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Photoelectric effect at wavelength 492nm for mercury

  • Thread starter JulienB
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  • #1
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Homework Statement



Hi everybody! I just did the photoelectric experiment this week, and I have a report to write about it. We used a mercury vapour lamp, a set of metal interference filters and a grey filter in order to find the intersection point (and therefore ##U_{g,max}## between the curves (with and without grey filter) for each wavelength as the current is changing.
The experiment went very well except for the filter ##492##nm, where the curves look kind of flat making the intersection point harder to find. Our advisor said there is an explanation that we must give in our report.

Let me know if you need more informations about the experiment. I am assuming it is a pretty standard one that many of you probably already performed, but I might be wrong.

Homework Equations



##e \cdot U_{g,max} = h \cdot \nu - e \cdot \phi_A##

The Attempt at a Solution



So the first thing I've done was to search for the spectral lines of mercury (see attached file), and what I found was that the peak is not very pronounced at ##492##nm. But what does that mean physically? I've googled it but could not find anything tangible yet. My guess would be that for some quantum reason I do not know, fewer electrons are released from the mercury atom at that wavelength than for others. That is still pretty vague, and maybe wrong... Any help?


Thanks a lot in advance for your answers.


Julien.
 

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Answers and Replies

  • #2
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Have you compared your results to the spectral lines and their intensity of mercury?
 
  • #3
ehild
Homework Helper
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Use those filters only which transmit the strong lines of mercury. The spectrum show the intensity of light mercury emits at different wavelengths. The mercury lamp does not emit electrons. The wavelengths of the peaks are determined by the change of the energy of the mercury atom between the electronic levels. The intensity is determined by the change of dipole moment during the transition. Low-intensity peaks, like at 492 nm, can be combination or overtones.
 
  • #4
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@fresh_42 and @ehild, thanks a lot for your answers. We just finished finding every curve's intersection (for six filters in total) and then we had to make an "Einstein's line" (see attached file) to find ##h##, ##\nu_0## and ##e \cdot \phi_A##. The value for ##492##nm is slightly off but still close to the line (we didn't do the linear regression yet), and does not seem to give any indication about why the curve looks a little bad. By the way we were not given a choice of what to measure; we HAD to take the ##492##nm as well.

The spectrum show the intensity of light mercury emits at different wavelengths. (...) The wavelengths of the peaks are determined by the change of the energy of the mercury atom between the electronic levels. The intensity is determined by the change of dipole moment during the transition. Low-intensity peaks, like at 492 nm, can be combination or overtones.
May I ask what a "combination of overtones" would be then? Does that mean there is no change of energy at ##492##nm?

The mercury lamp does not emit electrons.
That was a typo, but it gives me the opportunity to resume the experiment and see what I get/don't get: the mercury lamp emits photons that go through a filter in order to let only photons with a certain wavelength pass through. They then hit something (the diode?) and the atoms of that something release detectable electrons if their kinetic energy is higher than the counter voltage (not sure about that). If this description is correct, I don't get why the spectrum of mercury would matter at all since the electrons are not coming from the lamp.

Thanks a lot for all your help.


Julien.
 
  • #5
ehild
Homework Helper
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May I ask what a "combination of overtones" would be then? Does that mean there is no change of energy at ##492##nm?
There is photon emitted at 492 nm. The photon has energy of hc/λ, which is the energy difference between the initial and final states of the mercury atom. I worked with infrared spectra, so I am not sure "combination bands" are mentioned in UV-Vis spectroscopy. The level schema of mercury is quite complicated. The emitted lines are assigned to transitions between the levels.
levels2.gif

The figure shows the transitions between the energy levels of mercury, producing the strongest lines in the visible range.
Some transitions are allowed, some are forbidden, but even the forbidden transitions can appear with a two-step process, and then you can find that the frequency of a weak band is the sum or difference of those of two strong bands. Considering the strong lines of mercury, the frequency of the 492 nm band is close to the difference between the 254 nm band and the 546 nm band.

That was a typo, but it gives me the opportunity to resume the experiment and see what I get/don't get: the mercury lamp emits photons that go through a filter in order to let only photons with a certain wavelength pass through. They then hit something (the diode?) and the atoms of that something release detectable electrons if their kinetic energy is higher than the counter voltage (not sure about that). If this description is correct, I don't get why the spectrum of mercury would matter at all since the electrons are not coming from the lamp.

Thanks a lot for all your help.


Julien.
I think you got some reading material conserning the basics of the apparatus and the measurement, like this:
http://hepweb.ucsd.edu/2dl/pasco/Photoelectric Effect Apparatus, AP-8209.pdf
and you also need to know what is the photoelectric effect.
The photoelectric effect is observable on solid materials, mainly metals or metal oxide semiconductors. In metals and semiconductors, there are free electrons in the material, not connected to individual atoms, but at lower potential, as if in a well, and they need some energy called work function to escape from the material. The energy comes from the photons of a light source, illuminating the material. The energy of the photon has to by higher than the work function in order to produce photo electrons. In the experiment you performed, there was a photo diode, and the photo electrons came from the cathode of that diode. Mercury and the filters produced monochromatic light with different frequency/wavelength , so you could observe the relation between the photocurrent and applied voltage at different wavelengths.
 

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