Determining Planck's Constant

In this experiment we were using LEDs (Light Emitting Diodes) to determine Planck's Constant experimentally. We tested two light emitting diodes which were blue and yellow. They shine at different levels of intensity at different voltages because of the color of light that they emit. The values for the wavelength(λ) of blue and yellow light were estimated. The picture below is of the an LED with a diffraction grating in front of the lens. The only color seen through the diffraction grating is a deep blue color and a pale yellow color.

The energy was determined by the voltage output coming from the power supply. When the electrons pass through a potential,  they will have an energy in terms of electron volts(eV).

E_Blue = 1.95 eV λ_Blue = 450 nm
E_Yellow = 2.76 eV λ_Yellow = 590 nm

The experimental value of Planck's constant can be determined by plotting the values of λ vs 1/E and using the slope. The slope can be used to calculate Planck's constant.

 

The slope from the graph is m = 1.490225*10^-25.  h can be found through this equation, 

m = hc, where m is the slope and it is a relation between the wavelength and the energy, h is Planck's constant, and c is the speed of light.  m/c = h 

h = 4.967 * 10^-34 Js is the calculated value. The percent error of this value compared to the actual value is  25%. We have this large error because only two colors were used to determine the slope and the values of the wavelength were estimated instead of calculated. 

Light and Matter Waves

When light and matter waves pass through an object or an opening, they diffract. The intensity can be calculated with Huygens' Principle along with the ideas of the superposition principle. These different ideas will be simulated with python(x,y). These simulations will help us visualize how the waves behave. We will begin by observing how the electric field from a point source changes with a detector. The detector will detect through all of the points in the space being observed.


When the slit space(y) is 12mm and a wavelength(λ) of 2mm and there is one point source the will be seen in this form.

y = 12 mm, λ = 4 mm 1 point source

y = 12 mm, λ = 8 mm 1 point source

y = 12 mm, λ = 2 mm 2 point sources

y = 12 mm, λ = 4 mm 2 point sources

y = 24 mm, λ =4 mm 2 point sources

y = 12 mm, λ = 8mm 2 point sources 

y = 24 mm, λ = 8mm 2 point sources 

In the next set of pictures we changed the amount of sources and the number of detectors in the wave.

There are 50 sources (s) and 500 detectors (d)

This is when M = 5 and N = 7

This is when M = 50 and N = 500

This is when M = 51, N = 500, the slit distance = 1μm and the slit width = 5μm

This is when M = 51, N = 600, the slit distance = 0.5 μm and the slit width = 10μm 

This is when M = 51, N = 600, the slit distance is 50mm and the slit width is 25μm

This is when the length of the wave is 600nm

This is when the length of the wave is 700nm

This is when the screen distance is 5 mm

This is when the screen width is 50 mm 

This is when the screen width is 5 mm

There are different lines of code that help us program these different simulations. The code below helps us calculate our coordinate system that we used. 

N = 100

dX = screendist/N

dY = screendist/N
Xcoords = arange(0, screendist+2*dX, dX)

Ycoords = arange(-screendist/2, screendist/2 + 2*dY, dY)

[xd,yd] = meshgrid(Xcoord,Ycoord)

When we are graphing 2D images of our 3D graphs, it gives us a representation of how the graph behaves and any differential area. It is easier visually, to interpret a 2D graph versus a 3D graph. Some of graphs go to a zero point because this is when destructive interference occurs at its most intense amount. 

The maximums of the graphs change as you go away from the center. The further the detector is away, the harder it is to detect the waves. When the detector is close it can tell us a lot of information of how the waves behave.