Video: Maximum Wavelength Produced by a One-Dimensional Bound State of a Quantum Particle

An electron confined to a box of width 0.15 nm by infinite potential energy barriers emits a photon when it makes a transition from the first excited state to the ground state. Find the wavelength of the emitted photon.

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Video Transcript

An electron confined to a box of width 0.15 nanometers by infinite potential energy barriers emits a photon when it makes a transition from the first excited state to the ground state. Find the wavelength of the emitted photon.

We’re told in this statement that the box has a width of 0.15 nanometers, a value that we’ll call 𝐿. We want to solve for the wavelength of the photon that is emitted when the electron transitions from the first excited state to the ground state. We’ll call that wavelength πœ†.

To begin, let’s draw a sketch of the situation. We have a well of width 𝐿, which we imagine to be infinitely deep. Meaning, there is zero probability that an electron can escape the well. If we sketch in various energy levels an electron in the well can have, the values are discrete, quantized. There is a mathematical relationship that describes the energy levels of a particle in an infinite well. As is typically the case on a very small scale, those levels are quantized, denoted by the integer value 𝑛. The 𝑛th level energy of a particle in an infinite well equals β„Ž, Planck’s constant, squared times 𝑛 squared divided by eight times the mass of the particle times the width of the well, 𝐿, squared.

In this exercise, we’ll assume that β„Ž is exactly 6.626 times 10 to the negative 34th joule seconds. When we apply this relationship to our scenario, we notice the factor of 𝑛 squared in the numerator. Looking back at our diagram, we can label the various energy levels drawn by their integer value 𝑛, where the ground state is 𝑛 equal one and the first excited state is 𝑛 equal two. It is the transition of an electron from the 𝑛 equals two, or first excited state, down to the ground state, 𝑛 equals one, which corresponds to the smallest change in energy possible within this system.

That change in energy, Δ𝐸, of the electron is equal to 𝐸 sub two minus 𝐸 sub one. When we write this out in terms of our equation for 𝐸 sub 𝑛, the two terms simplify. So we can write Δ𝐸 equals β„Ž squared over eight π‘šπΏ squared times two squared minus one squared or times three. So that’s the amount of energy corresponding to a change from the first excited state to the ground state. This transition results in the emission of a photon. And if we recall that photon energy 𝐸 equals β„Ž times 𝑓, which for a photon equals β„Ž times 𝑐 over πœ†, then we can write that three β„Ž squared over eight π‘šπΏ squared equals β„Žπ‘ over πœ†. A factor of Planck’s constant β„Ž cancels from each side.

And if we rearrange this expression to solve for πœ†, we find that it is equal to eight times the mass of the electron times the width of the well, 𝐿 squared, times the speed of light 𝑐 all over three times β„Ž. We’ll assume that the mass of the electron, π‘š, is exactly 9.1 times 10 to the negative 31st kilograms, and that 𝑐, the speed of light, is exactly 3.00 times 10 to the eighth meters per second.

When we plug-in for the variables describing πœ†, being careful to use a value in meters for our length 𝐿, and enter all these terms on our calculator, we find that πœ†, to two significant figures, is 25 nanometers. That’s the wavelength of the photon that would be emitted when the electron transitions from the first excited state to the ground state.

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