# Some questions about the photoelectric experiment

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• rtareen
In summary: If so, that is the energy of an electron in a particular state of motion. If the energy were quantized, then the electron could only have certain specific energy values, not any arbitrary value. This is what quantization means in this context. So in a way, the energy of a single photon is quantized, but the overall energy spectrum is continuous because it is made up of individual photons with quantized energy values.
rtareen
TL;DR Summary
This is about the photoelectric experiment as it is presented in the introductory sequence. The book throws a lot of information at me at once and still leaves a lot of questions unanswered. I have attached the section for your reference.
Background: self-studying. Very confused. Here are some initial questions I have about the photoelectric experiment. Some more may pop up later.

1. The book says we know photons exist due to energy considerations (such as emission or absorption). They also say that this photon energy is quantized. But Einstein said that E = hf. So if the frequency spectrum is continuous, how can a photon's energy be quantized?

2. Why do the electrons so conveniently escape from the grasp of the nucleus instead of being pushed into it? Will we cover the answer later? (If this is too complicated please don't answer).

3. The photoelectric equation can be written ##hf = K_{max} + \Phi### . I know phi is the energy required for the electron to be ejected from the atom. But after it leaves, if we are not doing any experiment where there is a potential difference, the only energy the electron has is its kinetic energy. Where does the work function energy go after the electron escapes?

4. I think that, if ##hf = eV_0 + \Phi##, all photons will make it atleast to the anode? Is there ever a time where some electrons end up with less energy than others? In the beginning of the section they talked about "less energetic electrons" before they explained the experimental results. Was that just speculation based on the wave model?

5. What happens if ##hf > eV_0 + \Phi##? I know if the intensity increases the current increases. But what about the frequency?

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rtareen said:
1. The book says we know photons exist due to energy considerations (such as emission or absorption). They also say that this photon energy is quantized. But Einstein said that E = hf. So if the frequency spectrum is continuous, how can a photon's energy be quantized?
You skipped an important piece of what Einstein said, namely that E = hf is the energy of a single photon. The energy spectrum, say of a blackbody, is continuous because it contains individual photons of energies arbitrarily close to one another.
rtareen said:
2. Why do the electrons so conveniently escape from the grasp of the nucleus instead of being pushed into it? Will we cover the answer later? (If this is too complicated please don't answer).
In a metal, where the photoelectric effect is observed, one or maybe two atomic electrons are free to move within the metal (what do you think carries electric current in a copper wire?) and are not attached to any particular atom. This is covered in a condensed matter course at the undergraduate level under the topic of band theory.
rtareen said:
3. The photoelectric equation can be written ##hf = K_{max} + \Phi### . I know phi is the energy required for the electron to be ejected from the atom. But after it leaves, if we are not doing any experiment where there is a potential difference, the only energy the electron has is its kinetic energy. Where does the work function energy go after the electron escapes?
The free electrons in a metal are free to move within the metal but not free to escape the metal's surface. To do that, they need an extra oomph which is the work function. The work function is the price the electron has to pay to free itself from the metal. The initial photon energy is split into energy that is carried by the electron in the kinetic form and energy that stays in the metal essentially as heat.
rtareen said:
4. I think that, if ##hf = eV_0 + \Phi##, all photons will make it atleast to the anode? Is there ever a time where some electrons end up with less energy than others? In the beginning of the section they talked about "less energetic electrons" before they explained the experimental results. Was that just speculation based on the wave model?
What does V0 represent? If it is the stopping voltage, then all electrons (you wrote photons) will just barely make it to the anode assuming of course that monochromatic radiation is incident on the cathode. That's because they all have the same kinetic energy equal to the difference between the photon energy and the work function. If you throw 10,000 identical balls straight up in the air with the same kinetic energy, they will all reach the same maximum height.
rtareen said:
5. What happens if ##hf > eV_0 + \Phi##? I know if the intensity increases the current increases. But what about the frequency?
Increased frequency for the same work function means increased kinetic energy of the emitted electrons. If you increase the frequency and keep the retarding voltage the same, then the electrons will slam into the anode and there will be a current detected. The idea is to adjust the voltage at the threshold of having no current. Then you know that eV is the kinetic energy of the emitted electrons.

rtareen
kuruman said:
You skipped an important piece of what Einstein said, namely that E = hf is the energy of a single photon. The energy spectrum, say of a blackbody, is continuous because it contains individual photons of energies arbitrarily close to one another.

So you're saying the values the energy itself can take is not quantized? Then what is the "thing" that is quantized here? Just because the energies are separated from each other doesn't mean they're quantized. I could give the example of moving balls. 1 ball might be moving at 1 mph and the other at 2 mph. The kinetic energies they could possibly take are continuous,. The fact that their energies are appearing separately doesn't mean the balls or their energies are quantized. I know its only an introductory course but I feel I should understand something as fundamental as energy being quantized.

kuruman said:
In a metal, where the photoelectric effect is observed, one or maybe two atomic electrons are free to move within the metal (what do you think carries electric current in a copper wire?) and are not attached to any particular atom. This is covered in a condensed matter course at the undergraduate level under the topic of band theory.

The photoelectric effect only works if the material is a metal? This makes sense but I want to confirm.

kuruman said:
The free electrons in a metal are free to move within the metal but not free to escape the metal's surface. To do that, they need an extra oomph which is the work function. The work function is the price the electron has to pay to free itself from the metal. The initial photon energy is split into energy that is carried by the electron in the kinetic form and energy that stays in the metal essentially as heat.

Thank you.

kuruman said:
What does V0 represent? If it is the stopping voltage, then all electrons (you wrote photons) will just barely make it to the anode assuming of course that monochromatic radiation is incident on the cathode. That's because they all have the same kinetic energy equal to the difference between the photon energy and the work function. If you throw 10,000 identical balls straight up in the air with the same kinetic energy, they will all reach the same maximum height.

Thank you!
kuruman said:
Increased frequency for the same work function means increased kinetic energy of the emitted electrons. If you increase the frequency and keep the retarding voltage the same, then the electrons will slam into the anode and there will be a current detected. The idea is to adjust the voltage at the threshold of having no current. Then you know that eV is the kinetic energy of the emitted electrons

I know that increasing intensity will increase the number of electrons that are ejected into the circuit, and now I know that increasing the frequency will increase the kinetic energy. How does increasing the number of electrons that smash into the anode increase the current, and what does the kinetic energy do in terms of a circuit (how do we detect an increased kinetic energy).

rtareen said:
1. The book says we know photons exist due to energy considerations (such as emission or absorption). They also say that this photon energy is quantized. But Einstein said that E = hf. So if the frequency spectrum is continuous, how can a photon's energy be quantized?
The photon spectrum is continuous but the electromagnetic field consists of light quanta called photons.
When it comes to the emission and absorption spectrums, the energy levels of atoms is also quantized. When an excited electron decays (becomes de-excited), the energy that is released in the form of a photon is given by ##E=hf##. The same way, an photon of frequency ##f## can be absorbed if there is an energy level that is ##hf## higher than the current energy level of the atom.
rtareen said:
2. Why do the electrons so conveniently escape from the grasp of the nucleus instead of being pushed into it? Will we cover the answer later? (If this is too complicated please don't answer).
Think of a marble at the bottom of a shallow bowl. If you give it a hard enough push it will leave the bowl. If you give it a smaller push it will go around in the bowl for a while (being excited, having more energy than the "ground state at the bottom"). Even if it is in motion around the sides of the bowl, giving it a hard enough push will make it leave the bowl, even if you push it toward the center. What matters most is how much energy the marble gets, not the direction.
rtareen said:
3. The photoelectric equation can be written ##hf = K_{max} + \Phi### . I know phi is the energy required for the electron to be ejected from the atom. But after it leaves, if we are not doing any experiment where there is a potential difference, the only energy the electron has is its kinetic energy. Where does the work function energy go after the electron escapes?
The work funktion is the energy required to get the marble to the rim of the bowl. After the marble leave the bowl, the remaning energy will be kinetic energy.
rtareen said:
4. I think that, if ##hf = eV_0 + \Phi##, all photons will make it atleast to the anode? Is there ever a time where some electrons end up with less energy than others? In the beginning of the section they talked about "less energetic electrons" before they explained the experimental results. Was that just speculation based on the wave model?
Not all electrons emitted have the same energy and not all electrons are emitted in the same direction. The formula ##hf = eV_0 + \Phi## is about the most energetic electrons. There is a more elaborated answer in another forum here: https://physics.stackexchange.com/q...s-the-kinetic-energy-of-a-photo-electron-vary .
rtareen said:
5. What happens if ##hf > eV_0 + \Phi##? I know if the intensity increases the current increases. But what about the frequency?
If the frequency of the light increases, the kinetic energy of the (most energetic electrons) increases.

rtareen
rtareen said:
So you're saying the values the energy itself can take is not quantized? Then what is the "thing" that is quantized here? Just because the energies are separated from each other doesn't mean they're quantized. I could give the example of moving balls. 1 ball might be moving at 1 mph and the other at 2 mph. The kinetic energies they could possibly take are continuous,. The fact that their energies are appearing separately doesn't mean the balls or their energies are quantized. I know its only an introductory course but I feel I should understand something as fundamental as energy being quantized.
I think you need to understand the historical background a little bit. Before Planck, the belief was that a blackbody emits continuous electromagnetic radiation continuously, i.e. electromagnetic waves emitted by oscillating charges in the emitting blackbody had a continuous range of frequencies. Doing the math led to the ultraviolet catastrophe which in a nutshell predicted that at the higher frequencies, there will be more and more emitted energy per frequency interval. This contradicted the observed experimental results. Max Planck saw that the problem can be fixed by assuming that blackbody electromagnetic radiation is emitted in packets of energy that is proportional to the frequency of the radiation and calculated the proportionality constant which is now known as Planck's constant. Planck, however, was reluctant to go as far as saying that electromagnetic radiation is also absorbed in packets. It was Einstein's interpretation of the photoelectric effect that provided this view. In case you are wondering, the mathematical formulation of blackbody radiation is covered in a statistical mechanics intermediate-level course (so much to learn!)
rtareen said:
The photoelectric effect only works if the material is a metal? This makes sense but I want to confirm.
Confirmed.
rtareen said:
Thank you.
You are welcome.
rtareen said:
Thank you!
Ditto!
rtareen said:
I know that increasing intensity will increase the number of electrons that are ejected into the circuit, and now I know that increasing the frequency will increase the kinetic energy. How does increasing the number of electrons that smash into the anode increase the current, and what does the kinetic energy do in terms of a circuit (how do we detect an increased kinetic energy).
More photons of the same frequency (increased intensity) means more electrons per unit time are reaching the anode. More electrons per unit time reaching the anode means more charge per unit time goes into the anode. The definition of current is charge per unit time. So ##\dots##

rtareen

We know increasing intensity increases the number of photons that hit the metal, and increasing frequency increases the energy of photons that hit the metal. So intensity causes an increased number of ejected electrons and frequency increases their speed so both contribute to increased photocurrent (current is charge per unit time). Since the frequency determines the discrete energy of a photon, if the frequency is not enough, there will not be enough energy for an electron to reach the anode no matter how many photons hit the metal. But why don't the energies of the photons that excite the electrons add up and eventually eject one? Do the electrons lose the energy from a previous photon before the next photon strikes?

rtareen said:
The book

What book?

PeterDonis said:
What book?
I attached it to the original post.

rtareen said:
I attached it to the original post.

No, you didn't. You attached a PDF of an excerpt from it. That doesn't tell me what book it is--who wrote it, who published it, when was it published, is it an actual textbook, etc. You need to provide that information.

PeterDonis said:
No, you didn't. You attached a PDF of an excerpt from it. That doesn't tell me what book it is--who wrote it, who published it, when was it published, is it an actual textbook, etc. You need to provide that information.
Sears and Zemansky's University Physics 15th edition

rtareen said:
Sears and Zemansky's University Physics 15th edition

Ok, thanks!

PeterDonis said:
Ok, thanks!
Sorry 14th edition not 15th

Note that there's a mistake in almost all textbooks concerning the "work function" in the Millikan experiment. In
$$h f=e V+\Phi$$
##\Phi## is the work function of the anode note the cathode. For details, see

J. Rudnick, D. Tannhauser, Concerning a widespread error in the description of the photoelectric
effect, Am. J. Phys. 44, 796 (1976).
https://doi.org/10.1119/1.10130

Millikan, of course, got it right (and he did a large effort to experimentally clarify this issue):

R. Millikan, The Distinction between Intrinsic and Spurious Contact EMFS and the Question of
the Absorption of Radiation by Metals in Quanta, Phys. Rev. 18, 236 (1921).
https://doi.org/10.1103/PhysRev.18.236

rtareen said:

We know increasing intensity increases the number of photons that hit the metal, and increasing frequency increases the energy of photons that hit the metal. So intensity causes an increased number of ejected electrons and frequency increases their speed so both contribute to increased photocurrent (current is charge per unit time). Since the frequency determines the discrete energy of a photon, if the frequency is not enough, there will not be enough energy for an electron to reach the anode no matter how many photons hit the metal. But why don't the energies of the photons that excite the electrons add up and eventually eject one? Do the electrons lose the energy from a previous photon before the next photon strikes?

The frequency does not increase the current. The frequency do increase the speed of the electrons but the speed of the electrons is not correlated to the current. The charge per unit time depends on how many electrons that is emitted per unit time, not how fast the electrons is moving.

## 1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electrons are emitted from a material when light of a certain frequency is shone onto it. This effect was first observed by Heinrich Hertz in 1887 and was later explained by Albert Einstein in 1905.

## 2. How does the photoelectric effect support the concept of photons?

The photoelectric effect showed that light behaves like particles, known as photons, rather than just waves. This is because the energy of the emitted electrons was found to be dependent on the frequency, rather than the intensity, of the incident light. This supports the idea that light is made up of discrete packets of energy, or photons.

## 3. What is the work function in the photoelectric effect?

The work function is the minimum amount of energy required to remove an electron from the surface of a material. It is different for each material and can be thought of as the "binding energy" holding the electron to the material. In the photoelectric effect, the energy of the incident photons must be greater than the work function in order for electrons to be emitted.

## 4. How does the intensity of light affect the photoelectric effect?

The intensity of light does not affect the photoelectric effect, as long as the frequency of the light is above the threshold frequency. This is because the energy of the photons is what determines the energy of the emitted electrons, not the number of photons. However, increasing the intensity of light will result in a higher number of emitted electrons, as more photons will be hitting the material.

## 5. What is the significance of the photoelectric effect in modern technology?

The photoelectric effect has many applications in modern technology, such as in solar panels, photocells, and photomultiplier tubes. It also plays a crucial role in the development of quantum mechanics and our understanding of the dual nature of light. Additionally, the photoelectric effect has led to advancements in fields such as spectroscopy and photochemistry.

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