Planck's Constant from an LED

In this experiment we will use different colored LEDs or light emitting diodes to help us determine the value of Planck's constant. To do this we will have an apparatus that closely resembles the one we use in the Color and Spectra Experiment. However instead of a tube of gas we will use LEDs in their place. We will still use the rulers, diffraction gradient, and make measurements as before. In addition to calculating the LEDs wavelength we will also measure the voltage across each of the LEDs. We will then plot a graph of voltage vs. 1/wavelength to obtain a relationship between voltage and wavelength. This should result in a linear line with a definite slope. If we take this slope and multiply q and divide by c we should the obtain the value of Planck's constant which is 6.626*10^-34 J*s. A picture of the apparatus is bellow.

To begin the experiment we connect each of the LEDs one at a time in series with a 680 ohm resistor to a power supply. We then use the equation (lambda)=(D*d)/[sqrt(L^2+D^2)] to calculate the wavelength of each of the LEDs. L was 1.29 +/- 0.005 meters and d was 1.67*10^-6 +/- 0.015 meters. The values for D of each of the different colored LEDs are given below. Each of the values have an uncertainty of +/-0.01

Red: 0.54 m
Yellow: 0.49 m

Green: 0.42 m

Blue: 0.38 m

The wavelengths were then calculated with corresponding uncertainties of +/-1.15*10^-8 meters.

Red: 644.9 nm

Yellow: 593.0 nm

Green: 517.0 nm

Blue: 471.9 nm

Next we measure the voltage across each of the LEDs.

Red: 1.86 ohms

Yellow: 1.90 ohms

Green: 2.50 ohms

Blue: 2.66 ohms

We then plotted voltage vs. 1/wavelength and received the following graph.

The slope of our equation is 1.573*10^-6 meters. Multiplying by q and dividing by the speed of light we get a value for Planck's constant to be 8.756E-34 J*s.The true value for Planks constant is 6.626*10^-34. This gives us a percent error of 32%. This percent difference is very high but not surprising considering the apparatus we used and the inaccuracies of the measurements. Remember, in the previous experiment we used a graph to account for any discrepancy in the wavelength to reduce the error; in this experiment we were unable to do that. To reduce the error we would have to make more accurate measurements, account for the small angle approximation, and make sure the our rulers were place exactly perpendicular. Also I would have been useful to know the true wavelengths that each of the LEDs gave off so that we could have something to compare our data too. In this experiment we were also able to see that the red LED consumed the least amount of energy while the blue LED used the most amount of energy. This matches what the equation E=hc/(lambda) tells us.

Color and Spectra

In this experiment we used a diffraction gradient to measure the spectral lines of a white light source and that given off by an unknown gas. The purpose of the first part of this experiment is to make sure that we have to appropriate setup and that we can me accurate measurements with our apparatus. In the second part of the experiment we will measure the spectral line spacing and record their color. We would then match these lines with known spectral lines a sever different elements and determine what was our unknown gas.

To begin the experiment would first take a two meter long ruler and place it perpendicular to a one meter long ruler so that the two rulers formed an "L" shape. Where the two ruler met we placed our light source. At the opposite end of two meter long ruler we placed our diffraction gradient. We were careful to make it so the light source sat right at the end of the ruler so that the diffraction gradient was exactly two meters away (within uncertainty). In a dark room we turned on our light source and looked trough the diffraction gradient. We then made measurements where the visual spectrum laid along the one meter long ruler for the white light source. For the unknown gas we did the same except we noted where the spectral lines fell along the one meter long ruler and their color. The setup with the unknown gas is shown below.

In this experiment we will be using the equation (lambda)=(D*d)/[sqrt(L^2+D^2)] where lambda is our wavelength, L is 2 meters, D is the distance along the one meter ruler we will be recording, and d is the slit spacing in the diffraction gradient which is 1.67*10^-6 meters.

For the first part of the experiment were we used the light bulb we measured the length of the spectrum and where we saw the colors red, yellow, green, blue, and violet. We then got the following values. All measurements have an uncertainty of +/- 0.005 meters
Violet: 0.435 m
Blue: 0.495 m
Green: 0.585 m
Yellow: 0.65 m
Red 0.76 m
The entire visible spectrum ranged from 0.42 m to 0.855 m. Using the above equation we find that our spectrum ranged from 341 nm to 652 nm. The true visual spectrum ranges from about 380 nm to 750 nm. This indicates that our spectrum has been shifted to the left and compressed slightly. We will later plot a graph and use an equation to account for this.

In the next part of our experiment we replaced the light with a tube filled with an unknown  gas. A large voltage is run across the tube which allows it to glow. We then looked through the diffraction gradient and measured the spectral lines. The values are below with uncertainties of +/-0.0025 m
Violet: 0.5225 m
Green: 0.6075 m
Yellow: 0.645 m
Red: 0.665 m
Using the above equation we found the wavelength at each of our lines. All values have an uncertainty of +/-4nm
Violet: 422nm
Green: 485nm
Yellow: 513nm
Red: 642nm
Using a chart with spectral lines of known gases we compared the spectral lines of our unknown gas. We then determined that our lines most closely resembled the lines in mercury.

To help correct our values we then looked at the spectral lines of hydrogen and made our measurements. Hydrogen was used because it has fewer spectral lines and a plot could be graphed that corrected for the shift in the previous sections of the experiment. We obtained the following values for hydrogen.
Violet/Blue: 0.465 m
Green: 0.506 m
Orange/Yellow: 0.621 m
Red: 0.708 m
The corresponding wavelengths are as follows. All values have an uncertainty of +/- 11nm
Violet/Blue: 378 nm
Green: 410 nm
Orange/Yellow: 495 nm
Red: 557 nm
After plotting the values into excel we plotted our data to give us our correction factor which resulted in the equation (lambda)_actual=1.051*(lambda_exp)+4.538. Using this equation we were given our corrected wavelengths. All the values have an uncertainty of +/- 16 nm.
Violet/Blue: 402nm
Green: 435nm
Orange/Yellow: 525nm
Red: 590nm

After the calibration we found that  our results were fairly close to the true values for each of the corresponding colors with the largest percent difference being 5%. This falls well within our range of uncertainty and can be accounted for by the uncertainty of our measurements.

Modeling Wave Packets Using Python

 In this project we wrote a computer code in Python to help visualize wave packets. We also used Pylab to show the code as a series of graphs. In order to make our graphs we had to write our code in a series of steps. The first step was to create a sine graph. Next we manipulated that graph so that we could show several harmonics or several sine graphs of differing periodicity. After that was completed we then generated a Gaussian. After that was completed we then had to combine the sine functions and the Gaussian to generate a wave packet. This was very complicated to do and took us the most amount of time to resolve. When this was all completed we were able to generate normalized graphs that visually depicted wave packets. The code we used to do this is bellow. The code is "messy" so many lines were canceled out with ## so that everything would run like we wanted. However is is possible to obtain the sine functions and the Gaussian separately by canceling out the appropriate line and removing the ##s where necessary.

from pylab import*
harmonics= 20
center=harmonics/2
sigma=1
coeff=1/((sqrt(2*pi))*sigma)
gauss_list=[]
domain_list2=[]
L=10
knot= 2*pi/L
b_list=[]
kdomain=[]
##
##for k in arange(0,200,0.1):
##    bofk=(1/(knot))
##    b_list.append(bofk)
##    kdomain.append(k)
####plot(kdomain,b_list)
##




for x in range(1,harmonics):
    gauss=coeff*exp(-(x-center)**2/(2.*sigma**2))
    gauss_list.append(gauss)
    domain_list2.append(x)
  
##A_coeff_1=1
##wave_constant_1=1
##sine1_list=[]
##domain_list=[]
##for x in arange(-pi,pi,0.1):
##    sine1=A_coeff_1*sin(wave_constant_1*x)
##    sine1_list.append(sine1)
##    domain_list.append(x)
w=1
Fourier_Series=[]

for i in range(1,harmonics):
    sine_function=[]
    x=[]
    for t in arange(-pi,pi,0.01):
        sine_f=gauss_list[i-1]*sin(i*w*t)
        sine_function.append(sine_f)
        x.append(t)
    ##plot(x,sine_function)
    Fourier_Series.append(sine_function)

superposition=zeros(len(sine_function))
for function in Fourier_Series:
    for i in range(len(function)):
        superposition[i]+=function[i]
print kdomain
plot(x,superposition)
##plot(domain_list2,gauss_list)
##plot(domain_list,sine1_list)
show()

20.2 Quantum Mechanics: Particle in a Box

 In this activity we will learn about the probability of finding a particle in a box and how these 'particles' have a wave-like nature which can be defined by the de Broglie relation. We will also learn about the ground state energy of these particles.

Question#1: From your study of mechanical waves, what is the longest wavelength standing wave on a string of length L?

Answer: The longest wavelength a standing wave can have on a string that is L long is 2L.

Question #2: What is the momentum of the longest wavelength standing wave in a box of length L?

Answer: De Broglie stated that the momentum of a particle is plank's constant divided 2L.

Question #3: Assuming the particle is not traveling at relativistic speeds, determine an expression for the ground state energy.

Answer: If the particle is not traveling at relativistic speeds then we know that for the momentum, p=mv and for the kinetic energy,  K = 1/2mv^2. If we use De Broglie's equation for momentum (p = h/2L) we see that K = h^2/(8mL).

Question #4: If the size of the box is increased, will the ground state energy increase or decrease?

Answer: As L increase we can see from the equation above that the ground state energy will decrease.

Question #5: In the limit of a very large box, what will happen to the ground state energy and the spacing between allowed energy levels? Can this result explain why quantum effects are not noticable in everyday, macroscopic situations?

Answer: As L increases to macroscopic levels the ground state energy decrease to such a insignificant amount that it is almost impossible to detect by conventional methods for measuring energy.

Question #6: In the limit of a very massive particle, what will happen to the ground state energy and the spacing between allowed energy levels?

Answer: Since mass is also in the divisor in the above equation very massive objects also a negligible ground state energy.

Question #7: If a measurement is made of the particle's position while in the ground state, at what position is it most likely to be detected?

Answer: The particle is most likely to be found  in the center of the box.

Question #8: The most likely position to detect the particle, when it is in the ground state, is in the center of the box. Does this observation depend on either the mass of the particle or the size of the box?

Answer: No because the wave function is independent of the mass and the particle is always most likely to be found at the center of the box.

Question #9: The most likely position to detect the particle, when it is in the ground state, is in the center of the box. Does this observation hold true at higher energy levels?

Answer: No, for for example the most likely place to find the particle when it is in the first excited state is at L/4 and 3L/4.

Question #10: In the limit of large n, what will happen to the spacing between regions of high and low probability of detection? Does this agree with what is observed in everyday, macroscopic situations?

Answer: In the limit of a large n the probability density function gets "squished" together and appears to be constant. This means that the particle has become a "free particle" and has an equal probability to be found in all locations.

Programing in Python

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17.2 Modern Physics: Relativity of Length

In this online activity we learned how relativity distorts the length of objects as they approach the speed of light. We will be using length contraction equation L=L_0/sqrt[1-(v^2/c^2)]. As in the previous activity questions will be followed with a screenshot and the appropriate answers.

Question #1: Imagine riding on the left end of the light clock. A pulse of light departs the left end, travels to the right end, reflects, and returns to the left end of the light clock. Does your measurement of this round-trip time interval depend on whether the light clock is moving or stationary relative to the earth?

Answer: The round trip time depends on whether the clock is stationary or moving.

Question#2: Will the round-trip time interval for the light pulse as measured on the earth be longer, shorter, or the same as the time interval measured on the light clock?

Answer: The round-trip time will be measured to be shorter than the time measured relative to the earth.

Question #3: You have probably noticed that the length of the moving light clock is smaller than the length of the stationary light clock. Could the round-trip time interval as measured on the earth be equal to the product of the Lorentz factor and the proper time interval if the moving light clock were the same size as the stationary light clock?

Answer: Yes, if the loreanzt factor is known and the time required  for a moving clock to travel a given distance we can use the formula stated at the beginning of the activity to calculate the time required to reach the same distance relative to the stationary frame.

Question #4: A light clock is 1000 m long when measured at rest. How long would earth-bound observer's measure the clock to be if it had a Lorentz factor of 1.3 relative to the earth?

 Answer: The Lorentz factor is 1.3, and the length is 1000m at rest, so we use the formula given at the beginning of this activity to calculate the contracted length. We obtain 1000/1.3  = 769 meters.

17.1 Modern Physics: Relativity of Time

This online activity was used to help us understand the basic concepts of relativity. In these problems we will explore how space and time are distorted as different frames approach the speed of light. We will make use of the time dilation equation which is (delta)t' = (delta)t/sqrt[1-(v^2/c^2)]. Screenshots of each of the question a posted below with their corresponding answers.

Question#1: How does the distance traveled by the light pulse on the moving light clock compare to the distance traveled by the light pulse on the stationary light clock?

Answer: The distance traveled by the moving light clock is greater than the distance travel by the stationary light clock.


Question #2: Given that the speed of the light pulse is independent of the speed of the light clock, how does the time interval for the light pulse to travel to the top mirror and back on the moving light clock compare to on the stationary light clock?

Answer: Since the speed of light is constant, the moving light clock must travel a further distance and therefore must take more time to complete one cycle.

Question #3: Imagine yourself riding on the light clock. In your frame of reference, does the light pulse travel a larger distance when the clock is moving, and hence require a larger time interval to complete a single round trip?

Answer: In the moving frame of reference, the light pulse does not travel a greater distance, thus the time required for the light to make one cycle in the moving frame is the same time as the light to make one cycle in the stationary frame.

Question #4: Will the difference in light pulse travel time between the earth's timers and the light clock's timers increase, decrease, or stay the same as the velocity of the light clock is decreased?

 Answer: If the velocity of the light clock is decreased, the difference of earth's timer and the light clock's timers also decreases.

Question #5:  Using the time dilation formula, predict how long it will take for the light pulse to travel back and forth between mirrors, as measured by an earth-bound observer, when the light clock has a Lorentz factor (γ) of 1.2.

 Answer: The time dilation of a moving clock with a Lorentz factor of 1.2 will have a time 1.2 times that of the original time.

Question #6: If the time interval between departure and return of the light pulse is measured to be 7.45 µs by an earth-bound observer, what is the Lorentz factor of the light clock as it moves relative to the earth?

 Answer: The measured time is 7.45 µs, we used the simulator and found that the Lorentz factor is around 1.12, which is what about the same number we calculated.