Physics307L F07:People/Smith/Notebook/3
Contents
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)} h/e setup (refer to last year's lab manual for more information.)
 h/e apparatus (mfd. Pasco Scientific AP9638)
 Grating/lens assembly: 600 lines/mm, blazed for 500nm. 100mm focal length lens. (mfd. Pasco Scientific AP9369)
 Mercury vapor light, 115 volts (mfd. Pasco Scientific OS9286)
 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, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle KE_{max}} , 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)}Refer to lab manual
Measurements
^{SJK 02:58, 11 October 2007 (CDT)}
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.
 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle E = \nu h = KE_{max} + W_0 \,\!}
, where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle E\,\!}
is the energy of the photon, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \nu\,\!}
is the frequency of the incident photon, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle KE_{max}\,\!}
is the maximum kinetic energy of the ejected electron, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle W_0\,\!}
is the work function.
 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle KE_{max} = \frac{1}{2} m v^2} , where m is the (rest) mass of the electron and v is its velocity
 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \Rightarrow E = \frac{1}{2} mv^2 + W_0}
 The most energetic electrons will leave the surface of the cathode with energy Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle E = h \nu  W_0\,\!}
 In order for electrons to reach the anode plate, they must have energy Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle E = e\cdot V_{stopping}\,\!} upon leaving the cathode, where e is the electron charge and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle V_{stopping}} is the stopping potential.
 Thus, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle e \cdot V_{stopping} = KE_{max}  W_0 = h \nu  W_0}
 Or, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle V_{stopping} = \frac{1}{e} (h \nu  W_0)}
This is a linear relationship, and plotting V_{stopping} versus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \nu\,\!} and measuring the slope of the linear regression will yield Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{h}{e}} . Since we know what the charge of the electron is, we have just found Planck's constant.
By the same token, finding the "Yintercept" of the graph's linear regression will reveal the value of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{W_0}{e}} , and multiplying this by the elementary charge will give us the work function.
Conclusions
^{SJK 03:07, 11 October 2007 (CDT)} 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle 7.14817 \cdot 10^{34} \frac{kg\cdot m^2}{s}}
 This value is 7.88% different than the accepted value of Planck's constant, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle 6.626 \cdot 10^{34} \frac{kg\cdot m^2}{s}}
 The average measured work function is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle 2.51297 \cdot 10^{19} J} . 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)}
 Our Best Guess: 7.14817E34 +2.03352E36/4.06703E36
 (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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle J\cdot s} (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!
Links to other entries
Wednesday Labs  



