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?
We’re told in this statement that the box has a width 0.250 nanometers which we’ll call 𝐿. We want to solve for the wavelength of the photons emitted when the electron transitions from the second excited state to the ground state. Let’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. We’re told the electron starts at the second excited state where 𝑛 equals three, and it then transitions down to the ground state where 𝑛 is one. In the process, the electron emits a photon with a wavelength 𝜆. It’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 𝑛th energy level, 𝐸 sub 𝑛, equals 𝑛 squared times ℎ squared, where ℎ is Planck’s constant which we’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 𝑚 times the width of the box 𝐿 squared.
In our case, we have an electron that starts out at 𝑛 equals three and then moves to 𝑛 equals one. By the particle in a box energy equation, that equals nine ℎ squared over eight 𝑚𝐿 squared minus one ℎ squared over eight 𝑚𝐿 squared, which simplifies to ℎ squared over 𝑚𝐿 squared. Now that’s the change in energy of the electron, but what about the photon that’s emitted? We can remember that photon energy equals ℎ times the frequency 𝑓 or ℎ times 𝑐 over 𝜆. All the energy of the electron’s transition is delivered to the emitted photon. That means we can write: ℎ squared over 𝑚𝐿 squared is equal to the energy of the emitted photon ℎ 𝑐 over 𝜆. We can cancel out a factor of Planck’s constant from each side and then rearrange this equation to solve for 𝜆. And we find it’s equal to 𝑚, the mass of the electron, times 𝑐, the speed of light, times the width of the box, 𝐿 squared, divided by ℎ, Planck’s constant.
In this example, we’ll treat the mass of the electron 𝑚 as exactly 9.1 times 10 to the negative 31st kilograms and the speed of light 𝑐 as exactly 3.00 times 10 to the eighth meters per second.
When we plug in for 𝑚, 𝑐, 𝐿, and ℎ being careful to write our value for 𝐿 in units of meters, when we calculate 𝜆, we find that, to three significant figures, it is 25.8 nanometers. That is the wavelength of the emitted photon.