HW Question 1.57

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Phoebe Chen 4I
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Joined: Fri Sep 28, 2018 12:29 am

HW Question 1.57

Postby Phoebe Chen 4I » Mon Oct 22, 2018 6:09 pm

I am unsure how to solve this problem. First I found the energies of the wavelengths and thought I could use En = -hR/n^2 to find out the energy level number for each energy and see which number would come next. But I don't think that is the right way to solve it. Would appreciate help on this question!
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David S
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Joined: Fri Sep 28, 2018 12:15 am

Re: HW Question 1.57

Postby David S » Tue Oct 23, 2018 11:28 am

First, we must recognize that emitted radiation wavelengths in the Balmer series correspond to electron transitions from n_i > 2 to n_f =2.
Next, remember that a larger transition -> higher energy radiation ∝ smaller wavelength radiation.
This means that you are going to be calculating a wavelength for radiation emitted by an even larger energy transition than the energy transition that resulted in a wavelength of 410.2nm

Now, if you have some sort of table or reference for Balmer series, you can associate the given wavelengths with their corresponding electron transitions (e.g. http://astronomy.swin.edu.au/cosmos/B/Balmer+series) to find what n_i you need next.
Otherwise, you have to use the Rydberg equation, , to solve for n_i when given that n_f = 2 for Balmer series transitions.

Add 1 to the n_i you get for . This will be the n_i for the new electron transition you are concerned with.
Use the Rydberg equation again to solve for with your new n_i and n_f=2, and bam.

LedaKnowles2E
Posts: 62
Joined: Fri Sep 28, 2018 12:27 am

Re: HW Question 1.57

Postby LedaKnowles2E » Thu Oct 25, 2018 6:21 pm

Here's my work. What I did makes sense to me but I didn't know where to look for the solution so I don't know for sure that it's right. Hope it helps :)
1.57Solution.JPG

Ahmet_Dikyurt_3L
Posts: 35
Joined: Fri Sep 28, 2018 12:23 am

Re: HW Question 1.57

Postby Ahmet_Dikyurt_3L » Fri Oct 26, 2018 12:35 am

LedaKnowles4D wrote:Here's my work. What I did makes sense to me but I didn't know where to look for the solution so I don't know for sure that it's right. Hope it helps :)1.57Solution.JPG

Hi, I couldn't understand how n2 = 7
I also looked it up online, and they also give n2 = 7
I used n2 = 7 in my homework as well but could you explain it more? Thanks.

LedaKnowles2E
Posts: 62
Joined: Fri Sep 28, 2018 12:27 am

Re: HW Question 1.57

Postby LedaKnowles2E » Fri Oct 26, 2018 7:46 pm

Hi! I'll try to explain it as far as I understand it...

All the lines in the Balmer series result from electrons transitioning from various energy levels down to the n=2 energy level. As the electron drops from higher energy levels, more energy is released from the transition. The more energy, the higher frequency light, or shorter wavelength. The wavelengths of the lines given in the problem are in decreasing order, meaning that each consecutive one has higher energy, meaning the electron fell from a higher energy level. Since the last wavelength given is 410.2nm, and we calculated that that resulted from the electron falling from n=6, the next line in the series must have a shorter wavelength and higher energy, meaning a bigger energy jump -- thus, the next line in the series results from a n=7 to n=2 drop.

Hope that was clearer :)


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