# Photoelectric effect experiment: is current proportional to charging time?

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1. Nov 18, 2014

### quantumtimeleap

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 [Broken] ). 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!

Last edited by a moderator: May 7, 2017
2. Nov 18, 2014

### BvU

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. Nov 18, 2014

### andrevdh

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.

Last edited: Nov 18, 2014
4. Nov 24, 2014

### quantumtimeleap

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.