Planck's Constant from an LED

6.6262*10^-34 the value for planck's constant, we know that from the book and have used it millions of times, but to solve for where it came from, that when LED and color spectra comes in to place.





LEDs or A light-emitting diode  is a semiconductor light source. LEDs are used as indicator lamps in many devices and are increasingly used for other lighting. 


when using different colored LEDs we can find the spectra on where these LED hits and from the color and spectra equation we relate the wavelength to the amount of eVs that is being emitted. 


we had four different types of color

	Distance D (meters)	Voltage (Volts)	Wavelength (nm)
Yellow	0.65	1.92	590.43
Green	0.54	2.56	535.04
Red	0.75	1.79	660.93
Blue	0.532	2.61	445.76

with the values of waves length known, we can easily show that
E = hc/lambda = q_eV


since we know the charge of the electron the voltage 3V, and lamba and c are known values as well we can solve for planck's constant


the closes result we got from those wavelength was red which gave an vault of  6.23*10^-34 which is about right on the vault of planck's constant.


for the other vaules of h that we got it was anywhere going from a 1.6% error to about 10% error, this could have been measuring errors due to estimating and round off numbers. also when reading the color spectra it was hard to determine the center point of the spread, this might also affect the value of h found by the other wavelengths

Quantum Mechanics: Particle in a Box

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


wavelength = 2L


Question 2: The de Broglie Relation
What is the momentum of the longest wavelength standing wave in a box of length L?


p = h/2L
L = ph2


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


E= P^2/m         E= h^2/8mL .


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


if the size of the box was increased the ground state energy  would decrease 


Question 5: The Correspondence Principle: Large Size
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?


if the size of the box was small say on a scale where it can fit electrons in, then we can make the ground state energy almost close to zero.


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


yes if the particle is moving fast enough we have to taken in account of the rest energy as well as the kinetic energy.





Question 7: Ground State Probability
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?


to be in the center of the box

Question 8: Probability: Dependence on Mass and Size
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?


no at the ground state the particle is always most likely to be at the center of the box


Question 9: Probability: Dependence on Energy Level
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?


no when the energy leave ground state the probability of the particle is no longer more likely to be at the center it is then random, and can be at places where it matches the energy level present 

Modern Physics Relativity

Time Dilation 


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?


the time that it took for the moving object took longer then that of the stationary

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?

Run the simulation, which this time displays timers at rest on the earth's surface.

When the light clock is moving, the light pulse must travel along the hypotenuse of a right triangle whose legs are formed  Thus, this distance is larger than simply the distance between the mirrors.

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?

no when we are in the same reference frame as the light clock then we do not see the times being different

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?

To check you

if the velocity of the light clock is deceased then the travel time between the earth's timer and the light clock's timer is decreased


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.


i get about 8.7us for the new time


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?

t= t0gamma 

1.12 = gamma

Length Dilation 

Question 1: Round-trip time interval, as measured on the light clock
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?

if you are on the same frame as the moving object then that means that time or length dilation does not effect you, this measurement does depend if the light clock is moving relative to the earth.


Question 2: Round-trip time interval, as measured on the earth
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?

the time on earth would be longer then the interval measured on the light clock.

Question 3: Why does the moving light clock shrink?
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?

the light is taking less time around us because we see it moving faster being stationary .and  If time dilation is the law of the universe, then so must be length contraction.

Question 4: The length contraction formula

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?

l= l0/gamma

l = 736m 

Python Lab #2

from pylab import*

r=50

center = r/2

sigma=1

coeff=1/sqrt(2*pi)*sigma

gauss_list=[]

k=1

pii=3.14

for x in arange(0,r):

    gauss=coeff*exp(-(x-center)**2/(2*sigma**2))

    gauss_list.append(gauss)

fourier_series=[]

A=gauss_list

for i in range (1,r+1):

    sin_function_list=[]

    domain_list=[]

    for x in arange (-pii,pii,0.01):

        sin_function=A[i-1]*sin(i*k*x)

        

        sin_function_list.append(sin_function)

        domain_list.append(x)

    fourier_series.append(sin_function_list)

superposition = zeros(len(sin_function_list))

for function in fourier_series:

    for i in range (len(function)):

        superposition[i]=superposition[i]+function[i]

plot(domain_list,superposition)

show()

this python lab was a lot less about how to program and a lot more on how can programming help us understand waves better

the code above displaces a graph of different waves functions adding and subtracting forming one wave, the superposition of them.

we can see that we defined a gauss function which represents the graph and have a while loop making a few gauss functions and plotting the total of their superposition.

Python Lab #1

from visual import *

ball= sphere(pos= (-3,3,3), radius= .5, color=color.blue)

wallR = box(pos=(6,0,0), size =(.1,12,12), color=color.red)

wallS = box(pos=(-6,0,0), size =(.1,12,12), color=color.red)

wallT = box(pos=(0,6,0), size =(12,.1,12), color=color.red)

wallU = box(pos=(0,-6,0), size =(12,.1,12), color=color.red)

wallbak = box(pos=(0,0,-6), size= (12,12,.1), color=color.red)

dt = 0.6

g = .00982

ball.velocity = vector (.1,.1,.5)

bv= (arrow(pos=ball.pos,axis=ball.velocity,color=color.yellow))

ball.trail=curve(color=ball.color)

while(1==1):

     ball.pos=ball.pos+ball.velocity*dt

     rate(150)

     if ball.x>wallR.x-0.5:

         ball.velocity.x=-ball.velocity.x

     if ball.x>wallS.x-0.5:

         ball.velocity.x=-ball.velocity.x

     if ball.x>wallT.x-0.5:

         ball.velocity.x=-ball.velocity.x

     if ball.x>wallU.x-0.5:

         ball.velocity.x=-ball.velocity.x

     if ball.x>wallbak.x-0.5:

         ball.velocity.x=-ball.velocity.x

     if ball.z> 5.5:

         ball.velocity.z=-ball.velocity.x

     if ball.y>-5.5:

         ball.velocity.y=ball.velocity.y-g

     bv.pos = ball.pos

     bv.axis=ball.velocity

     ball.trail.append(pos=ball.pos)

there above i have posted this beautiful code.

from the first line i have accessed a lib that allows me to displace visual effects.

the first 10 lines of the code was to define the ball its position and initial speed.

the next 5 lines was made so that we can track the balls velocity with an arrow and the have a line follow the ball to show the balls trail

with in the whole loop we have statements to define the box, stating that if the position of the ball was to become too close to the the wall the velocity is shot backwards creating this "bounce" effect

when the last few lines stating the gravity of the system and how it effects the ball as it bounces, every time it bounce it loses  energy and is lowered 

Color & Spectra

when the subject of the light spectra is considered it is mostly referring to the visible spectra from 380 nm to about 780 nm, or what wavelength we humans can see.  there are of course are other spectra in the inferred and ultra violet and even beyond that, but we wont have to worry about that till another day

the diffraction lens we use is very small, naked to the human eye, we can however use diffraction pattern projected to find the spacing between them, but since it is a fixed lens we can assume that the spacing is just 1.6*10^-6 m

with the d spacing known we can use 

lambda = D/(L^2+D^2)^1/2 to find the wavelength of the spectra that we observe across the table.

for max D of .62m

and min D of .35m

we then found lambda of the max and min D to be 470 and 690 (nm)

As we measured each wavelength the purple was 393 nm, blue 446 nm, green 506 nm, yellow 507 nm, and red was 644 nm. with the different Ds

we took the data of our wave length and plotted it vs the theoretical data, which was fairly close to one another, we then used the different types of wavelength to obtain the element that was emitting  the light, and have found out that it was mercury.

using a high voltage difference we can excited electrons in to omitting light, from that light we can get a color spectra which then we can find the max and min wavelength, we then can use this data to determine unknown

elements.

this becomes very helpful in many projects when chemicals or subsistence are unknown.  by using the diffraction pattern of its color spectra we can found out what element it contain.although i am not quite sure if we can determine the type of element when substances are mixed.

Measuring a human hair

we are to accurately measure the thickness of a human hair, not by a ruler but by the idea of double slit.  We can use the distance between the hair and the board, the interference y_m of the slit as well as overtones.

we first make a hole in a 3x5 card and taped a human hair on to it, than we pointed a laser through the hole, we used the image to be projected oh to the white board, we now can measure the distance between the board, and the distance between the middle of the zero overtone and third.  

from what we learned in class about double slit, we can derive an equation 

 

y_m=L(lambda)(m)/d

but we are looking for d

d=L(lambda)(m)/y_m

where y_m is the distance between the zero overtone and the m we picked

L is the distance between the hair and the white board

d is the thickness of the hair

y_m we got to be about 6.5 cm _+ .023cm

L = 2m

we used the third overtone m=3

and the wavelength of the laser is 660nm

applying the data to the equation above we get d to be about 6.09 *10^-5 and 6.27*10^-5

from this picture we can clearly see the interference of the double slit experiment, we are getting a bright slit then a dark slit, bright  meaning the waves are converging, while the dark slits are diverging, we can also see that the bright slits are going to be brighter on the inside and as it gets closer to the out side it gets darker.

analysis

by taking looking at this experiment we can assume that there is a lot of mistakes that can be made during the processes of the experiment.  measuring the distance of y_m might be the biggest, but compare that number to that of L, the mistakes is minimum.  the focus of this experiments is to show that light can be shown as a wave rather then particles.    

Lens

from what we observed from the mirrors and lens labs we can find out where the focal length of the lens is measuring it or by using the eq

                                                              1/s+1/s`=1/f
the focal point of the lens is the point where the parallel beam refract to.  after the beams hits the surface
we measured f equals 14.97cm



by using a double convex mirror we can increase distance between the object and mirror by a factor of the focal length, to produce different images smaller and larger depending on how far the object is from the   double convex mirror.

d_0 cm	d_i cm	h_i cm	h_0 cm
79.88	23.45	1.3	3
59.88	24.5	1.75	3
39.98	28.7	1.95	3
29.99	44.5	5.2	3
22.96	332.5	53.6	3

 if we change the object distance to about .5 the focal length the image would become to large for us to see, so we would assume that as the lens become closer to the object the image we receive is going to get larger.
if we look through the lens at the object and view the image when the focal length .5, we the the image upright and big.


analysis
from the experiment we observed that the image is always inverted, along both the vertical and  horizontal axis, if this was a single convex mirror then the image would not have have been inverted.



the two lens are going to have a different length in R, if we take that in to considerations we get

1/R1+1/R2=1/s+1/s`


the slope of the negative inverse d_0 to the negative inverse d_i is about .565 this would be the degree M that the image is changing.  like we observed if the object was getting further away then the image would become smaller and smaller till it becomes to small to measure.

Concave and Convex Mirrors

In this experiment, we are to show the different effects of a convex mirror vs a concave mirror.

when we stand in front of  a convex mirror, the image appears to be smaller than the object itself, it does however stays upright, and is located about the same position inside and outside the mirror.

when the object is moved closer to the mirror the image becomes bigger, and when it is moved further away the image gets smaller.

for the convex mirror we gotten the object height to be around 3.1cm+_.034cm

the distance of the object about 6.2cm+_.45cm

the image height about .7cm+_.23cm

and the image distance to be 2.3cm+_.31cm

we are given an equation for M which equals h_i/h_0 which gets you M = 0.2258

there is no units for M it is just a scalar number

PartII

now we can look at concave mirrors, the image that appears in the mirror is larger, but it is inverted, and relative to the position of the mirror the object well seem closer.

moving the object closer to the mirror makes the image smaller, and upright and as for moving further away the image grows to infinite 

for the concave mirror we gotten the object height to be around 3.1cm+_.034cm

the distance of the object about 6.2cm+_.45cm

the image height about 1.7cm+_.23cm

and the image distance to be 2.3cm+_.31cm

again we get M to be about .463+_.012 (no units)

my sketch is a little off on the height of the image, but that could have been me measuring it wrong, but the difference wouldn't matter.

analysis

in this experiment we can clearly see how convex and concave mirrors would work on an object being placed in front of them, we also identified how we can see the image in the mirror by drawing lines to where the focal point, the lens of the mirror and the center of the sphere.  knowing that we can find the height of the image, with the magnification of the object.

                                                                 m= - s`/s
                                             from this we can find m then we can use m to find y
                                                                  m= y`/y