Video: 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 fourth and second excited states?

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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 fourth and second excited states?

This statement tells us that the box has a width of 0.250 nanometers, which we’ll call 𝐿. We want to solve for the wavelength of the photons that are emitted when the electron transitions between these two excited states. Let’s start by drawing a diagram of this box. In our scenario, we have a box, or an infinitely deep potential well, in which an electron exists starting out at the fourth excited state, or where 𝑛 equals five. We’re told the electron then transitions down to the second excited state where 𝑛 is equal to three, and in the process of making that transition, emits a photon with a wavelength πœ†.

To solve for that wavelength, we can recall the equation for the energy of a particle in a box. That energy, 𝐸 sub 𝑛, equals 𝑛 squared times β„Ž squared divided by eight π‘šπΏ squared, where 𝑛 is the energy level or principal quantum number of the particle, β„Ž is Planck’s constant which we’ll assume is exactly 6.626 times 10 to the negative thirty fourth joule seconds, π‘š is the particle’s mass, and 𝐿 is the width of the box.

In our case, Δ𝐸, the change in energy of the electron, equals its energy when 𝑛 equals five minus its energy when 𝑛 equals three, which is 25 β„Ž squared over β„Ž [eight] π‘šπΏ squared minus nine β„Ž squared over β„Ž [eight] π‘šπΏ squared or two β„Ž squared over π‘šπΏ squared. This change in the electron’s energy Δ𝐸 is equal to the energy of the emitted photon, which itself can be written: the energy of the photon 𝐸 equals β„Ž times frequency 𝑓 or β„Ž times 𝑐 over πœ†. So two times β„Ž squared over π‘šπΏ squared equals β„Ž 𝑐 over πœ†. We can see that a factor of Planck’s constant cancels from each side. And when we rearrange this equation to solve for πœ†, we see that it’s equal to π‘šπ‘πΏ squared over two β„Ž.

We’ll treat π‘š, the mass of the electron, as exactly 9.1 times 10 to the negative thirty first kilograms and 𝑐, the speed of light, as exactly 3.00 times 10 to the eighth meters per second. We can now plug in for π‘š, 𝑐, 𝐿 and β„Ž.

When we do so, being careful to use a value for 𝐿 in units of meters, and enter these values on our calculator, we find that πœ† Is 12.9 nanometers. That’s the wavelength of the photon emitted caused by a transition of an electron from the fourth to the second excited state.

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