The diagram shows the emission spectrum of hydrogen for the visible part of the electromagnetic spectrum. The table shows the binding energies for the different energy levels of a hydrogen atom. Between which two energy levels must an electron transition in order to produce the emission line marked 𝐷 on the diagram? Between which two energy levels must an electron transition in order to produce the emission line marked 𝐵 on the diagram?
Let’s begin with the first part of this question. We’ll clear some room by erasing the second part for now and come back to it later. Take a look at the diagram. These lines represent the wavelengths of photons produced by various electron energy level transitions. We’ve been asked about the line marked 𝐷, which is a tiny bit over halfway between 640 and 670 nanometers.
Now, the scale is pretty compact, so let’s have a closer look. And notice that it’s represented by these longer marks every 15 nanometers and these shorter marks every three nanometers, meaning there’s an uncertainty of half that size, or one and a half nanometers. And our reading from this diagram is an approximation, so it doesn’t have to be perfect, just reasonably within a nanometer or two. So this mark here is at 655 nanometers, and emission line 𝐷 is just slightly to the right of it. So we can say that the wavelength, which we’ll call 𝜆 sub 𝐷, is around 656 nanometers.
Now, we need to take this information that we got from the diagram, which is the wavelength of a photon, and match it to an electron energy level transition. To do so, let’s recall that an electron emits a photon when it moves from a higher to lower energy level. The energy of that emitted photon must equal the difference between the electron’s initial and final energy levels. We’ll call that amount Δ𝐸. Thus, if we know the energy of the photon, we can match it to a difference between two of these energy levels. Right now, we know the photon’s wavelength. And recall that we can relate the wavelength 𝜆 of a photon to its energy 𝐸 using the formula 𝐸 equals ℎ𝑐 divided by 𝜆, where ℎ is the Planck constant, 4.14 times 10 to the negative 15 electron volt seconds, and 𝑐 is the speed of light, 3.0 times 10 to the eight meters per second.
Now, let’s use this formula to find the energy of the emitted photon, which we’ll call 𝐸 sub 𝐷. And since we already know the values of ℎ, 𝑐, and 𝜆 sub 𝐷, we’re ready to substitute them in. We have the photon’s energy equals the Planck constant times the speed of light divided by the photon’s wavelength. But before we calculate, the wavelength should be expressed in plain meters not nanometers. Now, the prefix nano- means 10 to the negative nine, so we can make a substitution and write 𝜆 sub 𝐷 as 656 times 10 to the negative nine meters. Speaking of units, notice that we can actually cancel meters from the numerator and denominator. We can also cancel seconds and per seconds from the numerator, leaving only units of electronvolts, which is a good sign because that’s the unit of energy we’re using throughout this question.
Okay, now, let’s go ahead and calculate, and we found the energy of the emitted photon is 1.89 electronvolts. So we know this amount must match the difference between some two of these energy levels. And now, it’s our job to figure out which two energy levels they are. We can do this by picking two values from the table and calculating their energy difference Δ𝐸 until we find a Δ𝐸 that equals the photon energy, 1.89 electronvolts. First, let’s try energy level two to one. And we have that Δ𝐸 equals the energy at level two minus the energy at the ground state, which comes out to 10.2 electronvolts. This is way larger than what we’re looking for. And a transition from any higher energy level to the ground state would give an even larger energy difference.
So let’s just copy this down below the table, and we can make room to try another transition. This time, we’ll try energy level three to two. And the energy at level three minus the energy at level two gives a Δ𝐸 value of 1.89 electronvolts, which does match up with the emitted photon energy. And we know the electron couldn’t have transitioned to the second level from any level higher than three, or else the Δ𝐸 value would be too great. And just to be safe, we can consider whether the electron could’ve moved between two higher energy levels. Again, using this formula, we can verify that a move from level four to three would have a Δ𝐸 of 0.66 electronvolts. And a move from level five to four would have a Δ𝐸 of about 0.31 electronvolts.
Even when combined to consider a transition from level five to three, these Δ𝐸 values are just too small to account for that energy of emitted photon 𝐷. Therefore, we can be sure that in order to produce the emission line marked 𝐷, an electron must move from energy level three to two.
Now, we’ll keep this answer written down here and make room for the second part of the question.
Between which two energy levels must an electron transition in order to produce the emission line marked 𝐵 on the diagram?
Just like before, we should determine the energy of the photon that created this emission line. Line 𝐵 is just to the right of the first small mark that comes after 430 nanometers. That small mark denotes 433 nanometers, and emission line 𝐵 is closer to that mark than the next one to the right. And remember, this reading from the scale is an approximation, so it doesn’t have to be exactly perfect in order to find a legitimate answer to this question.
Keeping all this in mind, let’s say that emission line 𝐵 is at 434 nanometers. Once again, we’ll copy the photon energy formula and substitute in values for the Planck constant, the speed of light, and the photon wavelength. Remember, we need to write the wavelength in units of meters. So because a nanometer is 10 to the negative nine meters, we’ll write 𝜆 sub 𝐵 as 434 times 10 to the negative nine meters. Now, calculating, the energy of emitted photon 𝐵 is about 2.86 electronvolts. This must match the energy difference Δ𝐸 between some two of these energy levels in the table. Right away, we know that a transition from level two to one has a Δ𝐸 that’s far greater than 2.86 electronvolts, so that can’t be it.
Moving on, a transition from level three to two has a Δ𝐸 that we found to be 1.89 electronvolts, which is closer. But we actually need a greater energy difference like if the electron started at a higher energy level. So we can look at a transition from four to two. To find this Δ𝐸, we can just combine the energy differences between four and three and three and two. Now, adding these two values gives a Δ𝐸 of 2.55 electronvolts, which is better but still not large enough. So we can think about the electron starting from an even higher level.
If we consider a transition from level five to two, we can combine the Δ𝐸 from five to four with the one that we just found, four to two. This gives Δ𝐸 equals 2.55 electron volts plus 0.31 electron volts, which does equal 2.86 electron volts, the value we’ve been looking for. Therefore, we found that in order to produce the emission line marked 𝐵, an electron must move from energy level five to two.