Physics307L F07:People/Smith/Notebook/3
Experiment 3: Planck's Constant
Lab Partner: Kyle Martin
Purpose
To experimentally determine the ratio of Planck's constant over the charge of the electron, and thus identify Planck's constant.
Equipment
SJK 02:54, 11 October 2007 (CDT)
This is great that you put in the model numbers!
- h/e setup (refer to last year's lab manual for more information.)
- h/e apparatus (mfd. Pasco Scientific AP-9638)
- Grating/lens assembly: 600 lines/mm, blazed for 500nm. 100mm focal length lens. (mfd. Pasco Scientific AP-9369)
- Mercury vapor light, 115 volts (mfd. Pasco Scientific OS-9286)
- Filters
- Digital voltmeter (mfd. Keithly model 131)
Setup
Refer to lab manual.
Experiment 1: The Photon Theory of Light
From the lab manual: Photon theory of light says maximum kinetic energy, [math]\displaystyle{ KE_{max} }[/math], of photoelectrons depends only on the frequency of the incident light, not the intensity. Higher frequency implies greater energy. This experiment will investigate this claim.
Part A
Procedure
SJK 02:56, 11 October 2007 (CDT)
It's sensible to refer to a standard protocol like you do here. Just make sure that if you do something differently, you record it.
Refer to lab manual
Measurements
SJK 02:58, 11 October 2007 (CDT)
I can't remember if we talked about this or not...it looks like you are measuring the time to achieve the "full" stopping potential. You would get much better data if you measured the time to get to, say 50% of the stopping voltage...much like we measured the fall time as the time to get to 90% of the peak value during the oscilloscope lab.
| Yellow Line, Filtered | ||||||
|---|---|---|---|---|---|---|
| Time Required to return to recorded voltage (seconds) | ||||||
| Transmission % | Stopping Voltage | Trial 1 | Trial 2 | Trial 3 | Trial 4 | Trial 5 |
| 100 | 0.74V | 3.08 | 3.63 | 3.29 | 2.19 | 3.28 |
| 80 | 0.74V | 2.42 | 3.22 | 3.24 | 2.75 | 2.89 |
| 60 | 0.74V | 2.87 | 5.00 | 4.43 | 3.74 | 3.90 |
| 40 | 0.74V | 6.66 | 6.30 | 8.43 | 6.58 | 6.96 |
| 20 | 0.74V | 10.69 | 11.69 | 11.29 | 10.39 | 12.77 |
| Green line, Filtered | ||||||
|---|---|---|---|---|---|---|
| Time Required to return to recorded voltage (seconds) | ||||||
| Transmission % | Stopping Voltage | Trial 1 | Trial 2 | Trial 3 | Trial 4 | Trial 1 |
| 100 | 0.87V | 8.17 | 8.25 | 8.76 | 9.09 | 7.33 |
| 80 | 0.87V | 8.03 | 8.19 | 7.43 | 6.97 | 8.17 |
| 60 | 0.87V | 15.22 | 11.53 | 12.05 | 12.03 | 12.37 |
| 40 | 0.87V | 20.32 | 18.93 | 17.72 | 15.66 | 17.13 |
| 20 | 0.87V | 23.29 | 30.30 | 31.17 | 31.05 | 31.79 |
Part B
Examining different colors, no variable transmission filter.
Procedure
Remove transmission filters. Record stopping voltages for the different atomic emission spectral lines.
Measurements
| Color | Potential (V) |
|---|---|
| Yellow | 0.75 |
| Green | 0.87 |
| Violet | 1.50 |
| Violet 2 | 1.70 |
| UV 2 | 2.05 |
Experiment 2
Determination of h. Energy of light is proportional to the frequency. The coefficient of proportionality, h turns out to be one of the most important physical constants.
Procedure
Refer to the lab manual. Basically, measure the stopping potential of each color's first order maximum and compare it to the second order maximum. Take two measurements for each order.
Measurements
| First Order | Second Order | |||
|---|---|---|---|---|
| Color | Stopping Potential (V) #1 | Stopping Potential (V) #2 | Stopping Potential (V) #1 | Stopping Potential (V) #2 |
| UV | 2.07 | 2.05 | 2.06 | 2.1 |
| Violet 2 | 1.74 | 1.70 | 1.74 | 1.75 |
| Violet 1 | 1.52 | 1.50 | 1.58 | 1.58 |
| Green | 0.88 | 0.87 | 0.88 | 0.87 |
| Yellow | 0.76 | 0.75 | 0.71 | 0.71 |
Analysis
Theory
Noting, from the lab manual, that the frequencies of the Mercury atomic emission spectra are as follows:
| Color | Frequency (Hz) | Wavelength (nm) |
|---|---|---|
| Yellow | 5.18672E+14 | 578 |
| Green | 5.48996E+14 | 546.074 |
| Blue | 6.87858E+14 | 435.835 |
| UV 1 | 7.40858E+14 | 404.656 |
| UV 2 | 8.20264E+14 | 365.483 |
Also remember (or look up) the relationship between energy and frequency, and between energy and the photoelectric effect. I looked up the photoelectric effect on Wikipedia to refresh my memory.
- [math]\displaystyle{ E = \nu h = KE_{max} + W_0 \,\! }[/math], where [math]\displaystyle{ E\,\! }[/math] is the energy of the photon, [math]\displaystyle{ \nu\,\! }[/math] is the frequency of the incident photon, [math]\displaystyle{ KE_{max}\,\! }[/math] is the maximum kinetic energy of the ejected electron, [math]\displaystyle{ W_0\,\! }[/math] is the work function.
- [math]\displaystyle{ KE_{max} = \frac{1}{2} m v^2 }[/math], where m is the (rest) mass of the electron and v is its velocity
- [math]\displaystyle{ \Rightarrow E = \frac{1}{2} mv^2 + W_0 }[/math]
- The most energetic electrons will leave the surface of the cathode with energy [math]\displaystyle{ E = h \nu - W_0\,\! }[/math]
- In order for electrons to reach the anode plate, they must have energy [math]\displaystyle{ E = e\cdot V_{stopping}\,\! }[/math] upon leaving the cathode, where e is the electron charge and [math]\displaystyle{ V_{stopping} }[/math] is the stopping potential.
- Thus, [math]\displaystyle{ e \cdot V_{stopping} = KE_{max} - W_0 = h \nu - W_0 }[/math]
- Or, [math]\displaystyle{ V_{stopping} = \frac{1}{e} (h \nu - W_0) }[/math]
This is a linear relationship, and plotting Vstopping versus [math]\displaystyle{ \nu\,\! }[/math] and measuring the slope of the linear regression will yield [math]\displaystyle{ \frac{h}{e} }[/math]. Since we know what the charge of the electron is, we have just found Planck's constant.
By the same token, finding the "Y-intercept" of the graph's linear regression will reveal the value of [math]\displaystyle{ \frac{W_0}{e} }[/math], and multiplying this by the elementary charge will give us the work function.
Conclusions
SJK 03:07, 11 October 2007 (CDT)
I think the relationship should be linear, and I think if you think about why the rise time changes with intensity, you would agree eventually. See my comment above for possible problems with your methods...
- The relationship between intensity and rise time (from experiment 1A) appears to be a 2nd order polynomial relationship. I have graphed this in my Excel worksheet, as Chart 1 and Chart 2. The equations for the polynomial regressions are displayed on the graph. I don't know of what importance these equations may be, though. Is it even reasonable for this relationship to be polynomial? Or is an exponential relationship more plausible?
- My calculated value of Planck's constant, from our measured data, is [math]\displaystyle{ 7.14817 \cdot 10^{-34} \frac{kg\cdot m^2}{s} }[/math]
- This value is 7.88% different than the accepted value of Planck's constant, [math]\displaystyle{ 6.626 \cdot 10^{-34} \frac{kg\cdot m^2}{s} }[/math]
- The average measured work function is [math]\displaystyle{ 2.51297 \cdot 10^{-19} J }[/math]. This is on the order of an electron volt, but it is perhaps 1.5 times as large. Is this a reasonable value of the work function?SJK 03:09, 11 October 2007 (CDT)
03:09, 11 October 2007 (CDT)
Can you find out what the material is that is losing electrons, and then look up the work function? It's sort of interesting that you don't know the "real" value in this case...how confident are you in your number, not having the "real" value to guide you?
- Our Best Guess: 7.14817E-34 +2.03352E-36/-4.06703E-36
- (To get this estimate, I took the average of the 1st and 2nd order maxima stopping potentials. The error estimate was found by doing a linear regression on the largest and smallest values of the atomic emission spectral lines' stopping potentials. I don't know whether that was a good way to do it, and it looks a little weird since my error estimate is not symmetric; the deviation to the lower end of the estimate is larger than that to the high end - in fact, it seems to be exactly twice as large! This is rather odd.)
Koch comments
First, your answer would be much easier to read with only the significant digits. Something like:
(7.14 +0.02, -0.04) * 10^-34 [math]\displaystyle{ J\cdot s }[/math] (don't forget the units either!)
Does that make sense?
Second: No, there is no problem at all with having asymmetric error bars, although if you are assuming a gaussian random error, of course they have to be symmetric. In your case, I can see what you are trying to do...however, since it's not standard, when you report the result, you should immediately prior or after the value describe what the error range represents and how you calculated it (which is what you did).
Overall, excellent work! Good analysis with Excel, great charts. My main criticism is that you were one step shy of fully understanding the experiment. This would require a step back from the data analysis and asking yourself, "what am I measuring again, and how / why?" This is reflected in a lack of discussion of where your large systematic error may come from, as well as in the rise time data, which probably are not too correct. But otherwise, great job! You spent a lot of time taking good data and doing a great analysis!
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