anbuitachi

ASA Member
10+ Year Member
Oct 26, 2008
4,359
1,407
281
Utah
Status
Resident [Any Field]
Hi, i have a question.

so energy is quantized and lower energy can't bring e to the excited level? what if a photon that's higher in energy that the excited level transfers the energy to the electron? does that do anything or does the energy have to be exactly matching that of the excited level to make the electron move?

thanks
 
Oct 21, 2009
142
1
0
Status
Medical Student
Yes, a higher energy photon will make the electron jump energy levels. The energy has to be equal to or greater than the difference in energy levels.
 

anbuitachi

ASA Member
10+ Year Member
Oct 26, 2008
4,359
1,407
281
Utah
Status
Resident [Any Field]
So if it has more than n=1 and less than 2, it jumps to 1, but where do the rest of the energy go? reemitted as light?
 

RogueUnicorn

rawr.
7+ Year Member
Jul 15, 2009
9,746
1,595
181
Status
Resident [Any Field]
Turns into Kinetic energy for said electron
my quantum gets progressively fuzzier, but that can't be right, as electrons always move at the speed of light. what energy is not transferred is retained by the photon
 

MLT2MT2DO

10+ Year Member
Jan 21, 2008
2,600
386
281
Status
Resident [Any Field]
my quantum gets progressively fuzzier, but that can't be right, as electrons always move at the speed of light. what energy is not transferred is retained by the photon
As Kinetic Energy? Looks like I need to revisited this section as well.
 

BerkReviewTeach

Company Rep & Bad Singer
Vendor
10+ Year Member
May 25, 2007
3,884
655
281
so energy is quantized and lower energy can't bring e to the excited level? what if a photon that's higher in energy that the excited level transfers the energy to the electron? does that do anything or does the energy have to be exactly matching that of the excited level to make the electron move?
Think about what absorption spectra look like and you can deduce your answer. If what you describe is true, that you can't undershoot the level but you can over shoot it, then if n = 1 to n = 2 required orange light for instance, then the compound should absorb orange light once it reaches that threshold and then absorb everything above that because it can overshoot it. Absorption spectra don't look like that. They have distinct lines (widened in some cases by absorptions that also include rotational or vibration energy as well). Because they have a few distinct lines, the conclusion is that they can only absorb in a few distinct places, so the transition must be exact.

Another way of thinking about this can be applied to IR absorption in organic chemistry. Is a carbonyl 1700 cm-1 or is it 1700 and higher cm-1? Transitions are very specific, and can't be undershot or overshot (except with complete ionization--the photoelectric effect).

Turns into Kinetic energy for said electron
I think you might be mixing excitation and photoelectric effect together. To ionize an element, you can add a photon with the exact amount of energy as the work function (energy to ionize an electron from the surface of the material) or higher. When you add a photon of higher energy than the work function, the excess energy turns into kinetic energy for the ejected electron.
 

MLT2MT2DO

10+ Year Member
Jan 21, 2008
2,600
386
281
Status
Resident [Any Field]
I think you might be mixing excitation and photoelectric effect together. To ionize an element, you can add a photon with the exact amount of energy as the work function (energy to ionize an electron from the surface of the material) or higher. When you add a photon of higher energy than the work function, the excess energy turns into kinetic energy for the ejected electron.
Thank you, as always great explanation
 

anbuitachi

ASA Member
10+ Year Member
Oct 26, 2008
4,359
1,407
281
Utah
Status
Resident [Any Field]
I see so for energy levels it has to be exact but for photoelectric effect it can be exact or higher..
 
Oct 21, 2009
142
1
0
Status
Medical Student
From physics.about.com:
In Einstein's theory, a photoelectron releases as a result of an interaction with a single photon, rather than an interaction with the wave as a whole. The energy from that photon transfers instantaneously to a single electron, knocking it free from the metal if the energy (which is, recall, proportional to the frequency nu) is high enough to overcome the work function (phi) of the metal. If the energy (or frequency) is too low, no electrons are knocked free. If, however, there is excess energy, beyond phi, in the photon, the excess energy is converted into the kinetic energy of the electron:
Kmax = h nu - phi
Therefore, Einstein's theory predicts that the maximum kinetic energy is completely independent of the intensity of the light (because it doesn't show up in the equation anywhere). Shining twice as much light results in twice as many photons, and more electrons releasing, but the maximum kinetic energy of those individual electrons won't change unless the energy, not the intensity, of the light changes.
 

badmintondr

5+ Year Member
Dec 10, 2009
181
1
91
Status
Medical Student
Yup. Above is correct. If your a simpleton like me, I look at it as:
If the energy of the electron is less than the work function (the energy needed for it to be excited), then nothing happens.
If the energy is above the work function, then E= Work function + KE (or 1/2mv^2). So the particle would gain additional velocity and all that goodie goodie stuff.
 
Apr 6, 2010
154
3
0
Status
So if it has more than n=1 and less than 2, it jumps to 1, but where do the rest of the energy go? reemitted as light?
Correct. [Old photon energy level] - [electron energy level jump] = [new photon energy level]

When Econ is talking about "kinetic energy to the electron", he means the case where an electron is knocked completely out of the atomic shell. In that case, instead of emitting a photon, the electron retains the "extra" energy as kinetic energy. But if you're talking electron shell transfer, the "extra" energy is emitted as a photon.
 

BerkReviewTeach

Company Rep & Bad Singer
Vendor
10+ Year Member
May 25, 2007
3,884
655
281
Correct. [Old photon energy level] - [electron energy level jump] = [new photon energy level]
HUH!?!?!? There's a thought here that can be manipulated to be correct, but I think you might want to use the right terminology. There is no "new photon energy level" or "old photon energy level", as a photon simply has energy. Electrons exist in energy levels, not photons. Photons are absorbed during electronic excitation and emitted during electronic relaxation. The photon involved in the process has an energy equal to the transition energy. So you could make your equation correct by rewriting as follows.

deltaE = hf = Final electronic energy level - Initial electronic energy level

Because energy levels are quantized, the gaps between levels (transition energies) are also quantized. You absorb a photon with lamda = 412 nm only if the energy of that photon matches the energy difference between the initial energy level (ground state) and the final energy level (excited state). For that specific transition, 411 nm won't work as 413 nm won't work; only the exact wavelength works. They may work for another transition, but only one photon works for a specific transition.

When Econ is talking about "kinetic energy to the electron", he means the case where an electron is knocked completely out of the atomic shell. In that case, instead of emitting a photon, the electron retains the "extra" energy as kinetic energy. But if you're talking electron shell transfer, the "extra" energy is emitted as a photon.
You are correct that the photoelectric effect states that the kinetic energy of the ejected electron corresponds to the excess energy of the photon (energy beyond the work function), as has already been stated in three previous posts in this thread (see badmitondr, Econ, and mine).

But you are incorrect to say that if you try to excite an electron to a level using a photon of larger energy that it will emit the extra energy as a photon. This only occurs if the electron is exactly excited to a higher excited state and then it falls back to a lower excited state. In such a case for instance, you might first absorb a 511 nm photon to go to an excited state II, which is short lived and falls down to excited state I, emitting a photon of let's say 916 nm (IR range). In that case there are two transition energies, each with an exact photon of the corresponding energy. For a good example of this, look at how a simply ruby laser works.