## Sapling Problem #6

anikamenon1L
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### Sapling Problem #6

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 2H
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### Re: Sapling Problem #6

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_3G
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### Re: Sapling Problem #6

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 1H
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Joined: Wed Sep 30, 2020 9:42 pm

### Re: Sapling Problem #6

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
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### Re: Sapling Problem #6

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
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### Re: Sapling Problem #6

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
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Joined: Wed Sep 30, 2020 9:28 pm

### Re: Sapling Problem #6

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
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Joined: Wed Sep 30, 2020 10:04 pm

### Re: Sapling Problem #6

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!