Sunday, January 28, 2018

The Equations For the Neon Snail


The Neon Snail has been tested since before human existence, but I want to apply the patterns in a slightly different way. I want to direct neutrons into a hydrogen cloud but I must figure out, what ratios of neutrons vs. hydrogen atoms and how much energy will be produced.
So the I start with recipes based on the periodic table. The atomic number tells me how many hydrogen atoms to use in the recipes and the atomic mass tells me the total weight. Knowing the weight I can subtract the protons to find out how many neutrons I will need.

For example if I wanted to make gold I could use 79 Hydrogen Atoms and have a total mass 197, like the most stable isotope of gold. So 197-79 = 118 so using experiments the hole size can be adjusted to be just right.

How much energy will be produced by the reaction? This can be calculated using the famous e=mC squared equation. I capitalized C because it's a constant this improves the readability of problems, but be warned grammar in mathematics is subjective and teachers have their own way of wanting things written.

according to google
the speed of light =
299 792 458 m / s

Gold

Mass of original hydrogen atoms = 79*the mass of a hydrogen atom or 1.00794 u ± 0.00001 u according to google or 79.62726.

The mass of the neutrons is 1 u So the total is 79.62726 u+118 u or 197.62726 u so we subtract the mass of the final product 196.96654 u and get .66072 u

e=.66072 u*299 792 458 m / s

1.9807887e+13 or 1.9807887*(10 to the 13th power) atomic mass unit/m/s. Chemists use moles to make conversions easier so instead of using atomic mass units we can make a whole mole of gold and have 1.9807887e+13 g/m/s is system international units even...

That looks like a lot of energy, maybe we should make a batch of Osmium (the 76er) because it is very dense. 192 looks like a stable form. So we have 76 hydrogen atoms to 116 neutrons.

Osmium
Mass of original hydrogen atoms = 76*the mass of a hydrogen atom or 1.00794 u ± 0.00001 u according to google or 76.60344.

The mass of the neutrons is 1 u So the total is 76.60344 u+116 u or 192.60344 u so we subtract the mass of the final product 191.9614807 u and get 0.6419593 u

e=0.6419593 u*299 792 458 m / s

192454556.483 u/m/s or 1.924556.483*(10 to the 8th power.)

Neon
Maybe we want some Neon 20 that is 10 Hydrogen Atoms to 10 neutrons
Mass of original hydrogen atoms = 10*the mass of a hydrogen atom or 1.00794 u ± 0.00001 u according to google or 10.0794

The mass of the neutrons is 1 u So the total is 10.0794 u+10 u or 20.0794 u so we subtract the mass of the final product 19.9924401754 u and get .0869598246 u

e=.0869598246 u*299 792 458 m / s

26069899.5641 u/m/s or 2.660698995641*(10 to the 7th power.)

Isotopes of the Element Gold
https://education.jlab.org/itselemental/iso079.html

Isotopes of the Element Osmium
https://education.jlab.org/itselemental/iso076.html

The atomic mass of gold-197 is 196.96654 u
http://www.chegg.com/homework-help/questions-and-answers/1-atomic-mass-gold-197-19696654-u--much-energy-required-completely-separate-thenucleons-go-q852246

Osmium
https://en.wikipedia.org/wiki/Isotopes_of_osmium

Neon
https://en.wikipedia.org/wiki/Isotopes_of_neon

4 comments:

  1. Anyone see what is wrong with this yet? The equation is e=mC squared. So to get better answers needed to determine how many watts of power may be produced, I need to multiply by the speed of light again. Also a material that is heat resistant enough to with stand the reaction should be chosen based on these calculations. Of course to see what happens in real one would have to do experiments calculus is just a way of reducing waste by choosing things more likely to work.
    http://www.energyadvocate.com/si_basic.htm

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  2. I did a spreadsheet and noticed that http://www.science.co.il/elements/
    did not have exactly the same numbers I used here, but I got answers for all the elements that included c squared at least. Atomic mass units/m^2/s^2 just needs to be converted into Joules now and than I might finally know watt I'm talking about...

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  3. I mean Atomic Mass Units*m^2/s^2...If it's not easy it's provably wrong. OK just the avocado number * 1000 and I should finally have Joules...

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  4. Don't multiply by avogadro's number 6.0221409e+23 divide by it. amu are very small not very big.

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