Video: Photon Wavelength Produced by a Transition between One-Dimensional Bound States of a Quantum Particle

An electron is confined to a box of width 0.250 nm. What is the wavelength of photons emitted when the electron transitions between the second excited state and the ground state?

03:40

Video Transcript

An electron is confined to a box of width 0.250 nanometers. What is the wavelength of photons emitted when the electron transitions between the second excited state and the ground state?

Weโ€™re told in this statement that the box has a width 0.250 nanometers which weโ€™ll call ๐ฟ. We want to solve for the wavelength of the photons emitted when the electron transitions from the second excited state to the ground state. Letโ€™s start our solution by drawing a diagram. In this example, we have an infinitely deep box the electron is not able to escape out of. Weโ€™re told the electron starts at the second excited state where ๐‘› equals three, and it then transitions down to the ground state where ๐‘› is one. In the process, the electron emits a photon with a wavelength ๐œ†. Itโ€™s that wavelength we want to solve for.

To do that, we can recall that for a particle in a box, like we have here, the energy of the particle at the ๐‘›th energy level, ๐ธ sub ๐‘›, equals ๐‘› squared times โ„Ž squared, where โ„Ž is Planckโ€™s constant which weโ€™ll take to equal exactly 6.626 times 10 to the negative 34th joule seconds, all of which is divided by eight times the mass of the electron ๐‘š times the width of the box ๐ฟ squared.

In our case, we have an electron that starts out at ๐‘› equals three and then moves to ๐‘› equals one. By the particle in a box energy equation, that equals nine โ„Ž squared over eight ๐‘š๐ฟ squared minus one โ„Ž squared over eight ๐‘š๐ฟ squared, which simplifies to โ„Ž squared over ๐‘š๐ฟ squared. Now thatโ€™s the change in energy of the electron, but what about the photon thatโ€™s emitted? We can remember that photon energy equals โ„Ž times the frequency ๐‘“ or โ„Ž times ๐‘ over ๐œ†. All the energy of the electronโ€™s transition is delivered to the emitted photon. That means we can write: โ„Ž squared over ๐‘š๐ฟ squared is equal to the energy of the emitted photon โ„Ž ๐‘ over ๐œ†. We can cancel out a factor of Planckโ€™s constant from each side and then rearrange this equation to solve for ๐œ†. And we find itโ€™s equal to ๐‘š, the mass of the electron, times ๐‘, the speed of light, times the width of the box, ๐ฟ squared, divided by โ„Ž, Planckโ€™s constant.

In this example, weโ€™ll treat the mass of the electron ๐‘š as exactly 9.1 times 10 to the negative 31st kilograms and the speed of light ๐‘ as exactly 3.00 times 10 to the eighth meters per second.

When we plug in for ๐‘š, ๐‘, ๐ฟ, and โ„Ž being careful to write our value for ๐ฟ in units of meters, when we calculate ๐œ†, we find that, to three significant figures, it is 25.8 nanometers. That is the wavelength of the emitted photon.

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