The cloud chamber project was done with my partner, Yanting Li. Information regarding the project is linked below.
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To download the powerpoint presentation of the project, click the link below.
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Physics 4C csoemardy
Tuesday, June 11, 2013
Tuesday, May 28, 2013
Experiment 14: Determining Planck's Constant
Purpose
The colors of LED used in this experiment: red, yellow, green and blue.
Calculations from the measured data
Calculated planck's constant from the graph
The purpose of this experiment is to calculate the value of Planck's constant using the spectrum of different colored LED.
Experiment
The circuit shown above is built using 360ohms resistor and LED |
Experiment Setup |
The colors of LED used in this experiment: red, yellow, green and blue.
Spectra of red LED |
Spectra of yellow LED |
Spectra of green LED |
Spectra of blue LED |
Spectra of white LED |
The spectrum of white LED consists of rainbow color, however, each color is clearly defined as if there's a separation gap between each color.
Data and Analysis
Data from measurement:
Calculations from the measured data
Calculated Planck's constant:
LED Color
|
Wavelength(nm)
|
Voltage(V)
|
Calculated h (Js)
|
Error(%)
|
Red
|
607
|
1.890
|
6.12E-34
|
7.74
|
Green
|
514
|
2.579
|
7.07E-34
|
6.58
|
Yellow
|
570
|
1.920
|
5.75E-34
|
13.24
|
Blue
|
456
|
2.654
|
6.46E-34
|
2.63
|
Average
|
6.35E-34
|
4.26
|
From the data, we can see that the LED with longer wavelength has lower measured voltage.
The sky seem blue because blue light has higher frequency, making it easier to scatter. The sunset looks red because the red light has lower frequency, thus it concentrates in the direction of the sun.
Calculated planck's constant from the graph
The equation of the graph: y = 7.46E-7x+ 1.98E-7
Slope = 7.46E-7x = hc/e
Slope = 7.46E-7x = hc/e
h = 3.98E-34 J*s
% Error = 40.0%
% Error = 40.0%
Conclusion
The measured Planck's constant using graphical method has much higher percent error compared to when each point is calculated individually. This shows that in order to make the graphical method to be more accurate, we need more data points.
Experiment 13: Color and Spectra
The purpose of this experiment was to analyze the spectrum of white light and hydrogen gas.
Experiment
Part 1: White Light Spectra
Experiment Setup |
As shown in the picture below, the spectrum of white light looks like a rainbow with the color purple closest to the source of light.
By measuring the length of the range of color, we can calculate the range of wavelength of visible light.
According to calculation, the smallest wavelength a human eye can see is 392 +/- 11nm, while the largest is 744 +/- 34nm.
When compared to the theoretical value, we can formulate a calibration function:
λ_t = 1.07 λ_m - 16.7
Part 2: Hydrogen Spectra
The white light was then replaced by hydrogen gas.
From the spectrum of hydrogen gas shown above, we can see a bright red, teal, and purple stripes. The yellow smear seen is caused by other light in the classroom. The wavelength of each of these stripes are calculated:
The calculated wavelength of hydrogen gas should match the theoretical wavelength which comes from when the electron moves from n=3,4,5,6 to n=2. As shown above, three of the calculated wavelengths match closely to the theoretical values. We did, however, miss one of the stripe because the teal and purple stripe around it is too bright.
The energy of the transition associated with each line:
Line
|
Wavelength(nm)
|
ΔE (J)
|
ΔE (eV)
|
Red
|
423.07
|
4.70E-19
|
2.94
|
Green
|
490.48
|
4.06E-19
|
2.53
|
Blue
|
679.87
|
2.93E-19
|
1.83
|
Conclusion
The calculated values for this experiment match closely with the theoretical values, except for one of the wavelength that can hardly be seen because of the glare of the other two wavelengths.
Activphysics Classical Harmonic Oscillator
Purpose
The purpose of this experiment is to explore quantum mechanics in classical harmonic oscillation using a simulator.
Experiment
Potential Energy Diagrams
1. Range: -5cm -- 5cm
2. Because the kinetic energy of the particle is less than the potential energy outside of the well.
3. According to the data, there is a higher probability to detect the particle on the left region because the difference in the particle's kinetic energy and the well potential energy in the left region is smaller than it is on the right region.
4. The turning points move outward from the origin by a factor of radical two because U = (1/2) kx^2
5. The shape of the kinetic energy graph is a concave down parabola
6. At the turning points, because the speed of the particle at that point is 0.
Potential well
1.
E = n2 h2 / 8 m L2
= (1)2 (6.626 x 10-34 J s)2 / 8 (1.673 x 10-27 kg) (10 x 10-15 m)2E = 3.3 x 10-13 J
= 2.1 MeV
No, it's different compared to finite well.
2.
E = n2 h2 / 8 m L2
= 4 (2.1 MeV)
= 8.4 MeV
No, this value is different than the first excited state of the finite well model.
3. The wavelength of the wavefunction is larger in the finite well than it is in the infinite well because the wavefunction in the finite well can penetrate into the "forbidden" regions.
5. The penetration depth will decrease. As the mass of the particle increases, the chance to penetrate into the forbidden region decreases. Finally, when it has a large enough mass, it is consistent with the classical harmonic oscillator in macro scale.
Conclusion
The online tutorial helps understand quantum mechanics in classical harmonic oscillator.
The purpose of this experiment is to explore quantum mechanics in classical harmonic oscillation using a simulator.
Experiment
Potential Energy Diagrams
1. Range: -5cm -- 5cm
2. Because the kinetic energy of the particle is less than the potential energy outside of the well.
3. According to the data, there is a higher probability to detect the particle on the left region because the difference in the particle's kinetic energy and the well potential energy in the left region is smaller than it is on the right region.
4. The turning points move outward from the origin by a factor of radical two because U = (1/2) kx^2
5. The shape of the kinetic energy graph is a concave down parabola
6. At the turning points, because the speed of the particle at that point is 0.
Potential well
E = n2 h2 / 8 m L2
= (1)2 (6.626 x 10-34 J s)2 / 8 (1.673 x 10-27 kg) (10 x 10-15 m)2E = 3.3 x 10-13 J
= 2.1 MeV
No, it's different compared to finite well.
2.
E = n2 h2 / 8 m L2
= 4 (2.1 MeV)
= 8.4 MeV
No, this value is different than the first excited state of the finite well model.
3. The wavelength of the wavefunction is larger in the finite well than it is in the infinite well because the wavefunction in the finite well can penetrate into the "forbidden" regions.
4. The energy will decrease. The wavelength of the wavefunction is larger in the finite well compared to the infinite well, so on the same state, the energy of finite well is smaller than infinite well. As the depth of the well decreases, the energy decreases.5. The penetration depth will decrease. As the mass of the particle increases, the chance to penetrate into the forbidden region decreases. Finally, when it has a large enough mass, it is consistent with the classical harmonic oscillator in macro scale.
The online tutorial helps understand quantum mechanics in classical harmonic oscillator.
Wednesday, May 22, 2013
Activphysics Relativity
Purpose
The purpose of this activity is to understand the theory of special relativity such as time dilation and length contraction using a simulation.
Experiment
Time dilation
1. The distance traveled by the light pulse on the moving light clock is longer than the distance traveled by the light pulse on the stationary light clock.
3. In the moving frame of reference, the time interval for the light to travel to the top mirror and back is the same as the stationary clock.
4. As the velocity of the light clock decreases, the time interval difference decreases, because the value of gamma decreases also decreases. According to t' = t*γ , the time interval difference decreases.
4. L' = L/γ = 1000 / 1.3 m =769m
Conclusion
The online tutorial helps understand the idea of special relativity.
Experiment
Time dilation
1. The distance traveled by the light pulse on the moving light clock is longer than the distance traveled by the light pulse on the stationary light clock.
2. The time interval for the light pulse to travel to the top mirror and back is longer on the moving light clock compared to the stationary light clock.
3. In the moving frame of reference, the time interval for the light to travel to the top mirror and back is the same as the stationary clock.
5. t' = t*γ = 1.2 * 6.67 µs = 8.00µs
6. γ = t'/t = 7.45 / 6.67 = 1.12
Length contraction
1. The round trip depends on the velocity of the light clock.
2. The round-trip time interval for the light pulse as measured on the earth is longer than the the time interval measured on the light clock.
3. Because d = v*t. When the time intervals decreases, the speed of light has to stay constant, therefore for this equation to be consistent, the distance would decrease.
The online tutorial helps understand the idea of special relativity.
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