Photoelectric Effect [ENDORSED]

[Abbey Kerscher 3O]

During the lecture we discussed light acting in waves and in photons. I understood that you can increase the intensity of a large wavelength, but will not get ejected electrons. However, because the light is working as photons, does that mean that larger wavelengths don't have photons- since there are no ejected electrons when you shine the light on a certain metal? I'm super confused on the duality of light in waves and photons.

[Claire_Zhou_1A]

I think that light with larger wavelengths also has photon properties... It's just the light has low frequency since the wavelength is large and the energy of photon is too small to reach threshold energy. I don't know if I'm right.

[Christian Hardoy 3F]

Larger wavelengths do "have photons." A larger wavelength of light may not eject an electron for a particular type of metal because the energy of the incident light, a.k.a the photons hitting the metal, depends on the frequency according to the equation E = hv, where E is the energy of the photon, h is Planck's constant, and v is the frequency of the light. As Dr. Lavelle mentioned before, frequency and wavelength are related according to the following equation: c = λv, where c is the speed of light, λ is the wavelength of the light, and v is the frequency of the light. So basically, if you decrease the wavelength of light, you're increasing the frequency (the relationship between the two is from c = λv). If you're increasing the frequency, then you're increasing the energy (E) of the incoming photons according to E=hv because as v increases, so does the energy of the incoming photons. It is important to remember that light behaves as both a wave and a particle, it isn't just one or the other, even if we focus on one aspect particularly for one type of problem or experiment. In summary, larger wavelengths of light, like all light, behave as both photons and a waves, and if ejection doesn't occur it means that the wavelength isn't small enough, not that the light doesn't "have a photon."


*It is also important to note that in all of this calculation, the amplitude of light is absent, meaning that it does not play a role in the ejection of electrons from a metal.

Hope this helps!

[Brenton Hwee 2J]

I know that intensity and the number of photons are related, but is there any equation or way to numerically calculate those values? Or are the numerical values not relevant to what we're learning?

[Emma Cole 3F]

I understand that you need the threshold energy to in order to get the kinetic energy, but I don't really understand what it is. What does threshold energy mean in the photoelectric effect and where do you get it from?

[Christian Hardoy 3F]

The threshold energy when talking about the photoelectric effect is the energy required to remove an electron from an atom of a particular metal. The value of each threshold energy is unique to each metal. The values are determined experimentally I would guess. You might be asked to find the threshold energy if you're given enough information to be able to solve for it, but its a value that you'll be given otherwise. Its nothing you need to know off of the top if your head. Its like the melting point of a metal; each metal is unique and has relatively different properties and structure, producing different boiling points.

[JennaMinami1I]

I understand that you need the threshold energy to in order to get the kinetic energy, but I don't really understand what it is. What does threshold energy mean in the photoelectric effect and where do you get it from?

Hi! From my understanding, the threshold energy (also known as work function/Greek letter phi) is the minimum energy required to eject electrons, for a given metal (since the UV radiation is always reflected off of a metal).

[Abbey Kerscher 3O]

Larger wavelengths do "have photons." A larger wavelength of light may not eject an electron for a particular type of metal because the energy of the incident light, a.k.a the photons hitting the metal, depends on the frequency according to the equation E = hv, where E is the energy of the photon, h is Planck's constant, and v is the frequency of the light. As Dr. Lavelle mentioned before, frequency and wavelength are related according to the following equation: c = λv, where c is the speed of light, λ is the wavelength of the light, and v is the frequency of the light. So basically, if you decrease the wavelength of light, you're increasing the frequency (the relationship between the two is from c = λv). If you're increasing the frequency, then you're increasing the energy (E) of the incoming photons according to E=hv because as v increases, so does the energy of the incoming photons. It is important to remember that light behaves as both a wave and a particle, it isn't just one or the other, even if we focus on one aspect particularly for one type of problem or experiment. In summary, larger wavelengths of light, like all light, behave as both photons and a waves, and if ejection doesn't occur it means that the wavelength isn't small enough, not that the light doesn't "have a photon."


*It is also important to note that in all of this calculation, the amplitude of light is absent, meaning that it does not play a role in the ejection of electrons from a metal.

Hope this helps!

Thank you so much this totally helped me!!

[isabelle ruedisueli 1j]

Hi! On page 47 of the Course Reader, it says that as n approaches infinity, E approaches 0, and that (negative means bound e- has lower energy than free e-).
Can someone explain what "negative means bound e- has lower energy than free e-" means? What is a bound electron versus a free electron?
Thanks so much :)

[isabelle ruedisueli 1j]

Also, sidenote, do we have to do the quizzes in pen? and even the self-quizzes in pen?

[Da_Rhee_1O]

Yes, the quizzes should be done in pen. Dr. Lavelle mentioned himself that anything being graded must be done in pen. So I am not sure about the self-quizzes but for the last quiz without the answers provided, definitely do the problems in ink. Hope this helps!

[Christian Hardoy 3F]

Hi! On page 47 of the Course Reader, it says that as n approaches infinity, E approaches 0, and that (negative means bound e- has lower energy than free e-).
Can someone explain what "negative means bound e- has lower energy than free e-" means? What is a bound electron versus a free electron?
Thanks so much :)

A bound electron is an electron that is "bound" to the nucleus, as in it is part of an orbital and remains around the atom. A free electron is an electron that is off by itself; it isn't a "part" of any particular atom. For example, consider the photoelectric effect example that was talked about before. The atoms in the metal that is being bombarded with light all have bound electrons and the electrons that are ejected are free and are of a higher energy than the bound electrons because a photon of light had to be absorbed in order to eject the electron.
The phrase "negative means bound e- has lower energy than free e-" is also referencing the different energy levels that Dr. Lavelle was diagramming and talking about in lecture.
Here's a crude diagram for your reference: (the vertical line on the left represents energy)
|____________________________ n=4
|____________________________ n=3
|
|____________________________ n=2
|
|
|
|____________________________ n=1


( + ) <--- atomic nucleus

Anyway, when the atom is hit with a photon, the electron moves from one energy state to a higher one, lets say from n=1 to n=2, for example, because it absorbed the energy of the photon. This means that n=2 is a higher energy state than n=1 and n=3 is a higher energy state than n=2, etc. Therefore, it logically follows that the energy of a free electron, which is basically at n=infinity, would have a higher energy than the bound electron.

To explain the phrase, "negative means bound e- has lower energy than free e" you'll have to consider the equation E=-hR/n^2, where h is Planck's constant, R is the Ryberg constant, and n is the energy state.
This equation gives the energies of the different n states in the hydrogen atom, which is the atom we're talking about.

***It is important to note the negative sign on the term because it indicates that the energy's sign is negative, which tells us that this equation is talking about the energy of the electron in terms of energy lost, e.g. the energy emitted in the form of a photon, not in terms of energy added from the ground state, e.g. energy absorbed.*** Because the reference point is n=infinity and E=0 and not the ground state of the atom, any value of n, except n=infinity will be negative, and if n=infinity represents a truly free electron, and n=some smaller value like n=1,2,3,4,5,6... represents different feasible energy states where the electron is bound, then the energy of those states where the electron is bound will always be less than the energy of the free electron.

Does this make sense? I know its a bit long and convoluted...

To put it another, less mathematical way, you can think of a free electron as an that electron is very far away from the nucleus of an atom, like when n is some massive number or infinity. If an electron is very far from the nucleus then it doesn't really interact with the nucleus at all. This means that the absorption of a photon isn't necessary to eject the electron from the nucleus because if there is no attraction between the nucleus and the electron, then no energy is required to pull it away from the nucleus. To free a bound electron, an electron around an atom, you must put in energy. If energy must be put in to free an electron, and no energy is required to remove a free electron, then that means that the free electron already has that energy, meaning that it has a higher energy when compared to a bound electron.

Hope this helps!

[isabelle ruedisueli 1j]

I forgot to thank you for this explanation! Super helpful! You should be a TA:)