# User:Boleszek/Notebook/Physics 307l, Junior Lab, Boleszek/2008/10/20

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## A Short Introduction to Planck's Postulate

SJK 23:43, 1 November 2008 (EDT)
23:43, 1 November 2008 (EDT)
very captivating introduction! good citation below

It is arguably not an over exaggeration to say that Planck's constant is the most fundamental constant in quantum physics, because it arose from the earliest considerations of the quantization of energy. In response to the failure of the Rayleigh-Jeans formula to predict the finite energy density of a blackbody radiating at high frequencies (the ultraviolet catastrophe) physicist Max Planck decided to heuristically treat the energy as a function of frequency instead of representing it in terms of temperature. He could then treat the observed "hump" on the E density vs. fq graph as a version of the Boltzmann probability distribution. But Planck's great contribution came only when he realized that he could obtain the required cutoff (averageE → 0 as fq → ∞) if he modified the calculation leading from the probability distribution ((P(E)) to the average energy by treating the energy as if it were a discrete variable rather than a continuous one. With this insight he converted the integral of E*P(E) to a sum and found that the energy had to be directly proportional to the frequency. This proportionality factor would later bear his name with the minuscule value h=6.626068*10^-34 J-s.

Though Planck had been one of the first physicists to employ a discrete mathematics in describing radiation he was unsure whether or not energy was actually quantized or not. In a letter to R.W. Wood, Planck called his postulate "an act of desperation" (Quantum Physics, Eisberg & Resnick, pg. 21). It was not until later observations were made that the quantization of energy was thought to be a natural phenomenon and not just a clever mathematical interpretation. Among the most important of these observations was the fact that when light shined on a certain material electrons were emitted whose kinetic energies were independent of the light intensity and proportional to the light frequency. It was Albert Einstein, who was an early proponent of the reality of Planck's postulate even before Planck himself believed in it, who bravely made the step to arrive at a surprisingly simple linear formula (KE= hv + w) relating stopping potential to light frequency. The determination of the slope and y-intercept of this formula will be the aim of the following experimental procedure.

## Determination of Plank's constant

We will follow the procedure presented in Prof. Gould's manual to measure

1. the charge time of the capacitor in the h/e measuring apparatus
2. the maximum stopping potential

for various intensities and frequencies of light emitted by a mercury lamp.

## Equipment and Setup

1. Pasco OS-9282 Hg light source
2. Pasco AP-9368 h/e Apparatus
3. Tektronix TDS 1002 digital oscilloscope
4. Wavetek true RMS DMM

- We connect the output of the h/e apparatus to the DMM and Oscilloscope.

- Using wall socket (120V) as power source for lamp

- 2 9V batteries bower the h/e apparatus

Not much setup is required

## Procedure and Data Acquisition

### Experiment 1: Charge Time

• Part A.

Yellow band is centered on the opening and yellow filter is put in place. We cover this with the variable transmission filter using the 100% section.

The DVM represents the signal from the h/e apparatus. After charging the signal is constant (DC).

In order to avoid human error of alignment we take all measurements at different intensities (positions of the filter) for a single color.

SJK 23:45, 1 November 2008 (EDT)
23:45, 1 November 2008 (EDT)
I think the graph is the only place where you specify seconds as the units of time. If not for that, it would be possible to forget whether the units are seconds, milliseconds, etc. later in time

1ST TRY

Data 20%
Color         max(V)    time1  time2   time3   time4   time 5
Yellow       7.06      24.31   32.16   30.25    26.47   33.54
Orange 3
Green
Purple

Data 40%
Color         max(V)    time1  time2   time3   time4   time 5
Yellow        7.05      26.63  11.44   14.50   12.90   14.03
Orange 3
Green
Purple

Data 60%
Color         max(V)    time1  time2   time3   time4   time 5
Yellow        7.06      15.13   15.50   16.19   16.22   17.06
Orange 3
Green
Purple


We decide that the choice of voltage we had to reach (7.03V) was a bad one since it is on the far tail of the charging scale and the voltmeter tends to jump back and forth between numbers as it slowly reaches the saturated stopping voltage. We decide to use 7V as the cutoff.

2ND TRY

Redo measurements
Data 20%
Color         max(V)    time1  time2   time3   time4   time 5
Yellow       .706       33.94   27.00   31.15   28.59  31.06
Orange 3
Green
Purple
Data 40%
Color         max(V)    time1  time2   time3   time4   time 5
Yellow        .706      13.81   18.47   23.87    21.91   23.66
Orange 3
Green
Purple
Data 60%
Color         max(V)    time1  time2   time3   time4   time 5
Yellow        .706      22.34   19.81   22.03    23.72   23.03
Violet


The closeness of these values to the previous ones makes us skeptical, so we retake the 60% data

graph of charge time vs. intensity with error bars indicating the standard deviation of the mean at each point.

3RD TRY

Data 60%
Color         max(V)    time1  time2   time3   time4   time 5
Yellow       .706        6.66   5.25    6.00     5.06    5.75
Violet       1.46        7.06   7.66    7.31     6.97    7.19
Data 80%
Color         max(V)    time1  time2   time3   time4   time 5
Yellow       .706        4.37   5.19    4.13     4.25    4.03
Violet	      1.46        4.41   5.44    5.22     5.25    5.46
Data 100%
Color         max(V)    time1  time2   time3   time4   time 5
Yellow       .706       3.69    3.50    4.28    4.50     4.19
Violet       1.46       4.88    4.56    4.44    4.62     4.65
Data 40% redo
Color         max(V)    time1  time2   time3   time4   time 5
Yellow       .706       8.00   8.42     6.72    8.75     7.59
Violet       1.46       8.72   9.68     9.68    9.75     8.84
Data 20% redo
Color         max(V)    time1  time2   time3   time4   time 5
Yellow      .706        8.25    8.75    9.20    9.06     9.28
Violet       1.46      16.50   19.06   19.30   17.12    16.80


Darrel has noticed that the time measurements tend to be longer when his hand is left touching the h/e apparatus after pressing the discharge button. We decide to redo the measurements of 40% and 20% for yellow and are finally satisfied with the apparent linearity of our measurements. We proceed to make measurements using a violet strip. We choose 1.42V as our voltage to reach for each time delay measurement. We have noticed that, according to the procedure, what we see as violet the procedure terms “blue”. These measurements are shown in the table titled 3RD RUN just above. The preceding two data tables will not be used for analysis because they exhibit a noticeable systematic error consisting mostly of 1) an extra capacitance introduced by hand contact with the grounded h/e apparatus casing and 2) a failure to realign the light rays with the slit when the apparatus has moved.

• Part B

graph of stopping voltage vs. frequency. Note the wonderfully linear behavior. (Steve Koch: I'm assuming the lines are just connecting the points. Much better would be a line showing the linear fit -- this would highlight the wonderfully linear behavior too!)
         Violet    Blue1    Blue2    Yellow                      Green
Max(V)   1.49       1.71    2.044  .716 w/filter, .993/without .847 with, .972 without


I have noticed later on that these colors were labeled wrong. The first three should read Blue, Violet, UV. Comparison with measurements of stopping potential later on should confirm this correction. This behavior indicates to us that the frequency dependence of the stopping voltage is linear. Thus our experiment supports a quantum theory of light because a wave's energy is proportional to its intensity (and I recall that it is actually Intensity^2) but our measured voltages are not. If we assume that the quantum theory is correct then electrons are "knocked out" of atoms in the material because individual particle-like photons collide with them with an energy larger than their bonds. Then we would expect that even a single photon of sufficient frequency will impart enough energy on a bound electron to emit it with the same energy as a million photons would a million electrons. This expectation is confirmed by the observation that a lower intensity of light results in a lower emission current, which simply corresponds to lower number of emitted electrons, because the lower intensity light contains fewer particle-like photons.

### Experiment 2: Determination of h

We make measurements of stopping potential of the first and second order light bands according to the diagram on pg.38 of the procedure.

•	First Order 1st run
Color            max(V)
Ultra-Violet    2.030             As we turn the apparatus from UV to Violet we notice that V drops from
Violet          1.515             2 to about 1.7 in between and then down to 1.5 on Violet. Past violet V
Blue            1.707             rises again to 1.7. This is contrary to our expectations.
Yellow          0.720 with filter, 1.335 without
Green           1.234 with filter


We realize by closer inspection of the diagram in the procedure that we in fact were not measuring the first order light bands, so we redo the measurements at what we now believe to be the first order set of bands. This explains why the yellow line came before green and why “blue” had higher stopping voltage than “violet”.

4 graphs of stopping voltage vs. frequency with least squares fits which are used to determine h and ω by analytic and interpolation methods.
•	First Order  (redo) 1st run
Color           max(V)
Ultra-Violet    2.064
Violet          1.718
Blue            1.493
Green           .849 with filter, 1.007 without
Yellow          .717 with filter, 1.053 without
Now we move to second order bands.
•	Second  Order 1st run
Color           max(V)
Ultra-Violet    2.062
Violet          1.731
Blue            1.520
Green           1.242 with filter, 1.335 without
Yellow          0.735 with filter
And the process is repeated.
•	First Order 2nd run
Color           max(V)
Ultra-Violet    2.069
Violet          1.725
Blue            1.495
Green           0.850 with filter
Yellow          0.714 with filter
•	Second Order 2nd run
Color           max(V)
Ultra-Violet    2.061
Violet          1.725
Blue            1.515
Green           1.242 with filter
Yellow          0.735 with filter

SJK 23:52, 1 November 2008 (EDT)
23:52, 1 November 2008 (EDT)
It was really great how you guys figured out the source of the problem with the green 2nd order! You showed very good experimentalist's talents

For some reason as of yet unknown to us the green band results in a much higher stopping potential in the 2nd order than in the 1st. We speak to Prof. Koch about this anomalous behavior and come to the conclusion that there must be an unseen factor at play. When we scan the h/e apparatus across the green band we notice that the voltage remains high even when we are in an apparent “dark” region on either side of green. When we cover the slit with both green and yellow filters we find that the stopping voltage of the green line is 0.842V! This result is very close to the measurement of the first order green line so we choose to use it instead of the out of place 1.242V. A second measurement of the second order green stopping voltage yields 0.849V. We have reason to believe that whatever unseen light made the voltage rise to 1.79V has been filtered out by this combination of filters (we also notice that the yellow filter does most of the filtering). After rereading the procedure it would be appropriate to add that the overlap of higher order UV spectra with lower order green is alluded to in section 5.3.1 of Prof. Gould's manual

Darrell has the idea that his reading glasses might work as a good UV filter rather than the yellow filter (since his glasses have clear lens UV filters). When he placed a lens over the slit we did in fact witness a stopping voltage of 0.847V for the second order green band.

• DIFFERENCES IN MAXIMUM VOLTAGE AS INTENSITY OF LIGHT DROPS

With light shining upon the slit we measure the maximum voltage reached at consecutively higher intensities.

	Blue				        Green
20% - 1.487V    				0.846V
40% - 1.485V    				0.850V
60% - 1.489V    				0.852V
80% - 1.486V     				0.852V
100% - 1.505V a noticeable difference           0.856V


We find a trend towards increasing voltage as intensity increases, but the quantum theory of light predicts that stopping voltage should only increase as a function of increasing frequency, not intensity. Rather than assuming that we have found evidence against the photoelectric effect, we believe the rise in voltage with intensity occurs because the photodiode exhibits an imperfection consisting of a current leakage. At lower intensities the total current in the apparatus is lower than at higher intensities so the leakage constitutes a larger fraction of the lower intensity current and is therefore more noticeable to the DVM. Thus the voltage reading from the unfiltered light most accurately reflects the actual kinetic energy of the emitted electrons because the current leak is small compared to the emission current.

SJK 16:55, 2 November 2008 (EST)
16:55, 2 November 2008 (EST)
Good explanation.

## Data and Error Analysis

For convenience of viewing I have presented graphs next to their respective data sets above.

• CHARGE TIME

• Part A.

We collected five data points at each intensity for both yellow and violet ("blue" in the manual, but it sure looks violet). I choose to represent the data by averaging the 5 times for each intensity and plotting both colors on the same axes with error bars. The MATLAB calling sequence for error bar plots is errorbar(X,Y,E) where E is a predetermined error. I choose to determine E by calculating the standard deviation of the mean for 'each' intensity so I will have to plot each point individually such that each of the 5 intensities will have its own error present (I could make just one plot with E set to the average std of the mean so that the error bars are all the same length, but what's 20 more minutes for extra accuracy?SJK 23:56, 1 November 2008 (EDT)
23:56, 1 November 2008 (EDT)
I say it's 20 minutes well spent!
). After all this I calculate the mean of the standard deviations of the mean to provide an overall measure of error in the charge-time measurement. These procedures are undergone in the MATLAB code section entitled "Charge Time". The standard deviation of the mean is calculated according to the formulaSJK 00:00, 2 November 2008 (EDT)
00:00, 2 November 2008 (EDT)
You are missing another factor of 1/sqrt(N). As written, this is just the standard deviation, not SEM.
$\sigma_m= \sqrt{\frac{1}{N*(N-1)} \sum_{i=1}^N (x_i - \overline{x})^2}\,$

After calculations and plots are completed I notice that the intensity vs. charge time plot for violet exhibits an unexpectedly high charge time at 20% intensity along with the largest error bar, so it is reasonable to ignore that point. The overall standard deviation of the mean including this stray point is

0.5418s

Without it the standard deviation of the mean reduces to

0.4552s

This is the value that I choose to present in the summary.

Our results show that as the intensity of light increases the time required to charge the capacitor in the h/e apparatus decreases, though this relationship does not appear to be linear as we previously thought it would. This actually makes sense because a capacitor does not charge linearly over time, but quickly at first and slowly at the end, following a negative exponential determined by the physical properties of the material and the magnitude of current coming its way. Our graph actually appears to be approximating a negative exponential of some kind. Though I cannot be sure at the moment what is the exact formula for this graph I can comment that it follows very nicely from the quantum theory of light. Were we to imagine millions of photons colliding with correspondingly millions of electrons in a material we would expect more electrons to be emitted when more photons are incident upon them (as long as the photon energy is large enough). More electrons correspond to higher current, which in turn correspond to lower charge time because the capacitor becomes saturated more quickly. A skeptic might ask why it is that the charge times for each intensity are shorter for higher energy (yellow) light than lower energy (violet) light when we would expect that emission current should only be proportional to intensity, not energy. I would respond that a higher energy light causes electrons to be emitted at higher velocities, so even though a cross-section of the yellow and violet light induced electron beams would have the same density of electrons at the same intensity, the electrons from the yellow light travel from the photoelectric material to the capacitor in a shorter time than do those emitted by the violet light. I could also comment on the skeptic's incredulity in expecting an undergraduate student to rigorously defend quantum physics.SJK 00:03, 2 November 2008 (EDT)
00:03, 2 November 2008 (EDT)
Hey, the skeptic has his rights...I'd say the easiest explanation is that nobody said that the violet intensity is anywhere close to the yellow intensity. You don't actually know the #photons/second ... it's all just relative to "100%".

• Part B.

In the procedure we are asked to record the maximum stopping voltage reached during the charges and then comment on whether our findings support the wave or particle theories of light. Although we did follow this request it did not make much sense to us because we were only measuring the charge time for two frequencies. Instead of using limited data on which to base our conclusion we decided to make a separate set of measurements including each color at constant intensity. The results are plotted next to the corresponding data table above and the reasoning as to why they are in favor of the quantum theory of light is provided just below the table. Since these measurements were made for qualitative purposes only no error analysis is necessary here.

• DETERMINATION OF h

We have four sets of data to analyze in this section: Two sets for the first order spectrum and two sets for the second. It should be noted that after considering the possible reasons for the existence of an unexpectedly high stopping voltage for the green band of the second order spectrum (discussed under the corresponding table above) we redid those measurements and obtained values of .842V for the first run and .849V for the second run. These values will be used in the ensuing analysis instead of those in the table. Once again, the graphs of the data are presented next to their corresponding table.

In the procedure we are asked to make plots, perform linear least square fits, and determine h and ω_o for each of the four data sets. I comply with these requests and use the frequency values provided in p.38 of Prof. Gould's manual to perform the necessary calculations which are present in the MATLAB code section entitled "Determination of h". Since our plots represent e*Voltage vs. frequency we can deduce from the equation

eV = hν − ω0

that the slope is h and the y-intercept is ω_o.

Procedure for Error Propagation

It is only recently that I received a copy of Taylor's informative text on error analysis. I have chosen to apply the methods of Chapter 8 (Least Squares Fitting) to my data and so I have redone this section of the error analysis completely. It will help me as well as the reader to explain the steps that will be taken.
1.) I begin by assuming a normal distribution to the stopping potential data following the relation y = A + Bx, which is going to be very narrow distribution judging by the low deviation of the voltage data for each frequency. If we assume no substantial errors in the x variable the probability of obtaining the observed values of eV (voltage measurements scaled by e) will be proportional to the Gaussian:

$Prob_{A,B}(y_1,...,y_N) \propto \frac{1}{\sigma_y ^N} e^{-\chi^{2}/2}$

where $\chi^2 = \frac{\left(\sum_{i=1}^N (y_i - A - Bx_i)^2\right)}{\sigma_y ^2}$
and y = eV, A = − ω0, B = h for this experiment.

2.) Since the best estimates for A and B are those values at which ProbAB(y1,...,yN) is maximum, or identically for which Χ^2 is a minimum, the least square fits for A and B (-ω_0 and h) are found by taking the derivative of X^2 with respect to A and B separately and then solving the resulting two equations for A and B. In the end

$A = \frac{\sum_{i=1}^N x_i^2 \sum_{i=1}^N y_i - \sum_{i=1}^N x_i \sum_{i=1}^N x_iy_i}{\Delta}$
$B = \frac{N\sum_{i=1}^N x_iy_i - \sum_{i=1}^N x_i \sum_{i=1}^N y_i}{\Delta}$

where $\Delta = N\sum_{i=1}^N x_i^2 - \left(\sum_{i=1}^N x_i\right)^2$
I choose to perform this step using the MATLAB functions polyfit to find the coefficients (A & B) and polyval to evaluate them at at the data points for plotting.

3.) With the least squares coefficients determined I can proceed to calculating the errors of y (eV), A (-ω_0), and B (h). Taylor shows that the uncertainty for y (with x assumed to be without error) is

$\sigma_y= \sqrt{\frac{1}{N-2} \sum_{i=1}^N (y_i - A-Bx_i)^2}\,$

I use this definition but divide by an extra N to make it the SDOM. This formula is generally similar to the familiar SDOM formula quoted above, except that it has a factor of N-2 in the denominator. This difference results from the fact that, in calculating uncertainty, we divide by the number of independent measured values. When we have to calculate the average of one quantity before the uncertainty we are left with N-1 degrees of freedom, but when we have to calculate the average of two quantities (A & B) then we are left with N-2 degrees of freedom. Taylor says that there are good statistical reasons to divide by the number of degrees of freedom instead of the number of measurements, but does not justify it mathematically. Qualitatively, though, it makes sense, for were we to measure just two pairs of data (x1,y1) and (x2,y2) with N=2 we could always get a straight line through two points so the idea of a linear fit for two points should be undefined, in this case by a division by 0.

4.) With the uncertainty in y determined I finally calculate the uncertainties in the quantities that I am really interested in, A and B, or rather ω_0 and h. Taylor explains that from the two expressions for A and B above we can apply the general error propagation formula (summing partial derivatives times their respective uncertainties in quadrature) to arrive at the uncertainties for A and B (I repeat that I use the extra 1/N factor in my σ_y):

$\sigma_A = \sigma_y\sqrt{\frac{\sum_{i=1}^N x_i^2}{\Delta}}$
$\sigma_B = \sigma_y\sqrt{\frac{N}{\Delta}}$

where once more $\Delta = N\sum_{i=1}^N x_i^2 - \left(\sum_{i=1}^N x_i\right)^2$

5.) In obtaining my final results I calculate the standard error of the mean (stdm) for each set of data, then average these errors for the total stdm. In order to present my final result of Planck's constant with the correct value of error I must calculate the variance of a function. Luckily my function for h is simply h=slope where the slope is given by eV/fq (eV because I had to convert voltage into kinetic energy, of course). Since my data for V has a certain standard deviation, the variance (square of std) of h will be scaled by e/fq. I base these conclusions on the formula given in the Wikipedia page on error propagation: $\sigma_f^2 = a^2 \sigma_A^2\,$

SJK 01:46, 2 November 2008 (EST)
01:46, 2 November 2008 (EST)
I am having trouble following your matlab code (because I don't use matlab)...nevertheless I am fairly certain that your method for calculating the uncertainty is not correct. You are misunderstanding the error propagation method...there are well-defined ways to get the uncertainty of the slope from the linear fit.

No extra analysis is needed for ω_0 because it has the same error as the y-axis.

Here are the results for the four data sets:

h1 = 7.1921e-034, stdm = 3.5177e-034     ||     ω1 = 1.6097eV, stdm = 0.6622eV

h2 = 7.1812e-034, stdm = 3.5146e-034     ||     ω2 = 1.5954eV, stdm = 0.6617eV

h3 = 7.2338e-034, stdm = 3.5381e-034     ||     ω3 = 1.6246eV, stdm = 0.6661eV

h4 = 7.1433e-034, stdm = 3.4953e-034     ||     ω4 = 1.5807eV, srdm = 0.6580eV


The average h and omega are:

### ω_o = 1.6026eV, stdm = 0.6620eV

The value for ω_o certainly indicates a material with a low work function (perhaps even too low when compared with values of standard materials). There are a number of reasons why my value is so far from the accepted value h=6.626068*10^-34. These include accuracy with which we align the slit with the light rays, current leakage in the h/e apparatus, extra resistances influencing the measurements made by the DVM, and the possibility that the values of frequency given to us by the procedure which we used to make calculations were not the exact frequencies of the actual spectra shining upon our apparatus. Since the overshoot is consistently reported in all four trials for h I must conclude that there was some factor that has influenced our measurement such that the voltage readings vs frequency appeared to increase at a higher rate than they should (since the slope is directly related to h). My guess is that the current leakage in the h/e apparatus actually resulted in the DVM getting a noticeably lower reading than it should for the lower frequency measurements because, as was stated before, the leakage constituted a larger percentage of the total current at these lower frequencies. This resulted in the left end of the V vs. fq to slump down, thus increasing the slope. I have not verified this in anyway, so it is only a guess SJK 16:39, 2 November 2008 (EST)
16:39, 2 November 2008 (EST)
what would be a good way to test this hypothesis?
.

## References

Quantum Physics, Eisberg & Resnick, pg. 21 An Introduction to Error Analysis, John R. Taylor, Chapter 8