Sapling Problem #6

[anikamenon2H]

Question: The electron in a hydrogen atom is excited to the n= 7 shell and emits electromagnetic radiation when returning to lower energy levels. Determine the number of spectral lines that could appear when this electron returns to the lower energy levels, as well as the wavelength range in nanometers.

I'm not really sure how to solve this, especially how to figure out how many spectral lines there will be, does it correspond to how many energy transitions are possible?

[Vivian Chang 3L]

Yup, the number of spectral lines corresponds to how many energy transitions are possible, so in this case you can go from n=7 to n=6, 7 to 5, 7 to 4, 7 to 3, 7 to 2, and 7 to 1, so that would be a total of 6 possible spectral lines.

To find the wavelength range, you would find deltaE for the n=7 to n=6 transition for the lower number in the range because this is the smallest deltaE possible. Similarly, you would find deltaE for the n=7 to n=1 transition to get the higher number in the range because this is the highest deltaE possible.

Edit: I forgot to mention, after finding deltaE for both, you can use E=(hc)/lambda to find the corresponding wavelength.

[Kailani_Dial_2K]

This is how I approached this question. The ground state of is n=1 becuase that is the lowest state that the electron can go to. Electron Emissions Spectroscopy records the wave length of light emitted correct? And there is a difference is distance between levels ( what i mean by this is the distance/energy difference between n=1 and n=2 is not the same as the distance/ energy difference between n=6 and n=7). In the question it says that the e- emits EM radiation as it returns to lower energy levels. So i figured it like this. There will be a different wavelength if the e- returns from 7 to 6, 7 to 5, 7 to 4, 7to 3,7 to 2, and 7 to 1, so you know that 6 different wavelengths are possible if the electron starts at n=7. For the second part you have to find the smallest energy transition which would be from 7 to 6 and then the largest energy transition from 7 to 1 to figure out the range of wavelength. I hope this helps.

[Jacquelyn Challis 2B]

Okay for this question I got the answers: from 37.61nm to 12370nm and it is telling me that 37.61nm is wrong, but I don't understand why. Any help would be great!

[Reese_Gover2K]

Yea that's exactly it. The number of possible transitions is the number of spectral lines so 7->6 is a spectral line and 7->5 is another spectral line as so on until you reach 7->1 and get a total of 6 spectral lines.

[abby hyman]

Yes, the number of spectral lines is equal to the number of possible transitions from n=7 to n=1. To find the wavelength range you just have to find the wavelength corresponding to the transition from n=7 to n=6 and the wavelength corresponding to the transition from n=7 to n=1. These would be the minimum and maximum of your range.

[Will Skinner]

The first part of the problem you just find the number of energy transitions the electron can make from the excited state of n = 6. So we would have 6-5, 6-4, 6-3, and so on. That gives you the number of spectral lines. To find the wavelength range you calculate the energy difference of the first and last transitions and use those values to find their respective wavelengths.

[Kendall_Dewey_2D]

I was stuck on this question too- I found that it is easiest to solve by thinking of the number of spectral lines as equal to the number of possible transitions from n=7 to n=1. So, finding the wavelength range is the wavelength from to the transition from n=7 to n=6 and then the wavelength from to the transition from n=7 to n=1. When you get these 2, you have your range of wavelengths!

[Michael Crannell 1H]

This was very helpful on explaining the answers to the spectral line problem. My question was how do you know that N7-N6 is the smallest spectral range?

[706244927]

We know n=7 to n=6 transition has the smallest spectral range because energy levels get progressively closer together as they get further from the nucleus and in this problem, n=7 and n=6 are furthest from the nucleus and therefore closest together.

The reason for this is because as the distance between the electron and nucleus decreases, the electric potential energy increases, and the energy levels become more tightly spaced.

Because the frequency of the emitted photon is directly proportional to the energy difference between the two energy levels, the spectral lines become closer together as the energy levels become more tightly spaced meaning there would be a smaller spectral range.