# 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?

Weβre told in this statement that the box has a width 0.250 nanometers which weβll call πΏ. We want to solve for the wavelength of the photons emitted when the electron transitions from the second excited state to the ground state. Letβ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. Weβre told the electron starts at the second excited state where π equals three, and it then transitions down to the ground state where π is one. In the process, the electron emits a photon with a wavelength π. Itβ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 πth energy level, πΈ sub π, equals π squared times β squared, where β is Planckβs constant which weβ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 π times the width of the box πΏ squared.

In our case, we have an electron that starts out at π equals three and then moves to π equals one. By the particle in a box energy equation, that equals nine β squared over eight ππΏ squared minus one β squared over eight ππΏ squared, which simplifies to β squared over ππΏ squared. Now thatβs the change in energy of the electron, but what about the photon thatβs emitted? We can remember that photon energy equals β times the frequency π or β times π over π. All the energy of the electronβs transition is delivered to the emitted photon. That means we can write: β squared over ππΏ squared is equal to the energy of the emitted photon β π over π. We can cancel out a factor of Planckβs constant from each side and then rearrange this equation to solve for π. And we find itβs equal to π, the mass of the electron, times π, the speed of light, times the width of the box, πΏ squared, divided by β, Planckβs constant.

In this example, weβll treat the mass of the electron π as exactly 9.1 times 10 to the negative 31st kilograms and the speed of light π as exactly 3.00 times 10 to the eighth meters per second.

When we plug in for π, π, πΏ, and β being careful to write our value for πΏ in units of meters, when we calculate π, we find that, to three significant figures, it is 25.8 nanometers. That is the wavelength of the emitted photon.