Physics307L:People/Dougherty/Notebook/070924

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Cary M. Dougherty 16:05, 24 September 2007 (EDT)

Linh and i are going to attempt the Planck's constant lab

Goal

starting the first experiment, we want to determine in summary if it is more likely that light is in form of a wave or in photon form. the photon theory of light says that the energy of light is only determined by frequency and is independent of intensity. however, the wave thoery is only dependent on intensity. the first experiment is investigating this question.

the second experiment wants us to directly measure the constant h and the [math]\displaystyle{ W_0 }[/math], work function, along with the total kinetic energy of the photon. with the total kinetic energy and work function of the photon we can find Planck's constant times the frequency. [math]\displaystyle{ hv=KE_m+W_0 }[/math]

Setup

setting up the apparatus we make sure there are good batteries in the h/e device and then plug in the ight source and turn it on. we let it warm up for 2 min and then move the diffraction grating so it is the most focused light. we can the see each color in the spectrum and focus it into the h/e device. the entire apperatus is intact for the lab. the only thing we needed was to make sure the batteries were good in the h/e device and we also needed a volt meter to measure the stopping voltage or the electrons.

How it Works

the photons emitted by the mercury light source enter the h/e apperatus and hit a cathode tube of electrons. the electrons escape the tube with a minimum energy or the work force [math]\displaystyle{ W_0 }[/math]. leaving the tube with a max kinetic energy. then by applying a reverse voltage on the energy, we can reverse the energy back to zero recording its stopping potential. this stopping potential is then directly related to the kinetic energy of the electron. plotting voltage vs. frequency we can get h which is planck's constant.

Experiment 1 Procedure

the first experiment, we used the filters as told for each color and also the 100%-20% filter for each color. doing this we could measure how long it took for the stopping potentialto reach a percentage of maximum. doing this for each color we can plot it and see that indeed light acts as a photon and not a wave which depends on frequency and not intesity.

Experiment 1 Data

voltage on the voltmeter started at .04 V which will be added into our calculations


UV light

100% filter Stopping Potential: 1.86V measure at 90% about 1.70V

80% filter SP : 1.84V 90% of that 1.65V

60% filter SP : 1.81V 90% of that 1.63V

40% filter SP: 1.78V 90% of that: 1.60 V

20% filter SP:1.74 V 90% of that:1.57 V

Note: the neutral filter filters out some UV light

Intensity 100% 80% 60% 40% 20%
Time(s) 5.76 5.44 6.39 6.07 11.43
5.67 5.62 6.3 7.42 14.04
6.97 5.8 6.66 7.51 14.62
6.39 5.67 7.2 6.97 16.42
Avg Time 6.19 5.63 6.63 7.01 13.62
Blue -Violet

Stopping Potential: 1.75V Measure to 1.70V

Intensity 100% 80% 60% 40% 20%
Times(s) .9 1.26 1.08 1.44 2.88
.94 1.03 1.08 1.53 3.06
1.12 1.12 1.03 1.62 3.01
1.17 1.03 1.17 1.66 2.88
Avg Time 1.03 1.11 1.09 1.56 2.95
Green

Stopping Potential = .89V Measure at : .87V

Intensity 100% 80% 60% 40% 20%
Times(s) 1.21 1.44 1.66 3.28 5.76
1.35 1.25 2.25 2.74 4.90
1.66 1.44 1.93 3.10 4.81
1.26 1.44 2.38 3.24 4.60
Avg Time 1.37 1.39 2.055 3.09 5.01
Yellow

Stopping Potential : .73V Measure at: .70V

Intensity 100% 80% 60% 40% 20%
Times(s) 1.93 2.47 1.77 2.47 3.51
1.84 1.98 1.26 2.07 4.09
1.75 1.89 1.26 2.47 3.96
1.98 3.01 1.44 2.16 4.99
Avg Time 1.88 2.33 1.43 2.32 4.13

the yellow light seemed off because it went way up in 60% wen it shouldn't have. so we took the data again but kept our old data.

Stopping V =.75V Measure to:.74V

Intensity 100% 80% 60% 40% 20%
Time (s) 1.03 1.03 1.03 1.26 1.80
.9 1.17 1.08 1.26 2.38
.9 1.03 .99 1.21 1.66
1.3 .9 .94 1.39 1.66
.85 1.03 1.03 1.47 1.80
Avg Time (s) .996 1.032 1.014 1.318 1.86

Experiment 2 Procedure

the second experiment we only waited a few min for the stopping potential to reach a maximum (twice for each color), and then recorded that voltage. and by plotting the frequency which we got directly out of the manuel and the KE which is found by multiplying the stopping voltage we recorded by the charge of an electron. graphing the KE and the frequency we use a least square fit and we can discover the slope (h) and also the y-intercept ([math]\displaystyle{ W_0 }[/math]). this way discovering planck's constant and also the work function.

Experiment 2 Data

[math]\displaystyle{ KEmax = qVstopping }[/math]

Measuring Stopping Potentials

1st order

Yellow

.75V

.75V

at a frequency of 5.18672*10^14 Hz

at wavelength of 578 nm


Green

.87V

.87V

at a frequency of 5.48996*10^14 Hz

at a wavelength of 546.074 nm


Blue-Violet

1.70V

1.69V

at a frequency of 7.14358*10^14 Hz

at a wavelength of 420.246 nm


UV

(since we cant see UV, found the spot with highest Stopping Potential)

2.01V 2.03V

at a frequency of 8.20264*10^14 Hz

at a wavelength of 365.483 nm


2nd order

yellow

0.75 V

0.75 V


green

0.86 V

0.86 V

blue-violet

1.76 V

1.77 V

uv

2.07 V

2.06 V

1st order

a)

yellow

[math]\displaystyle{ KEmax }[/math] = 1.2*10^-19 J

green

[math]\displaystyle{ KEmax }[/math] = 1.392*10^-19 J

blue-violet

[math]\displaystyle{ KEmax }[/math] = 2.72*10^-19 J

uv

[math]\displaystyle{ KEmax }[/math] = 3.216*10^-19 J

b)

yellow

[math]\displaystyle{ KEmax }[/math] = 1.2*10^-19 J

green

[math]\displaystyle{ KEmax }[/math] = 1.392*10^-19 J

blue-violet

[math]\displaystyle{ KEmax }[/math] = 2.704*10^19 J

uv

[math]\displaystyle{ KEmax }[/math] = 3.248*10^-19 J

2nd order

a)

yellow

[math]\displaystyle{ KEmax }[/math] = 1.2*10^-19 J

green

[math]\displaystyle{ KEmax }[/math] = 1.376*10^-19 J

blue-violet

[math]\displaystyle{ KEmax }[/math] = 2.816*10^-19 J

uv

[math]\displaystyle{ KEmax }[/math] = 3.312*10^-19 J

b)

yellow

[math]\displaystyle{ KEmax }[/math] = 1.2*10^-19 J

green

[math]\displaystyle{ KEmax }[/math] = 1.376*10^-19 J

blue-violet

[math]\displaystyle{ KEmax }[/math] = 2.832*10^-19 J

uv

[math]\displaystyle{ KEmax }[/math] = 3.296*10^-19 J