### 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 weโll call ๐ฟ. We wanna solve for the wavelength of the photon thatโs emitted when the electron transitions from the second to the first excited state, weโll call that wavelength ๐.

Letโs 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, weโll first recall the discrete energy levels of a particle in a box, ๐ธ 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 ๐๐ฟ squared, where ๐ is the mass of the particle contained in the box, ๐ฟ is the width of the box, and โ is Planckโs 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 ๐ฟ, 4.18 nanometers. The electron in our box exchanges energy as it goes through its transition. That changing in energy, ฮ๐ธ, 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 ฮ๐ธ equals three squared โ squared over eight ๐๐ฟ squared minus two squared โ squared over eight ๐๐ฟ squared, or five โ squared over eight ๐๐ฟ squared. This change in energy that the electron undergoes is equal to the energy of the photon thatโs emitted. We can recall that photon energy ๐ธ equals Planckโs constant times the frequency ๐, or โ times the speed of light ๐ over the wavelength ๐. So we can write: the photon energy โ๐ over ๐ is equal to ฮ๐ธ. 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 itโs 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 Planckโs constant. Weโll 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. Weโre now ready to plug in for ๐, ๐, โ, and our given width ๐ฟ to solve for ๐. When we do, weโre 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 itโs equal to 11.5 times 10 to the negative six meters, or 11.5 microns. Thatโs the wavelength of the photon emitted by the electron making this transition.