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œ†.

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œ†.

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„Ž squared over eight ๐‘š๐ฟ squared, where ๐‘š is the mass of the particle contained in the box, ๐ฟ is the width of the box, and โ„Ž 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„Ž squared over eight ๐‘š๐ฟ squared minus two squared โ„Ž squared over eight ๐‘š๐ฟ squared, or five โ„Ž squared over eight ๐‘š๐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„Ž times the speed of light ๐‘ over the wavelength ๐œ†. So we can write: the photon energy โ„Ž๐‘ over ๐œ† 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œ†.

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‘š, ๐‘, โ„Ž, and our given width ๐ฟ to solve for ๐œ†. 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œ†, 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.

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