Atomic Spectroscopy N?

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BoparaiAdeshsurjit2F
Posts: 113
Joined: Wed Sep 30, 2020 9:51 pm

Atomic Spectroscopy N?

Postby BoparaiAdeshsurjit2F » Fri Oct 16, 2020 1:57 pm

in the Rydberg equation is the n^2 in the denominator always the initial energy level?

Lucy Wang 2J
Posts: 86
Joined: Wed Sep 30, 2020 10:09 pm
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Re: Atomic Spectroscopy N?

Postby Lucy Wang 2J » Fri Oct 16, 2020 2:06 pm

I think to find the change in energy, you do the final energy - initial energy.

Will Skinner
Posts: 84
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Re: Atomic Spectroscopy N?

Postby Will Skinner » Fri Oct 16, 2020 2:13 pm

No, the term n is the variable for energy level. Suppose an electron moves from n= 3 to n= 2 and you want to find the change in E. You would then plug the initial n into the Rydberg equation and do the same for the final n. Then you solve E(final)-E(initial) to find (delta)E. The term n^2 is not the initial energy level, it is part of the Rydberg equation in which you plug in the energy level to solve for the E at that level.

LeanneBagood_2F
Posts: 91
Joined: Wed Sep 30, 2020 9:32 pm

Re: Atomic Spectroscopy N?

Postby LeanneBagood_2F » Fri Oct 16, 2020 10:03 pm

BoparaiAdeshsurjit2I wrote:in the Rydberg equation is the n^2 in the denominator always the initial energy level?

As Will said in their post, the "n^2" in the denominator does NOT refer to the initial energy level. "n", unless otherwise specified, is just an energy level.
In the case of the example problem from today's lecture, we were given an initial and final energy level and we needed to find the difference in energy between the two levels. Because of this, we used the given equation twice - once for the initial energy level and again for the final energy level.
To find the actual change in energy you would do final - initial with the values you found.


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