Particle in a Box

We looked at problems/simulations on the Pearson physics website for section 20.2 LINKAGE

Answers to the questions:
1. λ=2*L
2. λ=h/p = h/(2L)
3. E=p^2/(2m) = h^2/(8m*L)
4. it will decrease
5. ground state ->0 & spacing goes to 0
6. Ground state energy and spacing is 0
7. the center of the box
8. It doesn't, it depends on L
9. Yes, probability of detection stays highest at the center
10. As n gets larger, regions of probability blur together.

Planck's constant from an LED

In this lab, we can use the principles on which an led operates to experimentally determine planck's constant.
We know the following relationships:
λ=h/mv
E=hf=hc/λ

We used 4 different color LEDs: Red, yellow, green, and blue. We ran the experiment the same exact way as the hydrogen and color spectra lab.

Color	Distance D (meters)	Voltage (Volts)	Wavelength (nm)
Yellow	0.67	1.88	592.43
Green	0.59	2.52	529.04
Red	0.735	1.82	641.93
Blue	0.525	2.64	475.63

Using the fact that E=hc/λ=q*V, we get that h = q*V*λ/c

Color	Wavelength (nm)	Voltage (Volts)	h
Yellow	592.43	1.88	5.94E-34
Green	529.04	2.52	7.11E-34
Red	641.93	1.82	6.23E-34
Blue	475.63	2.64	6.70E-34

From these results, only blue seems to give a value that was within scientifically acceptable bounds of the theoretical value at 1.12% error. Red also came in at a relatively low error of 5.98%. Both Yellow and Green had relatively high error. Coincidentally, their spectra also contained various colors which made it hard to pinpoint the maxima that corresponded to the color of the LED itself. This is likely due to the fact that the green LED had impurities and was contained additional wavelengths. The particular shade of yellow also had an orange tinge to it. Overall, it appears that the colors on the ends of the visible spectrum were much easier to identify.

Color and hydrogen spectrum

This lab uses a diffraction grating to analyze wavelengths of light from a specific light source.
Our setup was as shown.

We placed a light source at the origin of the 2 rulers, placed a diffraction grating at a distance from the bulb, and looked through it and measured the range of wavelengths.We use the following formula:

where λ is the wavelength, d is the slit spacing on the diffraction grating, D is the observed horizontal distance from the light source to the band of light, L is the distance from the diffraction grating to the light.
 We took readings at 2 distances to minimize error.

L (m)	Red(D)	Violet (D)
1.5	0.65	0.3
1.8	0.745	0.35

We were then given a gas discharge tube, and our job was to determine what type of gas was in the tube based on the spectral lines we observe.

We compared the spectral bands with a chart of different gasses' spectral lines, and they corresponded almost perfectly with mercury.

color	D (m)	Measured λ	Theoretical λ
Red	0.75	690	690
Yellow	0.61	589	580
Green	0.54	535	545
Violet	0.42	439	435

We plot the measured values against the theoretical values to get a correction value of 75 nm.
We were then given a hydrogen discharge tube, and measured the spectral lines of hydrogen.

D (m) λ (Measured) (nm) λ (Calculated) (nm) λ (Theoretical) (nm) Red (D) 0.7 655 654 656 Blue (D) 0.49 495 495 486 Violet (D) 0.43 447 446 434

The different spectral lines indicate the different energy levels that the hydrogen can transition between. For each of these transitions, the hydrogen emits a photon corresponding to a specific wavelength. This method can be used to analyze an unknown gas and determine it's composition based on the energy levels.

Visualizing Wave packets

This time we used python to help visualize wave packets, and better understand the wave function.
We used a python module called pylab to graph functions.
For my program, you can select the number of harmonics and the value for sigma, and compare different arrangements at once.

from pylab import *
harmonics,sigma = zeros(5),zeros(5)

#### Variables #####
harmonics[1] = 97
sigma[1] = 1
###
harmonics[2] = 97
sigma[2] = 5
###
harmonics[3] = 97
sigma[3] = 10

###
rng = 3.14 #range of x values to plot (-rng to +rng)
#############

def gauss(number,sigma):
    """harmonics, sigma"""
    gauss_list = [] #empty array
    coeff = 1/(sqrt(2*pi)*sigma)
    for x in range(0,number):
        funcval = coeff * exp(-(float(x) -float(number-1)/2 )**2/(2 * sigma**2))
        gauss_list.append(funcval)
    print gauss_list
    return gauss_list

def sinplot(start, stop, A, harms):
    """start, stop, amplitude, harmonics"""
    loop = int()
    superpos = []
    
    for n in arange( start, stop, 0.01 ): #Create list of zeros
        superpos.append(0)
    for i in range( 0, len(harms) ):
        plot_list = []
        domain = []
        loop = 0
        for x in arange( start, stop, 0.01 ):
            funcval = A[i] * sin( harms[i] * x )
            plot_list.append( funcval )
            domain.append(x)
            
            superpos[loop] = superpos[loop] + funcval #sum of funciton values
            loop += 1

        #plot( domain, plot_list) #show each harmonic
    plot( domain, superpos )

#sinamp = gauss(harmonics, sigma)
#print sinamp
#harmonic = range(1,harmonics+1)
try:
    for i in (1,2,3,4,5):
        sinplot(-rng, rng,
                gauss( harmonics[i],sigma[i] ),
                range( 1,harmonics[i]+1 )
                )
except:
    print("Graphing...")
    show()

The program generates this

The plot shows the sum of harmonics of a wave function, with each increasing harmonic having a decreasing amplitude that varies as a gaussian function, with a specific sigma value. The blue plot has a sigma value of 1, the green plot has a sigma of 5 and the red plot has a sigma of 10. This shows that by using the right amplitudes and numbers of harmonics, the superposition of waves can model a wave packet.
We can then use this to answer some questions

 a. The graph is a straight line
b.
c. 1*L
d. 2*L
e.
f. 6.63e-34 = h
g. 6.63e-34 = h

Modern physics: Relativity of Time

In this lab, we looked at some Modern Physics simulations and answered questions about them related to relativity of time in relation to distance and velocity.

 

Question 1: Distance traveled by the light pulse

   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?
Ans: The distance traveled is greater on the moving light clock than on the stationary clock

Question 2: Time interval required for light pulse travel, as measured on the earth
  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?
Ans: The time for the moving clock is greater than for the stationary clock

Question 3: Time interval required for light pulse travel, as measured on the light clock
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?
 Ans: No, from your frame, the distance is still 2x the distance between mirrors so it takes the same time.

Question 4: The effect of velocity on time dilation
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?
Ans: The difference will become smaller as the velocity of the light clock decreases

The relationship between the time interval measured by an observer and the proper time interval is:
Δt = γΔt_proper
where γ is related to the relative velocity between the observer and the clock measuring the proper time interval via
γ = (1 - v² / c²)^-1/2
Question 5: The time dilation formula
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.
Ans: This is simply γ*Δt_proper = 1.2*(6.67μs) = 8μs

Question 6: The time dilation formula, one more time
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?  
Ans: γ = Δt/Δt_proper -> 7.45/6.67 = 1.12

CD diffraction

In this lab we measured the width of the grooves in a CD by using the principles we know about laser diffraction

The distance between these groves is far too small for a the human eye to measure. With diffraction however, we can measure the distance between observed maxima on the diffraction pattern resulting from a laser (λ = 670 nm) reflected off of a CD. The grooves in a cd act as a reflection diffraction grating, so the distance from the center to the m'th maximum can be found as sin(θ) = mλ/d where d is the groove spacing.

Our data was as follows:

After plugging in all the numbers, we got an average d of 2.7*10^-5 meters
We had a very large error which probably due to the fact that our laser might not have been perfectly incident on the diffraction grating (ie, flat spot on the cd), and we are getting thin film interference from the plastic disk.

Measuring a human hair

Today we measured the thickness of a human hair by diffraction of light.
For a double slit diffraction of a light beam, light and dark interference spots are created on a surface some distance L from the slots.
If the slots are separated by a distance d,  the distance from the center bright spot and bright spot #m is y = λL*m/d
We treat the human hair as the separation between openings and if we shine a laser at the hair, we get a corresponding diffraction pattern on a distant surface.

We placed a whiteboard 1 meter away from the hair, to simplify calculations. our λ was 680nm for a red laser. We measured the distance between the center and the 4th bright spot and got y=4.8 cm
Rearranging and plugging in we got d = 680e-9*1m*4/.048 = 5.67e-5m = 56.7 μm

My hair is on the very lower border of human hair thickness according to google. Makes you think I have alopecia or something haha