An electron is confined to a box of width 0.250 nanometres. What is the wavelength of photons emitted when the electron transitions between the third and second excited states?
This statement tells us the width of the box the electron is confined to; it’s 0.250 nanometres. We’ll call that value 𝐿. We want to solve for the wavelength of photons emitted when the electron in the box transitions from the third to the second excited states. To begin our solution, let’s draw a diagram of this scenario.
We have a box — an infinite potential well — of width 𝐿 and we’re told that in this box an electron starting out at the third excited energy level, where 𝑛 equals four, then transitions down to the second excited level, where 𝑛 equals three. In the process of this transition, the electron emits a photon, a particle of light with wavelength 𝜆.
We can start solving for that wavelength by remembering that for a particle in a box like the electron is here, the 𝑛th energy level of that particle 𝐸 sub 𝑛 ist equal to 𝑛, an integer, squared times ℎ, Planck’s constant, squared divided by eight times the electron mass 𝑚 times the width of the box 𝐿 squared. Regarding this equation, we’ll assume that ℎ is exactly 6.626 times 10 to the negative 34th joule seconds and that 𝑚, the mass of the electron, is 9.1 times 10 to the negative 31st kilograms.
When you apply this relationship to the transition in our problem statement, we see that the change in energy of the electron Δ𝐸 equals the energy at the third excited level, where 𝑛 equals four, minus the electron energy at the second excited level, where 𝑛 equals three. This is equal to 16 ℎ squared over eight 𝑚𝐿 squared minus nine ℎ squared over ℎ [eight] 𝑚𝐿 squared or seven ℎ squared over eight 𝑚𝐿 squared. That’s the change in energy of the electron in the well.
That change in energy is equal to the energy of the photon that’s emitted during that transition. Photon energy 𝐸 is equal to ℎ times the frequency 𝑓 or ℎ times 𝑐 over 𝜆, where 𝑐 will treat as exactly 3.00 times 10 to the eighth meter per second. So photon energy ℎ 𝑐 over 𝜆 equals seven ℎ squared over eight 𝑚𝐿 squared. A factor of Planck’s constant cancels. And when we rearrange to solve for 𝜆, we find that it’s eight 𝑚𝑐𝐿 squared over seven times ℎ.
We can now plug in for 𝑚, 𝑐, 𝐿, and ℎ. When we plug those values in, being careful to use a value in units of metres for the width of the well 𝐿 and then enter them on our calculator, we find that to three significant figures the emitted photon has a wavelength of 29.4 nanometres.