A ground state hydrogen atom absorbs a photon of light having a wavelength of 94.96 nm. It then gives off a photon having a wavelength of 95 nm. What is the final state of the hydrogen atom?
I think I got 5 for an answer, but I am not sure if I did this right or not
5 answers
I ran through the calc quickly (and could have made an error) but I came up with 5 for where the electron was promoted to (from ground state of n = 1 to n = 5) but I did not get 5 for the final resting place. I will look for the error if you want to post your work.
well i got 5 for the first part but I thought that the delta E was the same for both because the wavelengths were pretty much the same, so i got 1 as the final but that's only because that made both of the equations exactly the same
You are absolutely right. And I should have notice that too but didn't until just a second before reading your latest post. The first photon sends it from n = 1 to n = 5 and the emission of energy then sends it back to n = 1. The first part of your observation is also right. Since the first photon and the second photon are ALMOST the same, that MUST be true because it emits as much energy on the way down as it took to get it up there (to n = 5) in the first place. :-).
I'm not sure what you mean by "made both equations exactly the same."
The second equation is
E = 6.626E-34*3E8/95E-9 = 2.0924E-18, then
2.09824E-18 = 2.180E-18*(1/x^2 - 1/25) which isn't the same as the first equation you solved.
x^2 = 1.00017E0 so x = 1.000
The second equation is
E = 6.626E-34*3E8/95E-9 = 2.0924E-18, then
2.09824E-18 = 2.180E-18*(1/x^2 - 1/25) which isn't the same as the first equation you solved.
x^2 = 1.00017E0 so x = 1.000
Ok I thought that before I submitted the question but I thought that seemed too easy! That you so much for your help though :)