# 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 weโll call ๐ฟ. We wanna solve for the wavelength of the photon thatโs emitted when the electron transitions from the second to the first excited state, weโll call that wavelength ๐.

Letโ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 ๐ equals three. The electron then transitions down to the ๐ equals two state losing energy, which is released as a photon, a particle of light, with wavelength ๐.

To start solving for that wavelength, weโll first recall the discrete energy levels of a particle in a box, ๐ธ sub ๐. The energy levels in this quantized system, described by the quantum number ๐, are equal to ๐ squared times โ squared over eight ๐๐ฟ squared, where ๐ is the mass of the particle contained in the box, ๐ฟ is the width of the box, and โ is Planckโ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 ๐ฟ, 4.18 nanometers. The electron in our box exchanges energy as it goes through its transition. That changing in energy, ฮ๐ธ, is equal to the energy of the electron at ๐ equals three minus the energy of the electron at ๐ equals two.

Using our particle in a box equation for these two energy levels, we see that ฮ๐ธ equals three squared โ squared over eight ๐๐ฟ squared minus two squared โ squared over eight ๐๐ฟ squared, or five โ squared over eight ๐๐ฟ squared. This change in energy that the electron undergoes is equal to the energy of the photon thatโs emitted. We can recall that photon energy ๐ธ equals Planckโs constant times the frequency ๐, or โ times the speed of light ๐ over the wavelength ๐. So we can write: the photon energy โ๐ over ๐ is equal to ฮ๐ธ. In this equation, factors of โ cancel from each side, and we want to then rearrange to solve for the wavelength ๐.

When we do, we see that itโ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 Planckโs constant. Weโll assume that the mass of the electron is exactly 9.1 times 10 to the negative 31st kilograms and that ๐ is 3.00 times 10 to the eighth meters per second. Weโre now ready to plug in for ๐, ๐, โ, and our given width ๐ฟ to solve for ๐. When we do, weโ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 ๐, we find itโs equal to 11.5 times 10 to the negative six meters, or 11.5 microns. Thatโs the wavelength of the photon emitted by the electron making this transition.