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  4. Does a fissionable nucleus weigh less than the sum of its components?
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Does a fissionable nucleus weigh less than the sum of its components?

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Offline Petrochemicals (OP)

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Does a fissionable nucleus weigh less than the sum of its components?
« on: 21/09/2017 01:55:39 »
E=MC2 is the mass sof something is equable to the energy contained within. Apparently this has nothing at all to do with fission as you have to add energy to remove neutrons protons etc. Thus the end products weigh more than when they where combined.

Unstable elements you gain energy from the splitting of them, so how does it work ?, I know you gain more energy out than you put in, but how ? Are some of the neutron/protons turned into quarks ? I should have thought this should have released far more energy through the chromodynamic mass than any nuclear reaction ? I do know the theorys of fission, neutron strikes an atom splits it and releases more neutrons in a chain reaction. I have read about fissile an fissionable materials, but i cannot find the clarity, the two theorys, E=mc2 and fission, seem opposed

[MOD EDIT - Title altered to turn it into a slightly clearer question. Please feel free to revise this if you are not happy with it.]
« Last Edit: 24/09/2017 15:35:33 by chris »
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Offline Kryptid

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Re: Does a fissionable nucleus weigh less than the sum of its components?
« Reply #1 on: 21/09/2017 05:52:46 »
The mass of an atomic nucleus is not merely the sum of protons and neutrons that are contained within it. The fact that the protons and neutrons are in a bound state means that they are in a lower potential energy state than when they are free particles. When you add in E=mc2, lower potential energy also means lower overall mass.

The mass of a uranium-235 atomic nucleus, for example, is 235.0439299 atomic mass units. The mass of a free proton is 1.007276466879 amu and the mass of a free neutron is 1.00866491588 amu. If you were to multiply a proton's mass by 92 (the number of protons in 235U) and a neutron's mass by 143 (the number of neutrons in the same isotope) and add them together, you'd get 236.908517923708 amu. This is higher than the actual mass if 235U by 1.864588023708 amu. This "mass defect" is the reason that nuclear reactions liberate usable energy.

Let's go with the actual reaction of a neutron and uranium-235 nucleus:

n + 235U → 92Kr + 141Ba + 3n

Now we calculate the mass change using (reactant mass) - (product mass):

(1.00866491588 amu + 235.0439299 amu) - (91.926156 amu + 140.914411 amu + 3.02599474764 amu)

= (236.05259481588 amu) - (235.86656174764 amu)

= 0.18603306824 amu

The products weigh less than the reactants. The difference in mass results in the changing of a small amount of potential energy held inside the reactants into the kinetic energy of the products (including not only the new elements and neutrons, but also gamma rays). Although it's true that the mass of a fast neutron is greater than that of its rest mass (which I used for the calculations), the principle is ultimately the same.
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Re: Does a fissionable nucleus weigh less than the sum of its components?
« Reply #2 on: 21/09/2017 13:05:34 »
Thanks kryptid .Could you clarify it a bit please, as by the calculations you did add up but it doesnt get me any closer to why. The mass defect , lower potential energy means lower overall mass, so that must mean that the split atom weighs more ? I know the sums check out, but i was just unable to find why.

The products usually weigh more than the original atom due to nuclear binding energy, how does it differ in fissile materials. The constituents of a carbon 12 weigh more than the carbon twelve atom, so removing a neutron means you have to add energy.

Fissionable materials like uranium 238 will not under go splitting and have to undergo processes to be turned into fissile materials,  so how does  fission material uranium 235 end up weighing less than its constituent parts ? Uranium I know is unstable and the atom decays naturally, qnd this obviously has something to do with it, but I would have thought a 238 was more ready to loose a neutron  than the 235 but apparently not.

« Last Edit: 21/09/2017 13:07:42 by Petrochemicals »
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Marked as best answer by Petrochemicals on 24/09/2017 21:41:45

Offline Kryptid

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Re: Does a fissionable nucleus weigh less than the sum of its components?
« Reply #3 on: 21/09/2017 21:56:33 »
Quote from: Petrochemicals on 21/09/2017 13:05:34
Thanks kryptid .Could you clarify it a bit please, as by the calculations you did add up but it doesnt get me any closer to why. The mass defect , lower potential energy means lower overall mass, so that must mean that the split atom weighs more ? I know the sums check out, but i was just unable to find why.

Yes, a uranium-235 nucleus (plus a neutron) weighs more than the resulting krypton-92 and barium-141 nuclei (plus 3 neutrons) added together. Different isotopes have different stability, which means different levels of binding energy per nucleon.

Quote
The products usually weigh more than the original atom due to nuclear binding energy

This is only true if the isotopes you are trying to break apart are very lightweight.

Quote
how does it differ in fissile materials.

The interesting thing about that is that you don't always get a net energy release by splitting an atom. It is highly dependent upon the isotope being used. As a general rule of thumb, lightweight nuclei (those that weigh less than iron) will release energy when they are fused together and absorb energy when they are broken apart. The opposite is true of isotopes that are heavier than iron. This is due to differences in the nuclear binding energy per nucleon. Iron-56 has the lowest mass per nucleon of all known isotopes, whereas nickel-62 has the highest binding energy of any known isotope. Fissile materials like uranium-235 are much, much heavier than iron or nickel isotopes, and as such can release energy when they are broken apart. What makes uranium-235 fissile is its ability to sustain a chain reaction by releasing neutrons of the right energy when it undergoes fission. Not all isotopes of uranium can do that. Uranium-238 can undergo fission if it is struck by a fast neutron, but it does not release more fast neutrons in the process to continue the reaction: any neutrons released are so slow that they are very unlikely to be absorbed by another nucleus.

Quote
The constituents of a carbon 12 weigh more than the carbon twelve atom, so removing a neutron means you have to add energy.

The reason is because the neutron is strongly bound within the nucleus, so some minimum activation energy is needed to liberate the neutron from the strong nuclear force's grip. In the case of removing a neutron from carbon-12 to create carbon-11:

12C → 11C + n

(12 amu) - (11.0114336 amu + 1.00866491588 amu)

= 12 amu - 12.02009851588 amu)

= - 0.02009851588 amu

In the case of this reaction, the products are less stable than the reactant. That's why this reaction requires net input energy while the fission of uranium-235 has a net release of energy instead.

Quote
Fissionable materials like uranium 238 will not under go splitting and have to undergo processes to be turned into fissile materials,  so how does  fission material uranium 235 end up weighing less than its constituent parts ? Uranium I know is unstable and the atom decays naturally, qnd this obviously has something to do with it, but I would have thought a 238 was more ready to loose a neutron  than the 235 but apparently not.

As I've already stated, uranium-235 weighs less than its constituent particles because nuclear binding energy makes it more stable than all of those nucleons would be if they were free. It's actually rather like the difference between a rock being held in your hand and that same rock on the ground. The rock on the ground is in a lower gravitational potentially energy state than when it is in your hand. The rock in your hand actually weighs a tiny bit more than when it is on the ground because it's in a higher potential energy state. Bizarre, but true.
« Last Edit: 21/09/2017 21:58:53 by Kryptid »
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Offline evan_au

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Re: Does a fissionable nucleus weigh less than the sum of its components?
« Reply #4 on: 21/09/2017 22:09:07 »
(Oops - overlap with Kryptid's reply!)
Quote from: Petrochemicals
The products usually weigh more than the original atom due to nuclear binding energy
For a nuclear fission power station, the mass of the reactants immediately before and the mass of the products immediately after the reaction is exactly the same. (The same is true for a nuclear fusion power station - if we can ever get it to work!)

However, the products of the nuclear reaction (subatomic particles and atomic nuclei) are traveling apart at speeds close to the speed of light, and so part of that mass is due to relativistic effects (mass appears greater when a particle is moving at a high velocity compared to you).

If you slow down the products until they are almost stationary in your lab, you then measure the "rest mass" of the particles, which is always less than than the "relativistic mass". So there is then some "missing mass", which can be used to generate power.

It is slowing down these high-speed particles from their initial relativistic speed to almost room temperature that delivers heat energy into a coolant (eg water), which can be used to turn turbines and generate electricity.

Iron-56 has the highest binding energy per nucleon. You can't release energy by fusing or fissioning Iron-56.
- When Uranium-238 undergoes nuclear fission, the products are "closer" to Iron-56, so energy is released.
- When Hydrogen undergoes nuclear fusion, the products are "closer" to Iron-56, so energy is released.

See: https://en.wikipedia.org/wiki/Nuclear_binding_energy#Nuclear_binding_energy_curve

PS: Can you give some of the (evidently) numerous examples where "The products usually weigh more than the original atom"?
Note: in power stations, we are talking about nuclear reactions releasing energy. It is believed that some of the nuclear reaction steps immediately before a supernova explosion in old stars actually absorb energy; fortunately, these examples are not common on Earth. I assume you are talking about reactions that release energy?
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Offline Petrochemicals (OP)

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Re: Does a fissionable nucleus weigh less than the sum of its components?
« Reply #5 on: 22/09/2017 05:24:36 »
Quote from: Kryptid on 21/09/2017 21:56:33

Yes, a uranium-235 nucleus (plus a neutron) weighs more than the resulting krypton-92 and barium-141 nuclei (plus 3 neutrons) added together. Different isotopes have different stability, which means different levels of binding energy per nucleon.

actually i meant  after the split .
Quote from: Kryptid on 21/09/2017 21:56:33

This is only true if the isotopes you are trying to break apart are very lightweight.
yep thats what i meant
Quote from: Kryptid on 21/09/2017 21:56:33

The interesting thing about that is that you don't always get a net energy release by splitting an atom. It is highly dependent upon the isotope being used. As a general rule of thumb, lightweight nuclei (those that weigh less than iron) will release energy when they are fused together and absorb energy when they are broken apart. The opposite is true of isotopes that are heavier than iron. This is due to differences in the nuclear binding energy per nucleon. Iron-56 has the lowest mass per nucleon of all known isotopes, whereas nickel-62 has the highest binding energy of any known isotope. Fissile materials like uranium-235 are much, much heavier than iron or nickel isotopes, and as such can release energy when they are broken apart. What makes uranium-235 fissile is its ability to sustain a chain reaction by releasing neutrons of the right energy when it undergoes fission. Not all isotopes of uranium can do that. Uranium-238 can undergo fission if it is struck by a fast neutron, but it does not release more fast neutrons in the process to continue the reaction: any neutrons released are so slow that they are very unlikely to be absorbed by another nucleus.

thats more like it
Quote from: Kryptid on 21/09/2017 21:56:33

The reason is because the neutron is strongly bound within the nucleus, so some minimum activation energy is needed to liberate the neutron from the strong nuclear force's grip. In the case of removing a neutron from carbon-12 to create carbon-11:

12C → 11C + n

(12 amu) - (11.0114336 amu + 1.00866491588 amu)

= 12 amu - 12.02009851588 amu)

= - 0.02009851588 amu

In the case of this reaction, the products are less stable than the reactant. That's why this reaction requires net input energy while the fission of uranium-235 has a net release of energy instead.


i see
Quote from: Kryptid on 21/09/2017 21:56:33

As I've already stated, uranium-235 weighs less than its constituent particles because nuclear binding energy makes it more stable than all of those nucleons would be if they were free. It's actually rather like the difference between a rock being held in your hand and that same rock on the ground. The rock on the ground is in a lower gravitational potentially energy state than when it is in your hand. The rock in your hand actually weighs a tiny bit more than when it is on the ground because it's in a higher potential energy state. Bizarre, but true.
bizarre indeed but thats what puzzled me and i couldnt find a thing on how or why, just lots and lots sbout a neutron to split an atom.
Quote from: evan_au on 21/09/2017 22:09:07
(Oops - overlap with Kryptid's reply!)
Quote from: Petrochemicals
The products usually weigh more than the original atom due to nuclear binding energy
For a nuclear fission power station, the mass of the reactants immediately before and the mass of the products immediately after the reaction is exactly the same. (The same is true for a nuclear fusion power station - if we can ever get it to work!)

Iron-56 has the highest binding energy per nucleon. You can't release energy by fusing or fissioning Iron-56.
- When Uranium-238 undergoes nuclear fission, the products are "closer" to Iron-56, so energy is released.
- When Hydrogen undergoes nuclear fusion, the products are "closer" to Iron-56, so energy is released.

See: https://en.wikipedia.org/wiki/Nuclear_binding_energy#Nuclear_binding_energy_curve

PS: Can you give some of the (evidently) numerous examples where "The products usually weigh more than the original atom"?
Note: in power stations, we are talking about nuclear reactions releasing energy. It is believed that some of the nuclear reaction steps immediately before a supernova explosion in old stars actually absorb energy; fortunately, these examples are not common on Earth. I assume you are talking about reactions that release energy?
The carbon in Kryptids reply or standard mass defect, and no i was not on about reactions of any particular energy direction, i was just asking about the fact thqt some elemets do require energy to be put in, and yet some like uranium release it. I did try reading the wiki page but couldnt find the part about this.
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Offline Toffo

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Re: Does a fissionable nucleus weigh less than the sum of its components?
« Reply #6 on: 24/09/2017 13:52:46 »
Quote from: Petrochemicals on 21/09/2017 01:55:39
E=MC2 is the mass sof something is equable to the energy contained within. Apparently this has nothing at all to do with fission as you have to add energy to remove neutrons protons etc. Thus the end products weigh more than when they where combined.

Unstable elements you gain energy from the splitting of them, so how does it work ?, I know you gain more energy out than you put in, but how ? Are some of the neutron/protons turned into quarks ? I should have thought this should have released far more energy through the chromodynamic mass than any nuclear reaction ? I do know the theorys of fission, neutron strikes an atom splits it and releases more neutrons in a chain reaction. I have read about fissile an fissionable materials, but i cannot find the clarity, the two theorys, E=mc2 and fission, seem opposed

Let's consider a stick of dynamite. It contains some negative binding energy. We can use the dynamite stick to neutralize some of the positive binding energy of a rock. The rock breaks to small rocks as we do that. The small rocks have a larger mass than the big rock. We decreased the binding energy of the rock, the mass of the rock increased, so binding energy has negative mass.

If binding energy has negative mass, then negative binding energy has positive mass. If a stick of dynamite has a positive mass, that means it has more negative binding energy than binding energy.

Dynamite sticks actually have a positive mass. So they have more negative binding energy than binding energy, so the net binding energy of a dynamite stick is negative, that's why it has positive mass.




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