How to convert C-12 / C-13 to C-1

There's no such thing as C-1. C has 6 protons.

 

Firstly, you won't get the equipment needed to do such nuclear reactions. It's expensive, complicated, dangerous and can be used to make nuclear weapons materials.

Secondly, you can't make Carbon-6, the lack of neutrons means it would be too unstable to exist. I doubt you can get Carbon-10, any decay process would favour Boron instead.

You still don't seem to have realised that if such processes were possible and simple, someone else would have realised it by now. You're showing that you haven't studied nuclear physics at all and are just stumbling about in the dark unfortunately.

 

You're doing a project that involves nuclear fission? And you're doing it without learning that a nucleus is?

I fear this is falling on deaf ears, but your approach to science is completely backwards. You start at the basics. You don't stumble upon them in the process of building a nuclear bomb!

 

Why is it weird? What's weird is expecting a light nucleus to exist with twice as many neutrons as protons. Though the heavy nuclei do it, it's highly unstable for light ones.

 

Jacquesl how about reviewing the basics.

Do you understand how to calculate the Q value for a decay or reaction?

Do you know that you need a positive Q value for a decay to be possible?

For instance calculate the Q value for the hypothetical reaction you are interested in.

12C ---> 6H+6n

It will turn out to be a quite large negative value. Thats the minimum energy you would have to put in for it to happen(ignoring whatever process you want to use).

 

The Q value is the energy difference betwen what you begin with and what you end with. Its very simple.

In the example above you would calculate it as

Q=[M(12C)-6M(H)-6M(n)]*C^2

Where M denotes atomic mass.

For a decay to be possible the Q value has to be positive, if its not positive that particular decay is not energeticaly favorable and you need to supply energy to it. Decays happen because the nuclei wants to progress to the configuration with the lowest energy.

 

After reading this whole thing... I still don't get whata C-1 is? I think I know what he means when he puts C-12 or C-13, but when I see C-1 I am very very confused. I think you need to pick up an introductory physics book and read the intro of nuclear physics. Then buy a nuclear physics book and work through it. I suggest the book by Krane as a nice starting point.

Like everyone else has said. Lets start from the basics. Cause you are not speaking science when you write something like C-1. And that makes it really hard on the rest of us.

 

I dont mean this in a offensive way. But you need to read up one some basic physics.

The Q value has nothing at all to do with electromagnetism. It is the net energy release from the decay. It is just E=mc^2 applied to the decay situation.

C at the end is the speed of light.
By M(12C) I mean the atomic mass of carbon 12. By 6M(H) I mean 6 times the atomic mass of hydrogen.

The energy can be of any unit depending on what units you put into the formula(this hold for every formula anywhere in physics).
Btw Coloumb and Watts are not even units of energy.

This above is mostly just jibberish. There is no logic behind it. You are mixing chemistry with nuclear physics in a way that makes no sense at all.

There is no such thing as C-1. You need to understand the notation.