fluorescence

[akimhaneul]

Why is there a difference between peak excitation wavelength and peak emission wavelength when a substance absorbs photon of certain energy? Why is the emission photon lower in energy?

Also is photoelectric effect a stronger phenomenon than fluorescence since the electron is actually emitted instead of going back down to ground state?

Thanks!

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[curvesetter]

Why is there a difference between peak excitation wavelength and peak emission wavelength when a substance absorbs photon of certain energy? Why is the emission photon lower in energy?

Also is photoelectric effect a stronger phenomenon than fluorescence since the electron is actually emitted instead of going back down to ground state?

Thanks!

I don't know where you're drawing this information from but I think for the purposes of the MCAT, it is okay to assume the two are equivalent (i.e. the energy required to bump an electron to a higher shell = the energy released when an electron drops down to the original shell). But, if you're curious, what you're describing is known as the Stokes shift. Essentially, when an electron absorbs energy, it can dissipate some of it as heat. This allows it to absorb more energy than is emitted when it jumps down.

As far as the photoelectric effect, I'm not sure if "stronger" is the correct word but you have the concept down perfectly. For the PE effect, you're essentially calculating the energy required to take an electron from its outermost shell (i.e. n = 5) and ejecting it (i.e. n = infinity). In fluorescence, electrons are jumping back down from non-infinitely-far-away shells which will emit a lower amount of energy when compared to the energy required for the PE effect.

 

[akimhaneul]

I don't know where you're drawing this information from but I think for the purposes of the MCAT, it is okay to assume the two are equivalent (i.e. the energy required to bump an electron to a higher shell = the energy released when an electron drops down to the original shell). But, if you're curious, what you're describing is known as the Stokes shift. Essentially, when an electron absorbs energy, it can dissipate some of it as heat. This allows it to absorb more energy than is emitted when it jumps down.

As far as the photoelectric effect, I'm not sure if "stronger" is the correct word but you have the concept down perfectly. For the PE effect, you're essentially calculating the energy required to take an electron from its outermost shell (i.e. n = 5) and ejecting it (i.e. n = infinity). In fluorescence, electrons are jumping back down from non-infinitely-far-away shells which will emit a lower amount of energy when compared to the energy required for the PE effect.

I'm being kinda nitpicky but do you know why there is some loss of energy as heat when the electron goes back down?


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[aldol16]

Why is there a difference between peak excitation wavelength and peak emission wavelength when a substance absorbs photon of certain energy? Why is the emission photon lower in energy?

I don't know where you're drawing this information from but I think for the purposes of the MCAT, it is okay to assume the two are equivalent (i.e. the energy required to bump an electron to a higher shell = the energy released when an electron drops down to the original shell). But, if you're curious, what you're describing is known as the Stokes shift. Essentially, when an electron absorbs energy, it can dissipate some of it as heat. This allows it to absorb more energy than is emitted when it jumps down.

Yeah, this concept way beyond the scope of the MCAT, but if you're interested, it has to do with quantum mechanics and what is known as the Franck-Condon principle. The Frank-Condon principle says that electronic transitions occur on a timescale that is much faster than nuclear translations, so we can, for all practical purposes, assume that nuclei are fixed during electronic translations. What that means is that in the following diagram, the energy functions are fixed along the x-axis and do not move:

Therefore, an electronic transition between the E0 ground state will go to an excited vibrational state of E1 and since it's an excited state, it's unfavorable for it to remain there for long. So it relaxes to the E1 ground state first before it jumps back down to the E0 state, releasing a photon in the process. This relaxation energy is dissipated in the form of heat. Since some of the initial excitation energy is dissipated in this way, the photon that is released in the E1 --> E0 transition will be lower in energy, or red-shifted.

 

[aldol16]

Also is photoelectric effect a stronger phenomenon than fluorescence since the electron is actually emitted instead of going back down to ground state?

Stronger in what sense? Yes, you need to put in more energy to completely knock an electron loose from a molecule than to simply put it in an excited state. You can excited a molecule using UV or even visible light. The photoelectric effect, on the other hand, gives rise to XPS, or x-ray photoelectron spectroscopy. X-rays are much stronger than UV or visible light rays and more energy is required to knock electrons completely loose.