## Sapling Week 2 & 3 HW Problem 5

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Faith St Amant 3D
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### Sapling Week 2 & 3 HW Problem 5

Hi everyone! I'm working on the Week 2 & 3 homework and am stuck on problem 5. The problem is:

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.
number of spectral lines: ________
wavelength range from: _______ nm to _______ nm

I'm having trouble understanding where to start with this, just simply thinking about it conceptually. Can anybody explain what to do?

Edward Tang 1k
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### Re: Sapling Week 2 & 3 HW Problem 5

I understand, the question was worded quite confusingly, but once you know what it's asking it's not that hard. Number of spectral lines would be equal to the number of times the electron emits energy, and the electron would do that for each complete energy level it moves down since energy is quantized. So from 7 to 1 you would have traversed 6 complete energy levels, therefore emitting 6 spectral lines.

Wavelength range is asking about the wavelength that would result from the energy difference between two levels. So the maximum wavelength would result from the transition with the least energy since energy is directly proportional to frequency, which is inversely proportional to wavelength. Therefore going from energy level n=7 to energy level n=6 would result in a spectral line with the longest wavelength, and going from n=7 to n=1 would give the shortest wavelength. This makes sense too because Lyman series where electrons return to n=1 has the lowest wavelength.

Then use the equation v(frequency)=R(1/(n2^2)-1/(n1^2)), where n1 represents the higher energy level (you could use n2 to represent that it's up to you, just know the higher energy level comes later), to find out the frequency associated with the energy transition, then use c=fv to find the wavelength.

I definitely made this sound more complicated than it is but basically number of spectral lines = numbers of energy level transition. And wavelength range is the wavelength corresponding to the energy change from each transition.

Xinying Wang_3C
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### Re: Sapling Week 2 & 3 HW Problem 5

Edward Tang 2E wrote:I understand, the question was worded quite confusingly, but once you know what it's asking it's not that hard. Number of spectral lines would be equal to the number of times the electron emits energy, and the electron would do that for each complete energy level it moves down since energy is quantized. So from 7 to 1 you would have traversed 6 complete energy levels, therefore emitting 6 spectral lines.

Wavelength range is asking about the wavelength that would result from the energy difference between two levels. So the maximum wavelength would result from the transition with the least energy since energy is directly proportional to frequency, which is inversely proportional to wavelength. Therefore going from energy level n=7 to energy level n=6 would result in a spectral line with the longest wavelength, and going from n=7 to n=1 would give the shortest wavelength. This makes sense too because Lyman series where electrons return to n=1 has the lowest wavelength.

Then use the equation v(frequency)=R(1/(n2^2)-1/(n1^2)), where n1 represents the higher energy level (you could use n2 to represent that it's up to you, just know the higher energy level comes later), to find out the frequency associated with the energy transition, then use c=fv to find the wavelength.

I definitely made this sound more complicated than it is but basically number of spectral lines = numbers of energy level transition. And wavelength range is the wavelength corresponding to the energy change from each transition.

Thank you. I was having the same question and this is really helpful. I am just wondering what does R in the equation v(frequency)=R(1/(n2^2)-1/(n1^2)) stands for, and does c=3.00*10^8m/s in this case ?

Faith St Amant 3D
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### Re: Sapling Week 2 & 3 HW Problem 5

@Edward thank you! Makes much more sense now. And @Xinying, R is the Rydberg constant: 3.29 x 10^15 Hz. Also, yes, c = 3.0 x 10^8 m/s.

Darren1j
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### Re: Sapling Week 2 & 3 HW Problem 5

The posts above ^ helped me figure this out. Basically, going from a higher level to a lower level causes spectral lines. To find the range of wavelengths, you must use the Rydberg equation. You use the n=7 to n=6 for the first wavelength and n=7 to n=1 for the second wavelength. The equation is v (frequency)= R( 1/n final squared - 1/n initial squared). So for one wavelength, the final would be n=1 and the initial would be n=7. The second wavelength would come from n=6 (final) and n= 7(initial).

Plug those into the Rydberg equation. Do it one time for n=7 to n=6, and do it again for n=7 to n=1. This will give you 2 frequencies'. You can use the equation c= Wavelength times frequency to find the wavelengths. (rearrange this to be Wavelength= c/frequency.)

Ellison Gonzales 1H
Posts: 63
Joined: Wed Sep 30, 2020 10:00 pm

### Re: Sapling Week 2 & 3 HW Problem 5

Darren1e wrote:The posts above ^ helped me figure this out. Basically, going from a higher level to a lower level causes spectral lines. To find the range of wavelengths, you must use the Rydberg equation. You use the n=7 to n=6 for the first wavelength and n=7 to n=1 for the second wavelength. The equation is v (frequency)= R( 1/n final squared - 1/n initial squared). So for one wavelength, the final would be n=1 and the initial would be n=7. The second wavelength would come from n=6 (final) and n= 7(initial).

Plug those into the Rydberg equation. Do it one time for n=7 to n=6, and do it again for n=7 to n=1. This will give you 2 frequencies'. You can use the equation c= Wavelength times frequency to find the wavelengths. (rearrange this to be Wavelength= c/frequency.)

Thank you, this is extremely helpful. Even though I understand what you are explaining, I was wondering why n=7 to n=1 isn’t just the wavelength range itself? Why do we do n=7 to n=6 as well and then n=7 to n=1?

Arnav Saud 2C
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### Re: Sapling Week 2 & 3 HW Problem 5

Ellison Gonzales 3F wrote:
Darren1e wrote:The posts above ^ helped me figure this out. Basically, going from a higher level to a lower level causes spectral lines. To find the range of wavelengths, you must use the Rydberg equation. You use the n=7 to n=6 for the first wavelength and n=7 to n=1 for the second wavelength. The equation is v (frequency)= R( 1/n final squared - 1/n initial squared). So for one wavelength, the final would be n=1 and the initial would be n=7. The second wavelength would come from n=6 (final) and n= 7(initial).

Plug those into the Rydberg equation. Do it one time for n=7 to n=6, and do it again for n=7 to n=1. This will give you 2 frequencies'. You can use the equation c= Wavelength times frequency to find the wavelengths. (rearrange this to be Wavelength= c/frequency.)

Thank you, this is extremely helpful. Even though I understand what you are explaining, I was wondering why n=7 to n=1 isn’t just the wavelength range itself? Why do we do n=7 to n=6 as well and then n=7 to n=1?

A range (in terms of this question) includes the lowest level to the highest level, which means that we need to get 2 different values.
Thus, we need to find the wavelength of the n=7 to n=1 transition, and then find the wavelength of the n=7 to n=6 transition. In terms of the wavelengths, the n=7 to n=6 transition has a larger wavelength while the n=7 to n=1 transition has a smaller wavelength. Based on this, we can create a range from the smallest possible wavelength (n=7 to n=1 transition) and the largest possible wavelength (n=7 to n=6).

Ellison Gonzales 1H
Posts: 63
Joined: Wed Sep 30, 2020 10:00 pm

### Re: Sapling Week 2 & 3 HW Problem 5

Arnav Saud 3K wrote:
Ellison Gonzales 3F wrote:
Darren1e wrote:The posts above ^ helped me figure this out. Basically, going from a higher level to a lower level causes spectral lines. To find the range of wavelengths, you must use the Rydberg equation. You use the n=7 to n=6 for the first wavelength and n=7 to n=1 for the second wavelength. The equation is v (frequency)= R( 1/n final squared - 1/n initial squared). So for one wavelength, the final would be n=1 and the initial would be n=7. The second wavelength would come from n=6 (final) and n= 7(initial).

Plug those into the Rydberg equation. Do it one time for n=7 to n=6, and do it again for n=7 to n=1. This will give you 2 frequencies'. You can use the equation c= Wavelength times frequency to find the wavelengths. (rearrange this to be Wavelength= c/frequency.)

Thank you, this is extremely helpful. Even though I understand what you are explaining, I was wondering why n=7 to n=1 isn’t just the wavelength range itself? Why do we do n=7 to n=6 as well and then n=7 to n=1?

A range (in terms of this question) includes the lowest level to the highest level, which means that we need to get 2 different values.
Thus, we need to find the wavelength of the n=7 to n=1 transition, and then find the wavelength of the n=7 to n=6 transition. In terms of the wavelengths, the n=7 to n=6 transition has a larger wavelength while the n=7 to n=1 transition has a smaller wavelength. Based on this, we can create a range from the smallest possible wavelength (n=7 to n=1 transition) and the largest possible wavelength (n=7 to n=6).

Thank you Arnav! This makes much more sense

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