Light emission and energy of states

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Homework Help Overview

The discussion revolves around the emission spectrum of thermally excited sodium atoms, specifically focusing on the energy difference between the excited and ground states, as indicated by the wavelength of the emitted photon at 589 nm.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between photon emission and energy state transitions, with some exploring the quantization of energy levels in atoms. Questions arise regarding the applicability of the Rydberg formula to sodium and the complexity of quantum mechanics in deriving energy levels.

Discussion Status

Participants have provided insights into the energy conservation during state transitions and the use of the formula E = hc/lambda for calculating energy differences. There is an ongoing exploration of the appropriate units for expressing energy differences, with some guidance offered on converting joules to electron volts.

Contextual Notes

Some participants express uncertainty about the constants and formulas applicable to different atoms, highlighting the complexity of atomic interactions and the need for clarity on energy level calculations.

sindre00
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Homework Statement



The emission spectrum of thermally excited sodium atoms practically consists of a single intensive line at 589 nm wavelength. What is the energy difference (in eV units) between the excited and ground states of the sodium atom?

Homework Equations


E = hc/lambda, we also know that electron electric charge = 1.6 * 10-19

The Attempt at a Solution


I know that the energy of a photon emitted from an excitation of an electron is E = hc/lamda and by dividing that result by the electric charge of the electron, i get the eV units, but I do not understand the ground state and excited states.
 
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The photon is emitted when the atom transits from the excited to the ground state. If the energy difference between the states is E, how much energy must the photon have in order for energy to be conserved?
 
The energy levels of an atoms are quantized.
That is the electrons can only occupy certain levels in the atom.
The electrons fill up from the bottom towards the top levels, but
when the atom is excited, as by an increase in the temperature
of the material in this case, some of the electrons can be raised
to higher energy levels leaving some lower energy levels unoccupied.
We then get photons being emitted by the atom when the electrons
return to the ground states - the unoccupied lower energy levels.
 
Ok. Thank you. So the atom emits an photon when it drops down from the excited state to a lower state, and absorb an photon when it goes from a lower to a higher energy state. I do know the different states of a hydrogen atom (calculated by the rydebergs formula) but is it possible for use that formula in conjuncton with other atoms like f.ex sodium?
Thank you.
 
No. The interactions of the various states come into play and changes the formula quite dramatically.
 
Ok. So does every atom have a constant number (which does not have to be calculated) for every energy level?
 
I would think that the quantum mechanics is so complex that such formulas cannot even be derived and
thus probably don't even exist.
 
Ok, thank you.

So according to my question, which formula would be the most appropriate to use for calculation of the energy difference in ground and exited state in a sodium atom?

Thank you
 
sindre00 said:
Ok, thank you.

So according to my question, which formula would be the most appropriate to use for calculation of the energy difference in ground and exited state in a sodium atom?

Thank you
The energy of the photon emitted is equal to the energy difference between the excited and ground states. You know the wavelength of the emitted photon. What is its energy?
 
  • #10
E = hc/lambda, hence
E= (6.6*10-34)*(3.0*108) / (5.89*10-7m)

So this will give me the energy difference of the ground state and excited state as the energy of the photon is equal to the energy difference?
 
  • #11
You can use the formula you gave above. lambda is the wavelength of the photon and E is its energy, which is also the
energy difference between the excited and ground state. This energy became available as the electron fell back to the ground
state from the excited state of the atom, that is it transited from a higher energy level to a lower energy level in the atom.
 
  • #12
sindre00 said:
E = hc/lambda, hence
E= (6.6*10-34)*(3.0*108) / (5.89*10-7m)

So this will give me the energy difference of the ground state and excited state as the energy of the photon is equal to the energy difference?
Yes, but in what units? Remember you have to give the energy difference in eV-s.
 
  • #13
That gives you the energy in joules since you have worked in S.I. units. Convert it to electron volts.
 

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