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### Rydberg Equation!

Posted: Mon Oct 16, 2017 10:10 pm
Hi, so I'm really confused with the Rydberg equation. In the book, the official equation is v = R(1/n^2 - 1/n^2), but when we derived it in class, a negative was applied to the equation (-R). I know we use E = -Rv/n on the actual equations sheet, but I'm still confused about the negatives as it's messing up a lot of supposedly simple problems. Do we rearrange the negative once we find the answer, or conceptually add it in or take away depending on the problem?

### Re: Rydberg Equation!  [ENDORSED]

Posted: Tue Oct 17, 2017 9:44 am
As I discussed in class use $E=-\frac{hR}{n^{2}}$ and the use Delta E = E(final) - E(initial).
Do this without changing any signs and it will work.

Stop by any of the 60 hours per week of office hours and peer learning sessions to work through this type of example.

### Re: Rydberg Equation!

Posted: Tue Oct 17, 2017 6:58 pm
If you use the formula as given on the sheet E = (-hR)/n^2, you should be able to do Delta E = Efinal - Einitial without any issues with negative signs. If you're solving backwards for an n value, you can still use this equation. You can rewrite Delta E = Efinal - Einitial as Delta E = (-hr)/n^2 - (-hR)/n^2. You can find delta E by calculating the energy of the photon - here you would need to consider the negative sign because while the energy of the photon is always positive, the change in energy may be negative so just consider whether the electron is going to a higher or lower (gaining or losing) energy.

### Re: Rydberg Equation!

Posted: Tue Oct 17, 2017 7:50 pm
The reason why your E is negative is because energy is being RELEASED! Similar to how in physics a negative value in front of acceleration represents something is going DOWN, here it is simply indicative of a release of energy.

You can calculate the change in E and use that for subsequent calculations.

### Re: Rydberg Equation!

Posted: Tue Oct 17, 2017 8:02 pm
The reason why $E = \frac{-hR}{n^{2}}$ is negative is because the energy levels of subsequent electron shells are relative to one another. To elaborate, an electron at shell n=$\infty$ would have zero energy (since there is no pull from the nucleus at that position). Since shell n=$\infty$ is technically the outermost shell that is the highest in energy, any shell lower than n=$\infty$ would have a negative energy (energy less than zero is negative). By this logic, n=1,2,3, etc. would all have negative energies.

### Re: Rydberg Equation!

Posted: Wed Oct 18, 2017 9:11 pm
Is it possible to end up with a negative frequency or will the answer always end up being positive?

### Re: Rydberg Equation!

Posted: Wed Oct 18, 2017 9:34 pm
You shouldn't be getting a negative frequency. The energy of light should always be positive. Changes in energy can be negative.

### Re: Rydberg Equation!

Posted: Sat Oct 21, 2017 12:45 pm
Electromagnetic radiation can't have a negative frequency because that would mean it oscillates negative times a second, which doesn't make sense; the least it could oscillate is zero times a second.

Frequency isn't like velocity, where a negative sign implies movement in the opposite direction, it's like absolute temperature, where negative Kelvin doesn't make sense.

### Re: Rydberg Equation!

Posted: Sat Oct 21, 2017 5:11 pm
Frequency can't be negative, but I've noticed that when you switch the two original n-values, one will result in a negative frequency and one will result in the same positive frequency. So whatever value you get, just take the positive form of it and continue.