### Question 1.15

Posted:

**Tue Oct 17, 2017 5:34 pm**Okay so I understand how to get that the two n's are n = 1 and n=3, but how are you supposed to know that it goes from n=1 to n=3 and not the other way around?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=17&t=22359

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Posted: **Tue Oct 17, 2017 5:34 pm**

Okay so I understand how to get that the two n's are n = 1 and n=3, but how are you supposed to know that it goes from n=1 to n=3 and not the other way around?

Posted: **Tue Oct 17, 2017 6:16 pm**

This is because in the formula v= R(1/n^2 - 1/N^2), if n was 3 and N was 1, the frequency would be a negative number.

1/9 - 1/1 = negative number.

1/9 - 1/1 = negative number.

Posted: **Tue Oct 17, 2017 6:18 pm**

is just another way of writing

When we convert the equation, the corresponds to , and corresponds to

We know that , the final energy level, and , the initial energy level, because "the electron during the emission of energy"- for an electron to emit energy, it has to move down energy levels. Because the electron is moving down, it must start from n=3, therefore the initial, and must end at the lower energy level, n=1

When we convert the equation, the corresponds to , and corresponds to

We know that , the final energy level, and , the initial energy level, because "the electron during the emission of energy"- for an electron to emit energy, it has to move down energy levels. Because the electron is moving down, it must start from n=3, therefore the initial, and must end at the lower energy level, n=1

Posted: **Tue Oct 17, 2017 10:50 pm**

@Julian Krzysiak 3L

No but so that's what's confusing me. I thought it should go exactly as you described, but in the solutions Manual it says that n=1 is initial and n= 3 is the final, which is the opposite of what you said :( and also opposite to what would make sense to me if the electrons are "emitting" energy.

Can anyone explain what's going on here?

Thank you!

No but so that's what's confusing me. I thought it should go exactly as you described, but in the solutions Manual it says that n=1 is initial and n= 3 is the final, which is the opposite of what you said :( and also opposite to what would make sense to me if the electrons are "emitting" energy.

Can anyone explain what's going on here?

Thank you!

Posted: **Wed Oct 18, 2017 12:23 am**

Okay based on my discussion section my TA said that n1 is considered the intial and n2 is the final. You would make the lowest n value (in this case 1) your initial that way the frequency always remains positive, which may explain why your initial n is 1 and your final n is 3. This might be confusing since it would be technically n=3 as your inital and n=1 would be final since it's emitting light. Hoped this helped!

Posted: **Wed Oct 18, 2017 6:48 pm**

So for this equation, we just order the energy levels to insure it's positive rather than doing final - initial energy level?

Posted: **Wed Oct 18, 2017 8:15 pm**

In most problems, it clearly states the order of this change, but if it does not you would use the equation: E= -hR/n^2

Stay away from V= R[1/n1^2 - 1/n2^2] because it will not clearly demonstrate what is truly happening in the problem.

Stay away from V= R[1/n1^2 - 1/n2^2] because it will not clearly demonstrate what is truly happening in the problem.