Atomic Excitation transition energy

[Sekonda]

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 de-excites to the ground state.

However, this is Sodium so I think we approximate the energy levels using the equation:

$$E=-Z^{2}\frac{13.61eV}{(n-\delta_{i})^{2}}$$

Such that we have

$$E_{initial}-E_{ground}=E_{wave}$$

which is

$$E_{initial}=\frac{hc}{\lambda}-Z^{2}\frac{13.61}{(3-1.35)^{2}}$$

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.

 

[mfb]

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.

 

[Sekonda]

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:

$$E=-\frac{13.61}{(n-\delta_{i})^{2}}$$

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 electron-electron interactions).

Though I'm unsure how to use the above equation to attain the ground state of Sodium with Z=11.

 

[mfb]

(provided we neglect electron-electron interactions)

Well, you cannot neglect them for sodium.
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.

 

[Sekonda]

Oh yeah, I was thinking that was a stupid/self-contradictory 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.

 

[Sekonda]

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:

$$E_{excited}=E_{initial}+E_{photon}$$

The photon energy is given by

$$E_{photon}=\frac{hc}{\lambda}$$

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

$$1s^{2}2s^{2}2p^{6}3s$$

and so the energy of the electron in the 3s orbital is given by

$$E_{3s}=\frac{-13.6}{(3-1.35)^{2}}$$

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?

 

[mfb]

So all we need to take is the energy difference between the electron orbitals that are changing?

Right.
You have the ground-state energy there, calculate the energy for p and d and find a difference of 4.8 eV.

Though I'm not sure if I'm correct in saying only one electron changes state during this transition?

That is correct.

 

[Sekonda]

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

Cheers,
SK

 

[Sekonda]

I think I am doing something wrong as I use this conservation of energy equation:

$$-\frac{13.6eV}{(n-\delta_{i})^{2}}+\frac{13.6eV}{(3-1.35)^{2}}=4.8eV$$

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

$$n-\delta_{i}=8.47$$

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

 

[mfb]

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

 

[Sekonda]

Maybe we just approximate due to using approximations in h-bar and 'c', as well as the rydberg constant. Perhaps it is just a 9s orbital -considering this :

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/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

 

[mfb]

A cross-check 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.

 

[Sekonda]

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

 

[Sekonda]

Woops... I got the signs the wrong way round, I was using the ground state as the initial state. Now I'm getting

$$n-\delta_{i}=1.18$$

Which gives me, almost, n=2 for the 'p' quantum defect - though I'm pretty sure this doesn't make any sense.

 

[mfb]

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

 

[Cthugha]

Do you know what I am doing wrong?

You are rounding 13.61 eV to 13.6 eV and 4.7931 eV to 4.8 eV. Considering the small numbers involved here, this introduces a considerable error. If you take the exact numbers, your result should be roughly 0.3 lower.

 

[Sekonda]

Yeah I was correct the first time, just going mad. Upon using more accurate values I attain

$$n-\delta_{i}=8.33$$

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.

 

[Cthugha]

Upon using more accurate values I attain

$$n-\delta_{i}=8.33$$

Then you are still not using values which are accurate enough. Using the values given in the exercise I get $$n-\delta_{i}=8.1286$$ which is a rather good match.

 

[Sekonda]

Indeed that would be, I could of sworn I was using the exact numbers - also depends on what you take for the Planck's constant. I was using 6.63 - though I'm guessing 6.626 would be better, I'll use a decent speed of light as well...

Thanks man!
SK

 

[Sekonda]

Using the Planck constant as 6.626*... and the speed of light as 2.99*... I get that

$$n-\delta_{i}=8.0094$$

so maybe it's a d-orbital of the n=8 level?

 

[Cthugha]

I have given you the result above. As you are still off, the values you use are still not exact enough. Maybe your conversion from J to eV for the h is not exact? You can already use the standard value of h=4.1356675...eVs directly. Also, there are numerous wavelength to energy converters around the internet, so you can crosscheck, whether your conversion is good or not.

You need at least two significant digits for the outcome to get an accurate result, so it is a good idea to use more than two significant digits for the constants you start with. You will spend much more time by tracking the mistakes in using numbers which are not exact enough than you safe by typing one or two numbers less.

 

[Sekonda]

Yeah I see now, I used more precise values for the constants and got your result. Though I'm not sure we'd get these constants to this degree of accuracy under examination conditions nor have we been told to expect to remember them to this degree! I'm guessing they were probably just looking for the correct application of physics.

I guess that'll do!


Thanks again,
SK