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### Rydberg

Posted: Wed Oct 18, 2017 10:53 pm
When using the Rydberg formula, do you do n initial minus n final or n initial minus n final? How do you know which one is subtracted from the other?

### Re: Rydberg

Posted: Wed Oct 18, 2017 11:01 pm
It depends - you can do 1/(n initial) minus 1/(n final) or you can do negative [1/(n final) minus 1/(n initial)]. It's the negative delta n, if that's easier for you to remember. Hope that helps!

### Re: Rydberg

Posted: Wed Oct 18, 2017 11:05 pm
When you use the Rydberg formula, always do nfinal minus ninitial. Dr. Lavelle used the example during class of temperature, where you use the final temperature minus the initial temperature to measure the change. The same can be said of money. If you start with 5 dollars and end up with 3 dollars, the difference is 3-5 which is -2 dollars.

### Re: Rydberg

Posted: Wed Oct 18, 2017 11:14 pm
Catherine Yang 3G wrote:When you use the Rydberg formula, always do nfinal minus ninitial. Dr. Lavelle used the example during class of temperature, where you use the final temperature minus the initial temperature to measure the change. The same can be said of money. If you start with 5 dollars and end up with 3 dollars, the difference is 3-5 which is -2 dollars.

I believe you have it backwards - the formula in the book says initial minus final, but you can do final minus initial only if you multiply it all by -1.

### Re: Rydberg

Posted: Thu Oct 19, 2017 11:37 am
n(initial)-n(final) gives me a negative frequency. Is this right?

### Re: Rydberg

Posted: Thu Oct 19, 2017 3:15 pm
Tatiana Hage 1E wrote:n(initial)-n(final) gives me a negative frequency. Is this right?

Correct!

### Re: Rydberg

Posted: Fri Oct 20, 2017 2:42 am
The frequency should always be positive, so subtract the bigger number from the smaller number to get a negative number and multiply it by the Rydberg constant times -1.

### Re: Rydberg

Posted: Fri Oct 20, 2017 2:49 am
Tatiana Hage 1E wrote:n(initial)-n(final) gives me a negative frequency. Is this right?

You should get a negative number because you are going from a higher energy state to a lower energy state. Quantitatively, you would be subtracting a not as negative number from a negative number. The negative result from this indicates that energy has left the system. For example, when an electron moves from n=4 to n=2, energy must be released because the high energy from the n=4 status must go somewhere. Hope this helps!

### Re: Rydberg

Posted: Fri Oct 20, 2017 10:05 am
When using Rydberg's equation, the frequency cannot be negative. To avoid this you put whatever the higher of the 2 energy levels for n1.

### Re: Rydberg

Posted: Sun Oct 22, 2017 3:54 pm
For future reference how does one deal with the hard to deal with negatives while showing work? Someone suggested just ignoring them, but I don't think a TA will accept the negatives just disappearing without any reason.

### Re: Rydberg

Posted: Tue Oct 24, 2017 11:29 am
Jean Mok 3K wrote:
Tatiana Hage 1E wrote:n(initial)-n(final) gives me a negative frequency. Is this right?

You should get a negative number because you are going from a higher energy state to a lower energy state. Quantitatively, you would be subtracting a not as negative number from a negative number. The negative result from this indicates that energy has left the system. For example, when an electron moves from n=4 to n=2, energy must be released because the high energy from the n=4 status must go somewhere. Hope this helps!

This makes sense now, thank you!

### Re: Rydberg

Posted: Mon Nov 06, 2017 1:50 pm
You can also use the equation given by Dr. Lavelle for hydrogen (En = -hR/n^2), as we seem to do examples generally with just hydrogen atoms. This rids of the issue of a negative answer with initial-nfinal