Neutron energy for nuclear fission?

[jumpjack]

How much energy does a neutron need to be able to split an atom and start nuclear fission?

 

[QuantumPion]

How much energy does a neutron need to be able to split an atom and start nuclear fission?

It depends on the nuclide. Fissile isotopes (e.g. U-235, Pu-239) fission easily with thermal neutrons (1 eV and lower), and the probability of fission increases with decreasing neutron energy (1/v). They can also fission with higher energy neutrons but with lower probability. There are some isotopes that have an energy cutoff where fission can only occur with a neutron of sufficient energy. Th-232 is an example, only neutrons greater than 5 keV can cause fission.

 

[jumpjack]

Could an electron ever have enough energy to be able to impact a nuclueus?

 

[QuantumPion]

Could an electron ever have enough energy to be able to impact a nuclueus?

An electron can be absorbed by some isotopes in a decay process called http://en.wikipedia.org/wiki/Electron_capture" . In this case the energy of the electron does not matter.

In order for an electron to collide with a free proton to form a neutron, the electron would need enough energy to overcome the mass difference between the two:

E = (mn - mp - me) * c^2
(1.674927e-27 kg - 1.672621e-27 kg - 9.109382e-31 kg)*(2.99e8 m/s)^2 = 778 keV.

Although in order for this reaction to work, a neutrino has to also be involved to conserve angular momentum. Therefore I don't think this reaction can normally take place, except in perhaps the core of a star or supernova.

The energy required for the electron to be absorbed by a different nucleus, say a bare Fe-56 nucleus, would be the mass difference between it and its product after the collision, Mn-56 (which I calculated to be 3.2 MeV in this case using the chart of the nuclides). Again, angular momentum still has to be conserved.

 

[jumpjack]

So, the final question is:
could electron receive this quantity of energy in z-machine?
http://en.wikipedia.org/wiki/Z_machine

They talk about tow BILLion °C ! How does it "convert" to MeV? Could 20.000.000 ampere give an electron enough energy to get it absorbed by a nucleus?

 

[Astronuc]

The thermal neutrons in a light water reactor have an average kinetic energy equivalent to 0.05 eV or about 300°C.

A temperature of 2 billion K is a paltry 172.35 keV.

2 MA is just current. One has to look at the potential or kinetic energy of the electrons.

 

[jumpjack]

2 MA is just current. One has to look at the potential or kinetic energy of the electrons.

I know. But I don't know how to calculate how much kinetic energy 20 MA give to an electron.

 

[QuantumPion]

I know. But I don't know how to calculate how much kinetic energy 20 MA give to an electron.

It's very simple. If you want to give an electron 1 eV of energy, you need a potential difference of 1 V. That's where the name http://en.wikipedia.org/wiki/Electronvolt" comes from.

 

[jumpjack]

It's very simple. If you want to give an electron 1 eV of energy, you need a potential difference of 1 V. That's where the name http://en.wikipedia.org/wiki/Electronvolt" comes from.

I think this does not answer at all my question:
how many eV do 20 mln Ampere give to an alectron?

 

[Astronuc]

I think this does not answer at all my question:
how many eV do 20 mln Ampere give to an alectron?

Ampere is a measure of current, which the number of charges passing through a given area per unit time. Current does not indicate anything about energy. One would need to know the potential or velocity of the charges.

 

[jumpjack]

Ampere is a measure of current, which the number of charges passing through a given area per unit time. Current does not indicate anything about energy.

Indeed I didn't say "current is energy", but "current gives energy" (sorry, don't know the proper english term).
I mean, as long an electron has a mass, "pulling" it will cause its kinetic energy to raise, and electricity pulls electron, and 20 MA are a lot of electricity, hence 20 MA should cause kinetic energy of electrons to raise very much.

Question is: is such energy high enough to allow an electron penetrating electrons cloud of other atoms, reaching the nucleus and splitting it?

If not, what is causing extra-energy to be produced in z-machine? Has it been clarified/explained/understood?

 

[Bob S]

Could an electron ever have enough energy to be able to impact a nuclueus?

Electrons under 5 MeV (5 million volts) can knock neutrons out of several elements, such as deuterium and beryllium. The electron-in, neutron-out energy threshold is essentially the same as the gamma,neutron reaction, except there is an extra vertex in the electron,neutron interaction, which reduces the yield (cross section).

Bob S

 

[jumpjack]

Electrons under 5 MeV (5 million volts) can knock neutrons out of several elements, such as deuterium and beryllium. The electron-in, neutron-out energy threshold is essentially the same as the gamma,neutron reaction, except there is an extra vertex in the electron,neutron interaction, which reduces the yield (cross section).

Bob S

Thanks.
Is it possible to calculate if 20 MAmpere can put into an electron such an energy? (in form of kinetic energy)

 

[Bob S]

Thanks.
Is it possible to calculate if 20 MAmpere can put into an electron such an energy? (in form of kinetic energy)

Because current is not volts, I will answer your question this way.

Two (2) amps into an automobile ignition system will create a 30,000 volt spark (30 keV electrons).

Bob S

 

[jumpjack]

Because current is not volts, I will answer your question this way.

Two (2) amps into an automobile ignition system will create a 30,000 volt spark (30 keV electrons).

Bob S

Does this mean that 20,000,000 amps would create a 300,000,000,000 (300 GeV) "spark"?!?

This would be 60,000 times the energy needed to "knock neutrons out of several elements".

 

[jumpjack]

BTW, I think "current is no volts" matches with "force is not speed".
But the former "causes" the latter.

 

[QuantumPion]

I think this does not answer at all my question:
how many eV do 20 mln Ampere give to an alectron?

The question is meaningless because amperes is a measure of current, not voltage. It's like asking how many miles per hour do you need to push a rock up a hill.

BTW, I think "current is no volts" matches with "force is not speed".
But the former "causes" the latter.

You are backwards. Voltage is the force that drives a current.

 

[jumpjack]

The question is meaningless because amperes is a measure of current, not voltage. It's like asking how many miles per hour do you need to push a rock up a hill.



You are backwards. Voltage is the force that drives a current.

Aren't you confusing (oer am I?) V with eV? The first is voltage, the second is energy.
An electron has a mass, hence it has kinetic energy 0.5 * m * v^2 .
Current moves electrons, as fast as high is the current.
Hence the more high is the current, the more high is the energy of electrons it moves.
Is this correct?

Wikipedia:
1 eV is equal to the amount of kinetic energy gained by a single unbound electron when it accelerates through an electric potential difference of one volt.

Voltage = electrical force that would drive an electric current between two points.

I guess the 20 MA in Zmachine are caused by a huge Voltage, which "drives an electric current", i.e. it causes electrons to move.

Wikipedia on Z-machine:
...the high electrical current vaporizes the wires, which are transformed into a cylindrical plasma curtain...

Wires are vaporized due to extremely high amount of energy which, by joule effect, vaporizes the wire.

Hence, those 20 MA do give electrons a very high energy.
We estimated this energy is around 300 GeV (I don't know how correct this calculation could be).
Are 300 GeV enough for an electron to pass through "electrons cloud" around a nucleus and split the nucleus itself?

 

[QuantumPion]

Hence the more high is the current, the more high is the energy of electrons it moves.
Is this correct?

No. Current has nothing to do with the energy of the electrons. Current is measured in charge over time. You can have a high current carrying relatively low energy or a low current carrying very high energy.

Hence, those 20 MA do give electrons a very high energy.
We estimated this energy is around 300 GeV (I don't know how correct this calculation could be).
Are 300 GeV enough for an electron to pass through "electrons cloud" around a nucleus and split the nucleus itself?

No. Current is a measure of the number of electrons flowing per second. Current does not "give electrons energy", the potential difference due to an electric field (i.e. voltage) does. You can have a current of 1 millionth of a picoampere (about 1 electron per second) and that 1 electron may or may not have enough energy to interact with the nucleus, depending on the voltage accelerating that electron.

I have no idea where you got the relation of 20 MA to 300 GeV. To create 300 GeV electrons (which would be very difficult to do due to pair production) you would need a voltage of 300 GV. If you wanted 20 MA of current at 300 GV you would need 6 billion gigawatts of power. For comparison, the electric generation capacity of the entire world is about 5000 gigawatts.

 

[jumpjack]

No. Current is a measure of the number of electrons flowing per second. Current does not "give electrons energy", the potential difference due to an electric field (i.e. voltage) does.

But as I said, to produce 20 MAmpere you'd need quite an high Voltage in the z-machine.

I have no idea where you got the relation of 20 MA to 300 GeV.

https://www.physicsforums.com/showpost.php?p=2975303&postcount=14" and 15 of this thread itself. But I'm just guessing (post #15) from what I read in post #14.

To create 300 GeV electrons (which would be very difficult to do due to pair production) you would need a voltage of 300 GV. If you wanted 20 MA of current at 300 GV you would need 6 billion gigawatts of power. For comparison, the electric generation capacity of the entire world is about 5000 gigawatts.

So, how many eV do electrons in z-machine have? All I know is that in z-machine 20 MAmpere are produced, and 2.000.000.000°C are reached.

 

[jumpjack]

If you wanted 20 MA of current at 300 GV you would need 6 billion gigawatts of power. For comparison, the electric generation capacity of the entire world is about 5000 gigawatts.

From wikipedia:

Originally designed to supply 50 terawatts of power in one fast pulse, technological advances resulted in an increased output of 290 terawatts, enough to study nuclear fusion. Z releases 80 times the world's electrical power output for about seventy nanoseconds

Your post: 6 billion GW = 6000 TW
Wikipedia: 290 TW

6000/290 =~ 20

Hence rather than 300 Gev, I guesss they would be 300/20 = 15 GeV .

edit:
my mistake, 6 billion GW = 6,000,000 TW
Hence 6,000,000/290 = 20,000
300/20,000 = 0.015 GeV = 15 MeV

 

[QuantumPion]

But as I said, to produce 20 MAmpere you'd need quite an high Voltage in the z-machine.

I'd assume so.

https://www.physicsforums.com/showpost.php?p=2975303&postcount=14" and 15 of this thread itself. But I'm just guessing (post #15) from what I read in post #14.

BobS was making a specious example to demonstrate to you the difference between current and voltage. A sparkplug uses only a small amount of power at low current, but at high voltage to create sparks for ignition.

So, how many eV do electrons in z-machine have? All I know is that in z-machine 20 MAmpere are produced, and 2.000.000.000°C are reached.

You can convert the energy of an average particle in a system using the equation E = 1/2 k T where k is Boltzmann constant and T is in kelvins.

From wikipedia:
Your post: 6 billion GW = 6000 TW
Wikipedia: 290 TW

6000/290 =~ 20

Hence rather than 300 Gev, I guesss they would be 300/20 = 15 GeV .

edit:
my mistake, 6 billion GW = 6,000,000 TW
Hence 6,000,000/290 = 20,000
300/20,000 = 0.015 GeV = 15 MeV

You only get 15 MeV electrons with 15 MV if you are accelerating free electrons in a vacuum tube. I don't know how to calculate the average energy of electrons in the conducting plasma generated by the z-machine but my guess is that it is just proportional to the temperature.