Photoelectric effect experiment: is current proportional to charging time?

In summary, the conversation discusses the basics of the photoelectric effect and the relationship between stopping voltage, frequency, and intensity of incident radiation. The experiment also includes measuring the charging time needed to achieve the stopping potential, and the question is raised whether the charging time is proportional to the current. It is concluded that the charging time is inversely proportional to the current, and the data suggests a more complex relationship than a simple linear one.
  • #1
quantumtimeleap
10
0
I'm not really sure where to put this question, but definitely this is just an introductory physics coursework. Let me refresh you first with the basics of the photoelectric effect.

We all know that in the photoelectric effect the stopping voltage is just the kinetic energy obtained by the electrons displaced by the photons that struck the photodiode. Usually we experimentally measure the maximum stopping voltage because we only know the work function of the metal, which is the minimum energy needed to displace an electron from the surface.

The main result of the experiment is that the stopping voltage is proportional to the frequency of the incident radiation, and not correlated whatsoever to its intensity. Instead, it is the current generated that should be directly proportional to intensity, since intensity is simply the amount of photons carried by a light beam (and more photons incident on the metal means more electrons displaced --> higher current).

Also, the experiment includes measuring the charging time needed to achieve the stopping potential. This is the apparatus we used: ( http://www.pha.jhu.edu/~c173_608/photoelectric/he9370.pdf ). I'm pretty sure most photoelectric effect experiments use something similar, such that the stopping voltage is achieved after a nonzero charging time in the circuits of the apparatus.

My question is this: Is the charging time proportional to the current? One thing for sure is that the current is correlated to the charging time. More electrons received by the anode by the displacing effect of the photons should mean a faster time to achieve the stopping potential. I'm afraid of my hunches, but I think the relationship is probably not linear even disregarding noise and random error. Can anyone give me a definite yes or no? Thank you in advance!
 
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  • #2
Long story, short answer: Current is charge per time, so: no, charging time is inversely proportional to current. As you remarked correctly: more current -> shorter time.
 
  • #3
I would think that it will be an exponential relationship since you are dealing with the charging of a capacitor. You can use your data to investigate it. Play around with the % transmitted and the charging time data.
 
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  • #4
thank you, everyone! the data did turn out exhibiting something more complicated than a simple linear relationship ## I = k \frac{1}{\Delta t} ## for a proportionality constant k.
 
  • #5


I can confirm that the charging time and current are not directly proportional in the photoelectric effect experiment. While it is true that the current is correlated to the charging time, the relationship is not linear.

The charging time is affected by various factors such as the intensity and frequency of the incident radiation, the work function of the metal, and the capacitance of the circuit. Therefore, it is not a simple linear relationship between charging time and current.

Additionally, the charging time also depends on the individual characteristics of the apparatus used in the experiment, such as the type of photodiode and the circuit design. This can introduce variations and uncertainties in the relationship between charging time and current.

In conclusion, while the current is correlated to the charging time in the photoelectric effect experiment, the relationship is not a direct proportion. It is important to carefully consider all the factors and variables involved in the experiment to accurately interpret the results.
 

1. What is the photoelectric effect experiment?

The photoelectric effect experiment is an experiment that demonstrates the emission of electrons from a metal surface when it is exposed to light of a certain frequency. This was first observed by Heinrich Hertz in 1887 and later explained by Albert Einstein in 1905.

2. How does the photoelectric effect experiment work?

In the photoelectric effect experiment, a metal plate is connected to a circuit and a light source is directed at the plate. When the light hits the plate, electrons are emitted from the surface and create a current in the circuit. The number of electrons emitted and the intensity of the current depends on the frequency and intensity of the light, respectively.

3. Is current proportional to charging time in the photoelectric effect experiment?

Yes, the current is proportional to the charging time in the photoelectric effect experiment. This means that the longer the light is shone on the metal plate, the more electrons will be emitted and the higher the current will be in the circuit.

4. What factors affect the current in the photoelectric effect experiment?

The current in the photoelectric effect experiment is affected by the intensity and frequency of the light, as well as the properties of the metal surface, such as its work function and the number of free electrons. Additionally, the distance between the metal plate and the light source can also impact the current.

5. What is the significance of the photoelectric effect experiment?

The photoelectric effect experiment played a crucial role in the development of quantum mechanics and provided evidence for the particle nature of light. It also led to the discovery of the photon and has practical applications in technology, such as photovoltaic cells used in solar panels.

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