Experiment 14: Determining Planck's Constant

Purpose

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

LED color	Wavelength(nm)	1/V (1/V)
Red	607	0.529
Green	514	0.388
Yellow	570	0.521
Blue	456	0.377

The equation of the graph: y = 7.46E-7x+ 1.98E-7
Slope = 7.46E-7x = hc/e

h = 3.98E-34 J*s

% 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

Purpose
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.



    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.

Conclusion

       The online tutorial helps understand quantum mechanics in classical harmonic oscillator.

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.

    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.

   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.

   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.

     4. L' = L/γ = 1000 / 1.3 m =769m

Conclusion
       The online tutorial helps understand the idea of special relativity.