Reply Sat 31 Jul, 2010 05:33 pm
Nuclear Fission:

Helium does not decay into alpha and beta particles like many other atoms such as Uranium 238. Theses alpha particles consist of two protons and two neutrons bound together into a particle identical to the Helium nucleus. Whereas, beta particles are high energy, high speed electrons or positrons. In this study we are utilizing beta particles that are high speed electrons. What would the effect be if Helium is synthetically decayed into alpha and beta particles? According to Einstein's equation E=MC^2 and the mass of Helium, alpha and beta particles, if such a decay occurs, a decrease in mass arises which results in a release of energy. This will release energy and heat. However, after the energy and heat is released, the beta particles will have left the containment and in the containment, alpha particles will bounce around with Helium atoms possibly generating perpetual static electricity on the outside of the containment which should be made of aluminum, that can be taped into AC and DC electricity.

Synthetic decay in Helium will produce one alpha particle and two beta particles. The following are their respective masses times 10^-27kg.
Helium: 6.64647824468[3]
Alpha Particles: 6.644646[1]
Beta Particles: .000911[2]
While the mass of the original Helium atom less one alpha particle and two beta particles, the resulting mass is .00010*10^-27kg. Such a decrease in mass results in a release of energy. Utilizing the E=MC^2, the release of energy is 9.04784*10^-22 joules. When this is converted into electron volts, the release is ..005647213 e V. Thus, it will take 177 atoms decaying to produce 1 electron volt.

We also need to look at the pressure inside the containment during the Helium synthetic decay. With an increase in pressure there would be a risk of the containment leaking. With a decrease in pressure the atoms and alpha particles will be further away from one another, limiting contact and decreasing the static electricity it produces. In this experiment, the pressure should be stable for a long time as the alpha particles will be attracted to the electrons of the Helium atoms. However, after a while if too many helium atoms are decayed, the pressure will eventually decrease as there are fewer atoms contained and subsequently less electrons for the alpha particles to attract and bounce off of.

Alpha particles are bipolar. One side has a positive and the other positive separated by a neutral middle. The positive sides have a maximum attraction to electrons at where the orbital levels are in Helium. Therefore, the bipolar alpha particle will attract the electrons in the surviving Helium atoms and they will(alpha particles) simulate the Helium atom. However, if too many Helium atoms are decayed, there will not be enough electrons to attract and the alpha particles will float to the bottom as it is subject to gravity. Nevertheless, if not too many Helium atoms are decayed alpha particles will simulate Helium atoms. Like Helium, the alpha particles will not be subject to gravity as Helium atoms are not.

What we are trying to generate is positive static electricity, which is the collection of electrically charged particles on the surface of a material. Some materials such as aluminum and lead are great at creating positive static electricity, whereas steel is neutral and gold is great for creating negative static electricity[4]. Lead is better at generating static electricity than aluminum, but it has a low melting point of 621.5 degrees F[5] compared to aluminum which is 1,220.666 degrees F[6]. Therefore due to aluminum being a great positive static electricity generator and a relatively high melting point, it is what should be chosen as the skin of the containment. The reason we want a higher melting point is that during the initial decay of Helium into alpha and beta particles, heat is released. With an aluminum containment shell, the Helium atoms and alpha particles inside will bounce around each other and the skin of the containment, generating positive static electricity on the aluminum.


This experiment involves the containment of pure Helium with a field of 2.5727 Me V of electricity entering inside an aluminum shell. This will release some of the Helium atoms into alpha and beta particles and will release heat. However, the head is temporary. We would not be looking to utilize the heat for energy. What we would be trying to do is create perpetual electricity.

The utilization of the field of at least 2.5727 Me V into the containment would have to be relatively slow so the containment is not melted. In addition, I theorized that there needs to be an amount of Helium always in the containment to keep the spinning of the Helium and alpha particles in the containment. Therefore, the may be a maximum percentage of Helium alpha beta decay compared to non-Helium alpha beta decay that produces the most perpetual static electricity. This point of maximum static electricity will occur when the pressure of the containment start decreasing. This means there are more alpha particles and less helium to have the alpha particles behaving similar to Helium atoms.

With Helium and alpha particles spinning around inside the containment skin of aluminum, static electricity may build up on the outside of the containment(Please not that a skin for the containment can not be steel as it is neutral to static electricity.) The alpha particles and Helium will perpetually spin in the containment, causing the creations of static electricity. This static electricity can be taped into AC and DC electricity to provide new nuclear power that is perpetual with no radio active waste.

[1] http://en.wikipedia.org/wiki/Alpha_particle
[2] http://en.wikipedia.org/wiki/Beta_particle
[3] http://en.wikipedia.org/wiki/Helium
[4] http://www.school-for-champions.com/science/static_materials.htm
[5] http://www.chemicalelements.com/elements/pb.html
[6] http://www.chemicalelements.com/elements/al.html
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Type: Discussion • Score: 4 • Views: 2,408 • Replies: 8
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contrex
 
  1  
Reply Sat 31 Jul, 2010 05:42 pm
Are you aged about 12?
jnthwrig2
 
  1  
Reply Sun 1 Aug, 2010 01:17 pm
@contrex,
I was wondering why this would or wouldn't work and if I neglected something important. Also, I'm interested if anyone would be interested in trying to get something published.
DrewDad
 
  2  
Reply Sun 1 Aug, 2010 01:34 pm
@jnthwrig2,
jnthwrig2 wrote:
What would the effect be if Helium is synthetically decayed into alpha and beta particles?

That's not a decay, that's simply ionizing the Helium.

jnthwrig2 wrote:
According to Einstein's equation E=MC^2 and the mass of Helium, alpha and beta particles, if such a decay occurs, a decrease in mass arises which results in a release of energy.

No. E=MC^2 means that if there is a decrease in mass, then there is a release of energy. It doesn't say anything about the "decay" of a Helium atom. There is no decrease in mass when a Helium atom is ionized, therefore there is no release of energy.
tsarstepan
 
  1  
Reply Sun 1 Aug, 2010 01:43 pm
@jnthwrig2,
jnthwrig2 wrote:

I was wondering why this would or wouldn't work and if I neglected something important. Also, I'm interested if anyone would be interested in trying to get something published.

You're trying to get something published ... a peer reviewed science article per se? And I'm taking at face value that this original research post in the initial post is yours, right?

Where do you expect to publish this? And should others truly consider it a professional work considering it's footnoted to Wikipedia of all sources?

Good luck with this potential Ig Nobel Prize winner....
engineer
 
  2  
Reply Sun 1 Aug, 2010 02:16 pm
@jnthwrig2,
In order to release energy, the products need to be at a lower energy state that the reactants. Ripping the electrons off of a helium atom will not produce two products (alpha and beta particles) at a lower overall energy, it will produce products at a higher energy level. It will take energy to do this, not produce it.
0 Replies
 
jnthwrig2
 
  1  
Reply Sun 1 Aug, 2010 03:37 pm
@tsarstepan,
The Ig Nobel nomination is awesome. Never heard of it before today.
0 Replies
 
jnthwrig2
 
  1  
Reply Thu 5 Aug, 2010 06:42 pm
@DrewDad,
Thank you for the corrects.
jnthwrig2
 
  1  
Reply Wed 11 Aug, 2010 11:26 am
@jnthwrig2,
Does anyone know how I can go about getting an experimental cold fission reactor built?
0 Replies
 
 

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