Wednesday 25 June 2014

deBroglie Relations and the scale of Quantum Effects, light waves as particles (b)

ii. a microwave operates at roughly $2.5 GHz$ at a max power
of $7.5 *10^9 erg/s$
How many photons per second can it emit? What about a 
low-power laser ($10^4 erg/s$ at $532nm$)?

Ans:
1.First for the microwave, 
The energy of one such microwave photon is $E = h v$
where the frequency v is $2.5*10^9 Hz$
so:
$E = 6.63*10^{-27}*2.5*10^9$

$E = 1.657*10^{-17}erg$

This is the energy, we have the power in erg/s which is $7.5*10^9erg/s$, to find the number of photons emitted we need to divide the power by the energy:

$\text{number of photons }=\frac{P}{E}$

$\text{number of photons }=\frac{7.5*10^9}{1.657*10^{-17}}$

$\text{number of photons }= 4.5265*10^{26}$

2.Now for the Laser,
Since now except the frequency, we have the wavelength we can substitute the formula of frequency in our energy formula:
$E=h*(\frac{c}{\lambda})$
where the frequency v is $2.5*10^9 Hz$ 
so:
$E = 6.63*10^{-27}*2.5*10^9$

$E = 1.657*10^{-17}erg$

$E=6.63*10^{-27}*(\frac{299,792,458}{532*\frac{1}{1*10^{9}}})$

$E=3.73*10^{-12}erg$

Now divide it with power:

$\frac{1*10^4}{3.73*10^{-12}} = 2.7*10^{15}photons$






Quantum micrometer experiment, how to find the width of your hair

HI! Yesterday I did an experiment to find the width of my hair using my green laser. The method is to hit the laser beam on the hair, the light passing above the hair will be like going through a slit, and the light passing below the hair will be like going though a slit. So in other words it is a type of double slit experiment. By using the effect of wave-particle duality of a photon, and the pattern of interference, we can calculate the distance between the two slits, or in other words the width of my hair. 

This is the setup of my so called lousy apparatus :P. I have supported the laser on the bucket using a stack of 4 coins, to adjust the beam to hit the hair. the hair is supported by a wire, which I have bended here and there for support. The wall where the interference pattern forms is 6.72meters away from the hair. 


This was the bucket with the laser and hair on it. the light flare ahead is where the beam hits the hair.



A close up of the hair and the beam hitting it.


So this was the interference pattern produced 6.72 meters away from my hair. The number of these fringes is 18, you can only see 15 in this pic, there are three more to the right. you have to count till where you cant even see anything. The length from the center fringe to the 18th fringe is 0.87meters, note all these values down.

To do this calculation there are two formulas which I had.

they are formulas to find the distance between the two slits, which will be the width of my hair.

the formula is $d = \frac{n l \lambda}{x}$
and
$d=\frac{n*\lambda*\sqrt{x^2+l^2}}{x}$ which is the same as the upper formula, but it uses the distance from the hair to the outer edge of the interference pattern. Im going to use both to find more accurate results.

in these formulas:
$\lambda$ = wavelength of laser in meters.= $532*10^{-9}m$
$n$ = the number of fringes. = $18$
$l$ = distance from hair to wall = $6.72m$
$x$ = distance from center of pattern to farthest countable fringe = $0.87m$

plugging these in the first formula we get:

$d = \frac{n l \lambda}{x}$
$d = \frac{18*6.72*532*10^{-9}}{0.87}$
$d = 0.000074m$

so using this fromula I get 74microns.

Now plugging in in the second formula we get:

$d = \frac{n*\lambda*\sqrt{x^2+l^2}}{x}$
$d = \frac{18*532*10^{-9}*\sqrt{(0.87)^2+(6.72)^2}}{0.87}$
$d = 0.0000746$

so using this formula I get 74.6microns.

Now I will take the average of these values I get 74.3 microns!!

For more accuracy, the lasers intensity should be high, distance between the hair and wall should be long, and the wavelength of the laser should be low.

violet lasers are best!

Monday 23 June 2014

deBroglie Relations and the Scale of Quantum Effects, Light waves as particles (a)

(a) Light Waves as Particles:
The Photoelectric effect suggests that light of frequency ν can be regarded as
consisting of photons of energy $E=h v$, where $h = 6.63*10^{-27}erg*s$

i. green light has a wavelength in the range of 532nm. What are the
energy and frequency of a photon of green light?

Ans:
First to find the frequency we can use the equation: $c=v \lambda$
putting in the values:
$299,792,458 = v*532$

$v = \frac{299,792,458}{532*\frac{1}{1*10^9}}$ mid calculation unit conversion from nm to m

$v =  5.45077*10^{14}Hz$

so the frequency is $545.077 THz$

Now we can use the above equation of $E=h v$ to find the energy of the photon
so:
$E = 6.63*10^{-27}*545.077*10^{14}$

$E = 3.614*10^{-10} erg$

AKA $36.14\text{ nano ergons}$


Wednesday 23 April 2014

Something on G-waves p2

Orbital Decay from Gravitational Radiation:

(Note: Remember those folds forming in the graph in Part one? Well those showed continuity...No strange effect in that...)

In part 2 we will use Power Radiated, which we found in Part one to find the rate of decrease of Orbit Radius with respect to time. In other words we will find $\frac{dr}{dt}$, the derivative or rate of change.
The formula for finding this stuff is \frac{\mathrm{d}r}{\mathrm{d}t} = - \frac{64}{5}\, \frac{G^3}{c^5}\, \frac{(m_1m_2)(m_1+m_2)}{r^3}\ .
The orbit decays at a rate proportional to the inverse third power of the radius. 
The variables are same as in part 1.
Well Im going to take $r=1.5*10^{11}m$ and $m2=6*10^{24}kg$ and $m1$ is our variable from $2*10^{29} \text{  to  } 3*10^{30}kg $:
A similar graph like in part 1. The unit for orbital decay is m/s. Earth with our sun is $1.1*10^{-20}m/s$ . 

Now lets take our 2 solar mass Neutron stars and plot their orbit decay with Radius as our variable:
So as the orbit radius decreases the decay increases exponentially.
Now for 3D:
We are going to only take our Neutron star case as graphs are similar.
So with relation to $m1,m2$ the graph is:
and for $m1,r$:





Sunday 30 March 2014

Something on Gravitational waves p1

Power Radiated by orbiting bodies:

As two or more bodies orbit each other, some amount of energy is released from the system via gravitational waves. This in turn causes an inspiral or decrease in orbit. So after losing all energy the two objects fall in too each other! This same is happening with the earth and sun system at a rate of 200W (I didnt bother to calculate this as it was done on Wikipedia ;) ) but as per wiki it will take 10^13 times the current age of universe for the earth to fall in and before that ever happens the sun will eat us in red giant phase(or will it...).

So in the formula of radiated power via G-waves we take two masses orbiting around each other $m_1,m_2$. For simplicity we can take standard keplarian orbits in x-y plane or in 2-D plane.
This is our formula: P = \frac{\mathrm{d}E}{\mathrm{d}t} = - \frac{32}{5}\, \frac{G^4}{c^5}\, \frac{(m_1m_2)^2 (m_1+m_2)}{r^5}
where $\frac{dE}{dt}$ is the derivative or rate of change of energy with respect to $t$.
$G$ is our gravitational constant and $G=6.673*10^{-11}m^3kg^{-1}s^{-2}$
$c$ is speed of light and $c=300000000m/s$
$r$ is the separation between the two orbiting bodies

(note: The answers will be minus because it is power lost!)

First of all I am going to make a 2-D graph in the x-y plane where y-axis will be $P$ and the x-axis will be $m_1$, we will take $r$ as to be constant, so we can take $r=1.5*10^11meters$. $m_2$ will also be constant and we can take that as earths mass so $m_2=6*10^{24}kg$

In this graph with mass $m_1$ ranging from $2*10^{30}kg$ to $3*10^{30}kg$ we can see that at mass $m_1=2*10^{30}kg$ and the contstant mass $m_2= 6*10^{24}kg$ or earths mass, the power lost by G-waves is 200W!!

If you notice that as $m_1$ increases with all other stuff constant then the power lost is increasing logarithamically as you can see in this graph:
you can see that at ten times the mass of sun how much the energy release by G-waves increase to 600,000W! Still not so much!
If we take two neutron stars with mass of sun at a distance of 2*10^8m away then see the graph as r increases or decreases.
This is a graph of Power radiated with respect to distance between objects. The mass is 1 Solar mass  for both objects and they are neutron stars. at $2*10^8m$ is about $10^{30}W!!!$ that means in a few days or months the stars would collide!
The graph might look flat or constant near 10^30W or as r increase but dont be fooled by that and look at the Power axis and see the scale ratio. the graph looks like this over their.
again dont be fooled that the power is gone positive, keep your eyes open and look at the axis,
it shows it in this way for understanding. 
And the graph will never go positive unless mass is negative!! which cant happen. 
TIME FOR MULTIVARIABLE GRAPH:

Now we are going to see the graph with respect to two variables. First with $m_1$ and $m_2$;
hmmm, not such an interesting graph but it has a lot of info in it.
The vertical axis is the Power, the rest two axis are the masses of the two bodies, As you can see that as  $m_1$ which is the top horizontal axis increases along with $m_2$ increasing the power radiated is exponentially increasing and then it suddenly does a strange thing and becomes a fold!
again, not to be fooled it looks like this at that fold:
But again note that these graphs axis ratios are not the same, so the graph looks different a bit.
Now we shall look at a 3D graph of our neutron star case. here the x and z axis will have equal rates of change( or axis ratio ). taking again with respect to $m_1$ and $m_2$:

basically its the same thing :(

Now taking the graph with respect to $m_1$ and $r$:
BTW: no this is not a waterfall model but it can be (Moral: different formulas make different real world models!)
Now that is interesting :)! This time I added some colour to show the heights in it.
As you can see another fold in these graphs too!! In the colour coded graph we can clearly see grey for the fold, so ill call it a well!(The water one)
The graph will be similar for neutron star case.

THIS WAS PART 1









, so $$  
. we will take this as simple keplarian orbit in x-y or 2-D space.

Sunday 2 March 2014

History repeats itself, my speech

Think about it, “Does history repeat itself?”                                   
History most certainly does repeat itself, or you can say, “It rhymes with the future. “. It's all based on power & money.  
You will notice when you Look at History. Germany was depressed financially during the 1930's. It took but one man (Hitler) to have millions of people look the other way while millions were murdered. Same for the killing of many scientists, they all were killed for opposing the laws of the church, people were in a way enraged against them in the blame that they were opposing religion, religion was used in history again and again by the people who used it for their own vested interests.
People will never learn, they will always be hostage to their extremist ideas they will always oppose science and technology in the name of religion.
I always ponder and feel so sad when I read the history. Pythagoras, a great mathematician of Greece, who gave us the well known Pythagorean Theorem. His ideas were rejected by extremists in Greece who later manipulated people, which led to his killing by an angry mob, a great mind wasted. But this didn’t stop and over the years history kept repeating itself. Centuries later Boltzmann suffered the same fate as his theories and ideas were ridiculed and mocked so much that he went into extreme depression which later forced him to take his own life.
It was after his death did people realize that his theories were correct, yet another waste of a brilliant mind.
History keeps repeating itself since it is human nature. Humans will forever make mistakes, out of jealousy, out of intolerance and lack of knowledge.  We just can't help ourselves.
Greed will always poison man’s soul, as it did it in the past, we will always be locked up in the abyss of hate. We have acquired technology and knowledge over the years but it did it not change us, instead it made us more cynical and unkind.
 Wars will always happen because not everyone is the same, and people have their own opinions on things. I find it funny that people think history doesn't repeat. Why is it that no matter how many times we're told not to do something, and we do it anyways? Because it's human nature.
I do not have the power to single-handedly change history. But I will be a public voice saying “No!” when I see these sorts of things transpiring, when I see signs of history repeating itself. And thanks to today's technology, each voice can be heard not only in its own location, but around the world.
Do not be among those who stand quietly by. Let your voice be heard.


Sunday 26 January 2014

Pencillium digitatum on oranges.


This is a microscopic pic of the mold that grows on oranges, it is from the genus pencillium but its not that pencillium which Alexander Fleming found killing staphyloccocus bacteria. this is pencillium digitatum! This fungi feeds on citric acid in some fruits.
It starts out as white dots and then becomes greenish blue. I had an orange will some greenish mold on it and I Put it on my roof under the shade next to the water tank. after a week I saw it and found out that the greenish mold had stayed the same, the orange was softer and some white mold started growing individually on other areas, today I saw it with some greeny spots in the middle.
The green one above is the previous mold and the white is the new one, notice small green patches.
I dont know why but whenI lifted the orange the it tore away the peel so this hole was made. And the black stuff is dirt.
BTW this pic was taken a few hours ago and now I see it and its darker green. you can rub this on some agar which I told how to make in the previous post, and you will get some neat colonies of the fungus.
My conditions at the place where the orange is placed is dark, damp(because of the water tank next to it(the black wall in the pic) has a small leak in it, you can see some wet ground in the pic.), cold.
The fate of the orange: Its very soft now and mushy from inside, when I squeezed the orange very slightly, small drops of orange juice fell out from hole, you can see it in the pick, the wet region around the torn skin.



Saturday 25 January 2014

How to make agar

Making agar to grow your cultures is like making jelly!!
the ingredients needed are:
1.a vegetable or chicken stock cube
2. 1/2 teaspoon of sugar
3. some gelatin
4. 250ml of water
That's about it!

instructions:
1. Wash your hands to remove any contaminations.

2. Take a deep pan, put it on the stove and turn up the heat on pan. pour all ingredients in it and stir well.
    let it boil for a few minutes until no lumps are left and then simmer for 30 mins.

3. pour the mixture in your petri dish or petri dish like think and cool in a fridge to let the liquid become jelly.

4. Keep the dishes upside down to stop moisture forming on it and store for later use.

So literally I told you a chicken or vegetable jelly recipee. Enjoy the jelly or grow stuff on it. 

Synthesis of chloramine and hydrazine.[TOXIC, HAZARD, NEVER DO IT]

This experiment is based on producing hydrazine( N2H4 ) in liquid form and producing chloramine( NH2Cl ) in gaseous form.
BTW this experiment is dangerous, I havent done it.

First we gather the equipment!
we will need
1. sodium hypochlorite( NaClO )
2. Ammonia solution( NH3 )
3. A flask
4. two stoppers with one pipe through it
5. a connecting pipe
6. a test tube
Now we set up our equipment:
The reaction will occur in the flask, the hydrazine will stay in the flask while the vaporous chloramine will go in the test tube due to the pressure created.
Now we do the reaction, thought the hydrazine will be polluted with many other chemicals and the chloramine will be with water vapour.