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### 1.57 Homework Conceptual

Posted: Tue Oct 04, 2016 5:23 pm
Hi,

For question 57 on the homework, involving lines in the Balmer series, it says in the solutions manual that "n[1]=2 so the fifth line in the spectrum should be n[2]=7. Then you plug into the Rydberg equation and find the fifth wavelength of the fifth line. What I dont understand is why you get [2]=7. Shouldnt it be equal to 6? N1=2, N2=3, N=5 and N6=6? Also when you plug into the Rydberg equation, why is the answer positive? Why is it that the form of the equation V=R(1/n1^2-1/n2^2) does not have a negative but in the formula E=-hr/n^2 does have a negative?

### Re: 1.57 Homework Conceptual  [ENDORSED]

Posted: Tue Oct 04, 2016 6:35 pm
To answer the first question, the answer is n=7 because you are looking for the next line in the series. So when you had solved and found n=6, you are using that as the baseline to find the wavelength of the next energy level.
To answer the second question, when you are going from a higher energy level to a lower energy level, you are losing the energy that is then transformed into a photon. You would keep the negative sign when looking at the change in energy so when plugging that value into the Rydberg equation { E sub n= (-hR/ n^2) } the answer will turn out positive, not negative.

### Re: 1.57 Homework Conceptual

Posted: Tue Oct 04, 2016 6:44 pm
So what you are saying in the second part of your answer is that deltaE becomes negative and you set that equal to the Rydberg formula which would be negative and the two negatives on both sides would become positive (mathematically dividing both sides by -1)?

### Re: 1.57 Homework Conceptual

Posted: Wed Oct 05, 2016 4:33 pm
Yes, the two negatives will cancel out when being divided. That will leave the answer as a positive value.