Conditions for H atom to absorb a photon

[SaintJude]

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What does this mean?

When a hydrogen atom absorbs a photon, its electron makes a transition to a higher energy state. This can only occur if the difference in energy between the initial and final states equals the energy of the photon

 

[MedPR]

What does this mean?

That energy is quantized and that the electron must absorb at least the threshold of energy transition in order to make that transition. If it takes 100kJ to excite an electron up 1 level, but you only give it 99kJ, the energy isn't going to move out of the current level. Also, I assume that if you gave it 110kJ of energy it would be excited up 1 level. So if what you're reading is saying that the photon must be exactly the energy necessary (e.g can't be more) then I don't know.

 

[theseeker4]

That energy is quantized and that the electron must absorb at least the threshold of energy transition in order to make that transition. If it takes 100kJ to excite an electron up 1 level, but you only give it 99kJ, the energy isn't going to move out of the current level. Also, I assume that if you gave it 110kJ of energy it would be excited up 1 level. So if what you're reading is saying that the photon must be exactly the energy necessary (e.g can't be more) then I don't know.

I actually made that same error a week or so ago on this forum, and turned out to be wrong. The light does need to be a specific wavelength to be absorbed by the electron and excited up one level. It is not simply a minimum value, but an exact value. The electron cannot absorb a higher-energy photon and be excited up, it has to absorb the exact energy needed for the jump. Therefore, the 110kJ energy photon wouldn't cause the jump of 1 level (though I believe it could cause a jump from level 1 to level 3 if the sum of energy differences from 1 to 2 and 2 to 3 was 110kJ).

 

[MedPR]

I actually made that same error a week or so ago on this forum, and turned out to be wrong. The light does need to be a specific wavelength to be absorbed by the electron and excited up one level. It is not simply a minimum value, but an exact value. The electron cannot absorb a higher-energy photon and be excited up, it has to absorb the exact energy needed for the jump. Therefore, the 110kJ energy photon wouldn't cause the jump of 1 level (though I believe it could cause a jump from level 1 to level 3 if the sum of energy differences from 1 to 2 and 2 to 3 was 110kJ).

So if it takes 100kJ to go from 1 to 2, the electron will not be excited to the n=2 level if it gets hit with a 110kj photon? Do you know what happens in a situation like that?

 

[SaintJude]

Ohhh, so is that why it actually takes more energy for a photon to be absorbed b/w n=1 & n=2? B/c the energy of the electron transition between the 2 lowest energy states is much bigger than that of the higher principle quantum numbers ?

And how interesting! The photon's energy must exactly equal the energy difference b/w the two orbits levels...

 

[theseeker4]

So if it takes 100kJ to go from 1 to 2, the electron will not be excited to the n=2 level if it gets hit with a 110kj photon? Do you know what happens in a situation like that?

In that situation, the photon of light would simply pass through/by as if the electron wasn't there. Just as a lower-energy photon would pass through. This phenomenon can be seen in absorption spectra, where specific wavelengths of light are absorbed corresponding to specific electrons making specific jumps in energy state upon the absorption of those photons.

Note that I am not 100% positive that an electron can jump two energy states when absorbing a single photon.

 

[theseeker4]

Ohhh, so is that why it actually takes more energy for a photon to be absorbed b/w n=1 & n=2? B/c the energy of the electron transition between the 2 lowest energy states is much bigger than that of the higher principle quantum numbers ?

And how interesting! The photon's energy must exactly equal the energy difference b/w the two orbits levels...

Correct, the lower the n of the quantum state, the greater the energy necessary to transition between them. Therefore the photon that will move the electron from n1 to n2 will have higher energy than the photon that will move the electron from n2 to n3.

 

[MedPR]

Ohhh, so is that why it actually takes more energy for a photon to be absorbed b/w n=1 & n=2? B/c the energy of the electron transition for the 2 lowest energy states is much bigger than that of the higher principle quantum numbers ?

Yes, because the electrons in the lower energy levels (1, 2, etc) are not shielded as well from the attractive force of the nucleus as electrons in the higher energy levels. You probably already know that, but this is something that somehow always manages to diffuse out of my brain every few weeks, so I wanted to take the opportunity to type it out!

To take this one step further the energy of a certain wavelength of light is calculated by E=h*f, where h is plancks constant and f is the frequency. Electrons emit light when they relax back down to a lower energy level. The wavelength of this light is longer (less energy) than the wavelength of the photon that excited the electron right?

Where did the energy go?

 

[SaintJude]

Correct, the lower the n of the quantum state, the greater the energy necessary to transition between them. Therefore the photon that will move the electron from n1 to n2 will have higher energy than the photon that will move the electron from n2 to n3.

Aha. Wow. Thank you! 🙂

 

[SaintJude]

Yes, because the electrons in the lower energy levels (1, 2, etc) are not shielded as well from the attractive force of the nucleus as electrons in the higher energy levels.

Thanks for that answer.

I also just read something that kind of sounded like it was related to your earlier comment about what the minimum energy & electron's kinetic energy

Kaplan:

If the energy of the photon is greater than 13.6eV then the condition for ionization is met. A photon above 13.6 eV will eject the electron from the atom, and the residual energy will be the kinetic energy of the electron .( I guess similar to the photoelectric effect.)

 

[theseeker4]

Thanks for that answer.

I also just read something that kind of sounded like it was related to your earlier comment about what the minimum energy & electron's kinetic energy

Kaplan:

If the energy of the photon is greater than 13.6eV then the condition for ionization is met. A photon above 13.6 eV will eject the electron from the atom, and the residual energy will be the kinetic energy of the electron .( I guess similar to the photoelectric effect.)

That is the photoelectric effect, correct. Photoelectric effect does require a minimum, not an exact value, of energy. The reason for this is the electron is not simply moving from one energy state to another, a discrete value, but is being ejected entirely from the atom's orbitals. Any energy over the minimum require for this contributes to the speed of the electron after ejection. Since there is no specific, discrete energy state the electron occupies after absorbing that photon, the photon simply needs a minimum energy for ionization.

 

[MedPR]

That is the photoelectric effect, correct. Photoelectric effect does require a minimum, not an exact value, of energy. The reason for this is the electron is not simply moving from one energy state to another, a discrete value, but is being ejected entirely from the atom's orbitals. Any energy over the minimum require for this contributes to the speed of the electron after ejection. Since there is no specific, discrete energy state the electron occupies after absorbing that photon, the photon simply needs a minimum energy for ionization.

And the equation for the resulting kinetic energy of the electron is KE=hf-workfunction, right? Just want to make sure I'm relating the right concept.

 

[SaintJude]

Yup

 

[MedPR]

Ok so why is the emitted energy of a relaxing electron less than the absorbed energy required for excitation? Where did the energy go?

 

[theseeker4]

Ok so why is the emitted energy of a relaxing electron less than the absorbed energy required for excitation? Where did the energy go?

What I read on Wikipedia, obviously not the most trusted source, is that the frequency of emission is identical to the frequency of absorption. Where have you seen that the energy was less? I don't have textbooks discussing the concept readily available.

 

[MedPR]

What I read on Wikipedia, obviously not the most trusted source, is that the frequency of emission is identical to the frequency of absorption. Where have you seen that the energy was less? I don't have textbooks discussing the concept readily available.

I've done passages in the past.. maybe it was a different absorption/emission topic though. I'm not really sure.

 

[theseeker4]

I've done passages in the past.. maybe it was a different absorption/emission topic though. I'm not really sure.

Maybe, I am not sure. I was under the impression that the frequency of the emission was the same as the absorption, and Wikipedia seemed to say the same thing. I have been wrong about this topic before, recently, so I would double check that one, lol.

 

[fluoropHore]

The emission frequency (energy) of a photon is less only in instances of phosphorescence/fluorescence. some of that energy is used in flipping the spin of the electron and it can no longer come back down to its original orbital (cause there can only be 1 out of 2 electrons with that spin per orbital- Pauli's exclusion principle) as phosphorescence or its used in vibrational energies in the case of molecules and emits IR photons before it comes back down (fluorescence).

 

[SaintJude]

Note that I am not 100% positive that an electron can jump two energy states when absorbing a single photon.

Also, just read that it is possible for an electron to move directly from the grounded state to a third or fourth excited state. Provided that the energy of the incoming light equals exactly the transition energy levels b/w the electron's final and original orbital levels.

Thanks for everyone's comments!

 

[MedPR]

The emission frequency (energy) of a photon is less only in instances of phosphorescence/fluorescence. some of that energy is used in flipping the spin of the electron and it can no longer come back down to its original orbital (cause there can only be 1 out of 2 electrons with that spin per orbital- Pauli's exclusion principle) as phosphorescence or its used in vibrational energies in the case of molecules and emits IR photons before it comes back down (fluorescence).

That's what it was. Thanks!