An electron in a long, organic molecule used in a dye laser behaves approximately like a quantum particle in a box with width 4.18 nanometres. Find the wavelength of the emitted photon when the electron makes a transition from the first excited state to the ground state.
This statement says that the width of the box is 4.18 nanometres, which we’ll call 𝑤. We want to solve for the wavelength of the photon emitted when the electron goes from the first excited state to the ground state; we can call that wavelength 𝜆. Let’s start by drawing a diagram.
Saying that a particle is in a box is another way of saying it is confined in a space with zero probability of escaping that space. We’re told that this electron starts at the first excited state, where 𝑛 equals two, and makes a transition down one energy level to the ground state, where 𝑛 is one. In the process of making this transition, it emits a photon of wavelength 𝜆.
To solve for 𝜆, we can begin by remembering the particle in a box quantized energy levels. These energies are given in terms of the quantum number 𝑛. If we take that number and square it, multiply it by Planck’s constant ℎ squared, and divide all that by eight times the mass of the particle times the length of the box 𝐿 and square that, we get 𝐸 sub 𝑛 — the energy of the particle in the box at the 𝑛th energy level.
In our scenario, we can say the change in energy the electron experiences Δ𝐸 is equal to 𝐸 when 𝑛 equals two minus 𝐸 when 𝑛 equals one. Based on our particle in a box energy equation, we can also write this as two squared ℎ squared over eight 𝑚𝐿 squared minus one squared ℎ squared over eight 𝑚𝐿 squared, which simplifies to three ℎ squared over eight 𝑚𝐿 squared.
Now, this change in energy that the electron experiences when it drops down one energy level is equal to the energy of the photon that’s emitted during this drop. Recall that photon energy 𝐸 is equal to ℎ times 𝑓, the frequency of the photon, or ℎ times 𝑐 over 𝜆. Using this expression for 𝐸 sub 𝑝 in our equation, we can see that a factor of ℎ cancels out of both sides.
We want to rearrange this equation to solve for 𝜆, the wavelength. When we do that, we see it’s equal to eight times the mass of the electron in the well times the speed of light times the width of the well 𝐿 squared divided by three times Planck’s constant. We’ll assume in this problem that the mass of the electron is exactly 9.1 times 10 to the negative 31st kilograms, the speed of light 𝑐 is 3.00 times 10 to the eighth meters per second, and Planck’s constant ℎ is 6.626 times 10 to the negative 34th Joule-seconds.
Knowing these constants and that the capital 𝐿 in our equation is represented by the width of the well 𝑤, we’re ready to substitute 𝑛 and solve for 𝜆. When we do that, we’re careful to express the width of the well in units of metres to match the SI units in the rest of the expression. Now, we enter these values on our calculator and find that 𝜆 is 19.2 times 10 to the negative six meters or 19.2 microns. That’s the wavelength of the photon emitted by this electronic transition.