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Atomic Excitation transition energy 
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#1
Dec3112, 08:17 AM

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Hey,
Here's the question Now I wanted to check if my thought process was correct and thus my formulation. Some sodium atom is initially excited, emits EM wave with stated λ and then deexcites to the ground state. However, this is Sodium so I think we approximate the energy levels using the equation: [tex]E=Z^{2}\frac{13.61eV}{(n\delta_{i})^{2}}[/tex] Such that we have [tex]E_{initial}E_{ground}=E_{wave}[/tex] which is [tex]E_{initial}=\frac{hc}{\lambda}Z^{2}\frac{13.61}{(31.35)^{2}}[/tex] For Z=11 and since we write the ground state of sodium as 1s{2}2s{2}2p{6}3s{1} I used n=3 and δi=δs. I got the initial state energy as 600eV though have no idea if I've done this correct. 


#2
Dec3112, 10:20 AM

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I think you should not mix so many different calculation parts here:
 Which energy difference does the transition have (in eV)?  How do the energy levels for s,p,d with n=3 look like? Do you see the calculated difference between two levels here? With the quantum defects reducing the effective n, I think you should work with the shielded charge for each shell instead of the full charge of the nucleus. The photon has an energy of a few eV, I would not expect any energy level of 600 eV involved. 


#3
Dec3112, 10:55 AM

P: 209

The difference in energy between the excited (initial) and ground (final) state I believe is just the energy associated with the emission which is about 4.8eV using hc/λ
Though I'm unsure how to determine the ground state of Sodium, I have the formula: [tex]E=\frac{13.61}{(n\delta_{i})^{2}}[/tex] Which is true for Hydrogen, for an atom of proton number Z I think we just multiply this value by Z^2 (provided we neglect electronelectron interactions). Though I'm unsure how to use the above equation to attain the ground state of Sodium with Z=11. 


#4
Dec3112, 11:30 AM

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Atomic Excitation transition energy
As first approximation, you can consider all electrons in "lower" shells and neglect electrons in the same shell. For sodium, this gives 11 protons and 10 electrons to consider, so the remaining charge is 1. The inner electrons do not provide a perfect shielding, and this leads to the quantum defects numbers, which are just a numerical way to take this into account. 


#5
Dec3112, 11:34 AM

P: 209

Oh yeah, I was thinking that was a stupid/selfcontradictory thing to say... Still confused on how to compute the ground state energy of Sodium  I wasn't sure if I have to consider each orbital of s and p in the electronic configuration.



#6
Jan413, 12:29 PM

P: 209

Right,
So I think I'm right in saying that the excited sodium atom state is equal to the sum of the photon and the ground state of sodium: [tex]E_{excited}=E_{initial}+E_{photon}[/tex] The photon energy is given by [tex]E_{photon}=\frac{hc}{\lambda}[/tex] and is about 4.8eV The photon energy is equal to the difference in the ground state and excited state energies of sodium. The groundstate configuration of sodium is [tex]1s^{2}2s^{2}2p^{6}3s[/tex] and so the energy of the electron in the 3s orbital is given by [tex]E_{3s}=\frac{13.6}{(31.35)^{2}}[/tex] So all we need to take is the energy difference between the electron orbitals that are changing? Though I'm not sure if I'm correct in saying only one electron changes state during this transition? 


#7
Jan413, 12:37 PM

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You have the groundstate energy there, calculate the energy for p and d and find a difference of 4.8 eV. 


#8
Jan413, 02:32 PM

P: 209

Marvellous, thanks for confirming this mfb  despite probably stating it before. I'm a bit "slow".
Cheers, SK 


#9
Jan413, 02:46 PM

P: 209

I think I am doing something wrong as I use this conservation of energy equation:
[tex]\frac{13.6eV}{(n\delta_{i})^{2}}+\frac{13.6eV}{(31.35)^{2}}=4.8eV[/tex] The first term being the excited orbital energy, second term the ground state orbital energy and final term the energy of the photon. Rearranging I find that [tex]n\delta_{i}=8.47[/tex] And neither the quantum defects for p or d give an integer 'n' or even close. Do you know what I am doing wrong? Thanks, SK 


#10
Jan413, 03:47 PM

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Hmm, interesting. Looks like you need a different n, with an unknown quantum defect.
I don't know. 


#11
Jan513, 05:47 AM

P: 209

Maybe we just approximate due to using approximations in hbar and 'c', as well as the rydberg constant. Perhaps it is just a 9s orbital considering this :
http://hyperphysics.phyastr.gsu.edu...um/sodium.html The only transition to the ground state capable of making an emission of 4.8eV is n>7, so perhaps what I have done is correct. But I prefer to doubt myself! Thanks, SK 


#12
Jan513, 08:01 AM

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A crosscheck with this database: It does not have any spectral line with 258.67nm with the ground state as lower state. The closest one is 259.383 (or 259.393) which corresponds to 7p > 3s.
It has a line with 258.631nm, but I am not sure how to interpret the notation of the excited state, and the lower state is not the ground state. 


#13
Jan513, 11:45 AM

P: 209

Maybe I'm missing out something, or maybe the question hasn't been thought out  though I find this unlikely and much more likely I've made an error  though I'm struggling to see where.
Oh well, can only hope a question like this doesn't pop up again. Let me know if you find the problem and thanks for walking me through this mfb. Thanks, SK 


#14
Jan513, 11:53 AM

P: 209

Woops..... I got the signs the wrong way round, I was using the ground state as the initial state. Now I'm getting
[tex]n\delta_{i}=1.18[/tex] Which gives me, almost, n=2 for the 'p' quantum defect  though I'm pretty sure this doesn't make any sense. 


#15
Jan513, 01:15 PM

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No, this time your sign is wrong, it was right before.



#16
Jan513, 03:23 PM

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#17
Jan513, 03:31 PM

P: 209

Yeah I was correct the first time, just going mad. Upon using more accurate values I attain
[tex]n\delta_{i}=8.33[/tex] There still is no delta for this, well at least given  I'm guessing I've made a mistake elsewhere, or it's a bad question or we just round it, but I reckon the former. 


#18
Jan513, 09:21 PM

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