5 microns. That’s the wavelength of the photon emitted by the electron making this transition.