Light emission and energy of states

In summary: E= (6.6*10-34)*(3.0*108) / (5.89*10-7m) In summary, the emission spectrum of thermally excited sodium atoms practically consists of a single intensive line at 589 nm wavelength. The energy difference between the excited and ground states of the sodium atom is 6.6*10-34 eV.
  • #1
sindre00
5
0

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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
No. The interactions of the various states come into play and changes the formula quite dramatically.
 
  • #6
Ok. So does every atom have a constant number (which does not have to be calculated) for every energy level?
 
  • #7
I would think that the quantum mechanics is so complex that such formulas cannot even be derived and
thus probably don't even exist.
 
  • #8
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
 
  • #9
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.
 

1. What is light emission?

Light emission is the process by which energy is released in the form of electromagnetic waves, specifically in the visible range. This can occur naturally, such as with bioluminescence in certain organisms, or artificially through the use of light sources like light bulbs or lasers.

2. How is light emission related to energy states?

Light emission is directly related to the energy states of atoms or molecules. When an atom or molecule absorbs energy, its electrons are excited to higher energy states. As these electrons return to their ground state, they release the excess energy in the form of light emission.

3. What factors affect the energy states of atoms or molecules?

The energy states of atoms or molecules can be affected by various factors, including temperature, pressure, and the presence of other atoms or molecules. These factors can influence the amount of energy absorbed or released, and therefore impact the frequency and intensity of light emission.

4. How is light emission measured?

Light emission can be measured using various techniques, such as spectroscopy or photometry. These methods involve detecting and analyzing the different wavelengths of light emitted by a substance and can provide information about the energy states and transitions of the atoms or molecules within it.

5. What are some practical applications of understanding light emission and energy states?

Understanding light emission and energy states is crucial in many fields, including chemistry, physics, and engineering. Some practical applications include developing new lighting technologies, studying the properties of materials, and designing efficient energy sources. It also plays a crucial role in the study of the universe, as light emission from stars and other celestial bodies provides important information about their composition and behavior.

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