I conducted an experiment today where I had to use a photocathode of unknown material (model: Daedalon Corporation Photoelectric Effect EP-05) and study the photoelectric (PE) effect to make a calculation for Planck's constant $h$ as well as the material's work function $\phi$. The number I calculated is within the correct order of magnitude, but I think better accuracy is possible through this experiment.

I think my error was due to:

(a) systematic error, or

(b) a fundamental misunderstanding on my part regarding how to define the stopping voltage $V_s$ in the experiment which my instructor will not help clarify.

If anyone has a good understanding of these concepts and is willing to help me decipher which it is (and why) I will be very grateful.

This was our PE device:Daedalon Corporation Photoelectric Effect EP-05

It is equipped with an ammeter with $\mathrm{nA}$ precision and we also connected a voltmeter with $\mathrm{mV}$ precision. The whole experiment was performed in the dark with maximum efforts to prevent any stray light from affecting our results so I doubt this is where the error comes from. I turned the voltage knob to maximum (approx $3.174 \ \mathrm{V}$) and zeroed the ammeter in the middle to make it easier to read. I first used a mercury lamp with a blue filter. I shined it at the photocathode, the current reading jumped up, then I turned the voltage knob down until the current reached zero again and recorded that value as our stopping voltage $V_s$. Is this correct? I asked my professor and he said yes but I thought that we must normalise this value with respect to something.

I repeated this measurement 5 times, recalibrating the zero position each time because of how sensitive the current is. I then repeated the experiment with a green filter and once more with no filter for mercury's UV spectrum line. I then repeated it with a green laser, a blue laser, and a red laser (while still using the appropriate filters).

For each wavelength, I took the mean of the 5 voltage data and then plotted that against $1/\lambda$. Since these are photoelectrons, $1 \ \mathrm{V}$ should correspond to $1 \ \mathrm{eV}$ of energy. So by finding the linear regression, it should be $V_s=\frac{hc}{\lambda}-\phi$. Here is my data:


The linear regression is $y=(705\pm83 \ \mathrm{eV\!\cdot\!nm})x+(-0.55\pm0.18 \ \mathrm{eV})$. Therefore, by my calculation, $hc=705\pm83 \ \mathrm{eV\!\cdot\!nm}$. The accepted value is $hc=1\,240 \ \mathrm{eV\!\cdot\!nm}$. This also implies a very low work function of $0.55\pm0.18 \ \mathrm{eV}$. According to this paper I found in the Journal of Physics: https://iopscience.iop.org/article/10.1088/1361-648X/aa79bd/meta, the lowest work functions discovered are around $0.7-0.8 \ \mathrm{eV}$. Is this the best accuracy I can hope to achieve with this data or have I completely misunderstood/missed something? Any help would be greatly appreciated.

  • $\begingroup$ If you are getting the right order of magnitude for $\hbar$ with only one day's work, you are doing reasonably well. There are tons of difficult to control systematics in an experiment like this, where a very small current is involved. $\endgroup$
    – Buzz
    Commented May 22, 2019 at 1:35
  • $\begingroup$ Determination of Planck’s Constant Using the Photoelectric Effec $\endgroup$
    – Farcher
    Commented Oct 25, 2021 at 8:07

1 Answer 1


For the benefit of others who might stumble on this:

We have used this equipment at SUNY Geneseo Physics for many years, with dozens of students using a dozen of the apparatus each year. With very high consistency, we get results very similar to those described above. Ten PhD physicists in the department have tried to resolve the mystery, with no success. There must be something about the equipment that causes this, but we can't figure out what it is. But we do note that the detector in this equipment is a vacuum tube that is not actually designed to be used with a reverse potential difference, and one electrode is a thin wire rather than a surface. Maybe there is something about the geometry?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.