Recent questions in Nuclear Fission

Nuclear physicsAnswered question

Belen Booker 2022-12-13

Nuclear fission is an exothermic process whereas nuclear fusion is an endothermic process.

energy stored by waves in the ocean

temperature difference at different levels in the ocean

energy stored by waves in the ocean

temperature difference at different levels in the ocean

Nuclear physicsAnswered question

Cesar Mcguire 2022-05-20

What Happens to electrons after Alpha Decay and Nuclear Fission?

Where do the electrons go? In alpha decay do 2 electrons follow the alpha particle and make stable Helium or does the larger daughter nucleus become an anion?

Also what do the electrons do in the mixture of fission and alpha decay? With Beryllium-8 it decays into two alpha particles and 4 lonely electrons or do you get two Helium atoms?If someone could add an entry to Holocron it would be greatly appreciated, thanks.

Where do the electrons go? In alpha decay do 2 electrons follow the alpha particle and make stable Helium or does the larger daughter nucleus become an anion?

Also what do the electrons do in the mixture of fission and alpha decay? With Beryllium-8 it decays into two alpha particles and 4 lonely electrons or do you get two Helium atoms?If someone could add an entry to Holocron it would be greatly appreciated, thanks.

Nuclear physicsAnswered question

lurtzslikgtgjd 2022-05-19

Is there any loss of potential energy during nuclear fission?

We know that energy is created during nuclear fission and there is a loss in mass. But every body possess potential energy even when it is at rest (where height = radius of earth). So during nuclear fission, due to loss in mass, should there be loss in potential energy of the body as P = mgh ?

We know that energy is created during nuclear fission and there is a loss in mass. But every body possess potential energy even when it is at rest (where height = radius of earth). So during nuclear fission, due to loss in mass, should there be loss in potential energy of the body as P = mgh ?

Nuclear physicsAnswered question

Yaritza Oneill 2022-05-19

Nuclear fission equation

Should not the energy conservation physics equation be

$E=-m{c}^{2}\text{instead of}E=m{c}^{2}$

because energy appears simultaneously with mass disappearance in splitting as

$|\mathrm{\Delta}E|=-|\mathrm{\Delta}m|{c}^{2}?$

Also what in Einstein's derivation dispenses with the factor $\frac{1}{2}$ from the usual kinetic energy

$KE=\frac{1}{2}m{v}^{2}?$

Sorry about the elementary question, it remained with me for longtime.

Should not the energy conservation physics equation be

$E=-m{c}^{2}\text{instead of}E=m{c}^{2}$

because energy appears simultaneously with mass disappearance in splitting as

$|\mathrm{\Delta}E|=-|\mathrm{\Delta}m|{c}^{2}?$

Also what in Einstein's derivation dispenses with the factor $\frac{1}{2}$ from the usual kinetic energy

$KE=\frac{1}{2}m{v}^{2}?$

Sorry about the elementary question, it remained with me for longtime.

Nuclear physicsAnswered question

lifretatox8n 2022-05-19

Is fission reaction considered natural or artificial?

As I learned, nuclear fission doesn't occur without the control of a human made nuclear reactor, by hitting a neutron to a fissile isotope. Thus, the fission reaction is considedred as a part of 'ARTIFICIAL REACTIONS' category.

But, I've just noticed in a wikipedia article (Natural nuclear fission reactor), that nuclear chain reactions had occurred on Earth about 2 billion years ago, what proves that fission existed naturally before.

Another additional question I was wondering is: Is there any chance that a natural nuclear chain reaction can re-occur on Earth?

As I learned, nuclear fission doesn't occur without the control of a human made nuclear reactor, by hitting a neutron to a fissile isotope. Thus, the fission reaction is considedred as a part of 'ARTIFICIAL REACTIONS' category.

But, I've just noticed in a wikipedia article (Natural nuclear fission reactor), that nuclear chain reactions had occurred on Earth about 2 billion years ago, what proves that fission existed naturally before.

Another additional question I was wondering is: Is there any chance that a natural nuclear chain reaction can re-occur on Earth?

Nuclear physicsAnswered question

datomerki8a5yj 2022-05-19

In nuclear reactions, the difference between the binding energy of the reactants and that of the products is converted to the thermal energy of the products.

Is the Coulomb force that actually gives kinetic energy to the product particles both in the nuclear fission and in the nuclear fusion?

Is the Coulomb force that actually gives kinetic energy to the product particles both in the nuclear fission and in the nuclear fusion?

Nuclear physicsAnswered question

Jay Barrett 2022-05-19

Calculating energy released in nuclear fission

Consider the neutron induced fission $\text{U-235}+n\to \cdots \to \text{La-139}+\text{Mo-95}+2n$, where $\dots $ denotes intermediate decay steps.

I want to calculate the released energy from this fission. One way would be to calculate the difference of the binding-energies ($B$):

$\mathrm{\Delta}E=B(139,57)+B(95,42)-B(235,92)\approx 202,3\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

(Btw. I didn't use the binding energies from a semi-empirical binding energy formula but calculated them directly via mass defect).

Another way is:

$\mathrm{\Delta}E=(m(\text{U-235})+{m}_{\text{Neutron}}-m(\text{Mo-95})-m(\text{La-139})-2{m}_{\text{Neutron}}){c}^{2}\approx 211,3\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

Which one gives the correct result? Why?

You notice that $57+42\ne 92$, if that was the case, it would be equal, but I don't clearly see where the difference physically comes from and what to add or subtract (and why) from to the first or from the second term to get the other result. How to make this clear?

A slightly other point of view: What different questions do both calculations answer?

Consider the neutron induced fission $\text{U-235}+n\to \cdots \to \text{La-139}+\text{Mo-95}+2n$, where $\dots $ denotes intermediate decay steps.

I want to calculate the released energy from this fission. One way would be to calculate the difference of the binding-energies ($B$):

$\mathrm{\Delta}E=B(139,57)+B(95,42)-B(235,92)\approx 202,3\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

(Btw. I didn't use the binding energies from a semi-empirical binding energy formula but calculated them directly via mass defect).

Another way is:

$\mathrm{\Delta}E=(m(\text{U-235})+{m}_{\text{Neutron}}-m(\text{Mo-95})-m(\text{La-139})-2{m}_{\text{Neutron}}){c}^{2}\approx 211,3\phantom{\rule{thinmathspace}{0ex}}\mathrm{M}\mathrm{e}\mathrm{V}$

Which one gives the correct result? Why?

You notice that $57+42\ne 92$, if that was the case, it would be equal, but I don't clearly see where the difference physically comes from and what to add or subtract (and why) from to the first or from the second term to get the other result. How to make this clear?

A slightly other point of view: What different questions do both calculations answer?

Nuclear physicsAnswered question

Stoyanovahvsbh 2022-05-18

Binding energy in nuclear fission

As far as i know, in a nuclear reaction we "go" from a binding energy ${B}_{1}$ to a binding energy ${B}_{2}$ with ${B}_{2}$>${B}_{1}$ because a bigger binding energy means more stability for the nucleus.

If we consider the nuclear fission $n{+}^{235}U{=}^{141}Ba{+}^{92}Kr+3n$ the binding energies ${B}_{1}$ and ${B}_{2}$ are the sum of all the binding energies of components before and after the reaction?

As far as i know, in a nuclear reaction we "go" from a binding energy ${B}_{1}$ to a binding energy ${B}_{2}$ with ${B}_{2}$>${B}_{1}$ because a bigger binding energy means more stability for the nucleus.

If we consider the nuclear fission $n{+}^{235}U{=}^{141}Ba{+}^{92}Kr+3n$ the binding energies ${B}_{1}$ and ${B}_{2}$ are the sum of all the binding energies of components before and after the reaction?

Nuclear physicsAnswered question

Alisa Durham 2022-05-18

The energie distribution in nuclear fission

When a nuclear fission occurs, the released energy is split between the reaction products:

${K}_{1}=\frac{{m}_{2}}{{m}_{2}+{m}_{1}}E,$

with $E$ the released energy, ${m}_{2}$ the mass of one of the products and ${m}_{1}$ of the other.

When calculating the total energy ${E}_{1}={K}_{1}+{m}_{1}{c}^{2}$, and doing the same for the product of mass ${m}_{2}$, these two energies aren't equal.

The energy released in a fission reaction isn't split equally between the products.

Why is this?

When a nuclear fission occurs, the released energy is split between the reaction products:

${K}_{1}=\frac{{m}_{2}}{{m}_{2}+{m}_{1}}E,$

with $E$ the released energy, ${m}_{2}$ the mass of one of the products and ${m}_{1}$ of the other.

When calculating the total energy ${E}_{1}={K}_{1}+{m}_{1}{c}^{2}$, and doing the same for the product of mass ${m}_{2}$, these two energies aren't equal.

The energy released in a fission reaction isn't split equally between the products.

Why is this?

Nuclear physicsAnswered question

llunallenaipg5r 2022-05-18

Nuclear fission problem

If during a nuclear fission U(235) decomposes into Xe(140), Sr(94) and n(1), how is it possible that the original U(235) has bigger mass than the three resulting nuclei together? Should it not be the case that the U(235) has a deeper bonding energy so the lower mass; this way its binding energy is released during the fission and the resulting elements have smaller mass defect?

If during a nuclear fission U(235) decomposes into Xe(140), Sr(94) and n(1), how is it possible that the original U(235) has bigger mass than the three resulting nuclei together? Should it not be the case that the U(235) has a deeper bonding energy so the lower mass; this way its binding energy is released during the fission and the resulting elements have smaller mass defect?

Nuclear physicsAnswered question

Jaylene Duarte 2022-05-18

Nuclear fission mechanism: neutron capture

Consider this nuclear fission reaction:

${}^{235}\mathrm{U}+{}^{1}\mathrm{n}\to {}^{236}\mathrm{U}\text{(excited)}\to {}^{92}\mathrm{K}\mathrm{r}+{}^{141}\mathrm{B}\mathrm{a}+3{}^{1}\mathrm{n}$

I have not understand why a thermal neutron (${}^{1}$n should be captured by the nuclei of ${}^{235}$U.

What are the conditions under which this neutron is captured?

My reasoning is the following: the energy that the neutron must have to be captured by the ${}^{235}$U should be equal to the ${S}_{\mathrm{n}}$ (separation energy for a neutron) of ${}^{236}$U which I found is 6.34 MeV, higher than the energy of the thermal neutron. So my reasoning must be wrong but I cannot understand where I make mistakes...

I looked on the Povh Rith Particles and Nuclei but it does not explain the reaction well. Could anyone give me a reference or explain here the fission reaction ?

Consider this nuclear fission reaction:

${}^{235}\mathrm{U}+{}^{1}\mathrm{n}\to {}^{236}\mathrm{U}\text{(excited)}\to {}^{92}\mathrm{K}\mathrm{r}+{}^{141}\mathrm{B}\mathrm{a}+3{}^{1}\mathrm{n}$

I have not understand why a thermal neutron (${}^{1}$n should be captured by the nuclei of ${}^{235}$U.

What are the conditions under which this neutron is captured?

My reasoning is the following: the energy that the neutron must have to be captured by the ${}^{235}$U should be equal to the ${S}_{\mathrm{n}}$ (separation energy for a neutron) of ${}^{236}$U which I found is 6.34 MeV, higher than the energy of the thermal neutron. So my reasoning must be wrong but I cannot understand where I make mistakes...

I looked on the Povh Rith Particles and Nuclei but it does not explain the reaction well. Could anyone give me a reference or explain here the fission reaction ?

Nuclear physicsAnswered question

Eve Dunn 2022-05-18

Control of Nuclear Fission

What is it precisely which prevents current technology in nuclear fission from controlling the size of fission products; in other words, why is it not feasible, presuming that such an approach would permit the translation of binding energies to heat in nuclear fission reactions, to create lighter more manageable fission products, thus rendering that fission technology relatively safe? Further, if this constraint is due to the electrical neutrality of the neutron, what prevents the induction of a manageable charge or manipulation of the neutron magnetic dipole moment?

What is it precisely which prevents current technology in nuclear fission from controlling the size of fission products; in other words, why is it not feasible, presuming that such an approach would permit the translation of binding energies to heat in nuclear fission reactions, to create lighter more manageable fission products, thus rendering that fission technology relatively safe? Further, if this constraint is due to the electrical neutrality of the neutron, what prevents the induction of a manageable charge or manipulation of the neutron magnetic dipole moment?

Nuclear physicsAnswered question

rockandriot0odjn 2022-05-15

Nuclear Feedback Loop (Fusion and Fission)

What are the main factors that prevent a feedback loop being created by using a hybrid method of fission and fusion. Fusion building up to fissionable materials, and fission breaking down till fusion is possible. Ex: numbers of neutrons, cross section for interaction, the gap between elements, amount of energy needed, probability of correct species being generated.

Further, would it be possible to at least get a few occurrences of the fission/fusion cycle.

What are the main factors that prevent a feedback loop being created by using a hybrid method of fission and fusion. Fusion building up to fissionable materials, and fission breaking down till fusion is possible. Ex: numbers of neutrons, cross section for interaction, the gap between elements, amount of energy needed, probability of correct species being generated.

Further, would it be possible to at least get a few occurrences of the fission/fusion cycle.

Nuclear physicsAnswered question

rockandriot0odjn 2022-05-14

Calculate kinetic energy of neutrons in nuclear fission

How can I calculate the mean kinetic energy of an emitted neutron in a nuclear fission. Take for example the fission of U-235 to Ba-141 and Kr-92.

A calculation which just shows that the mean energy will be in the range of "fast" neutrons (> 1MeV) would be enough for me.

How can I calculate the mean kinetic energy of an emitted neutron in a nuclear fission. Take for example the fission of U-235 to Ba-141 and Kr-92.

A calculation which just shows that the mean energy will be in the range of "fast" neutrons (> 1MeV) would be enough for me.

Nuclear physicsAnswered question

ureji1c8r1 2022-05-14

Energy released in nuclear fission

In induced fission of U-235, neutrons are bombarded at the U-235, producing U-236. This U-236 then undergoes fission:

U-235 + n --> U-236 --> Ba-141 + Kr-92 + 3n

As far as I understand, the energy released in fission is gained as kinetic energy of the products, and also released as gamma photons/beta particles and neutrinos when the products decay. My confusion lies in calculating the energy released, as I do not think the method in the textbook is correct.

The absorbed neutron loses nuclear potential energy. This causes the binding energy of U-236>U-235, meaning the rest mass of U-236 is less than the rest mass of U-235 + n. This increase in binding energy is then used to deform the nucleus into a double-lobed drop allowing the two fragments to separate due to electrostatic repulsion.

The two fragments formed have a greater binding energy per nucleon than U-236, and hence the binding energy of the fragments is greater than U-236. (In turn causing the mass of the products to decrease). However this increase in binding energy is gained as kinetic energy of the products/released as gamma photons etc.

Mo1 = Mass(U235+n)

Mo2= MassU236

Mo3 = Massfissionproducts

B1=binding energy of U235

B2=binding energy of U236

B3 = binding energy of fission products

The energy released in fission is due to the increase in binding energy B3-B2. The increase in binding energy B2-B1 is used to deform the nucleus - it is not 'released'. Therefore the energy released in the fission process = (Mo2-Mo3)c^2.

However my textbook states that the energy released in the fission process =(Mo1-Mo3)c^2. I don't understand this as they are including the energy to deform the nucleus, (Mo1-Mo2)c^2, when this is not actually released.

Any help is greatly appreciated!

In induced fission of U-235, neutrons are bombarded at the U-235, producing U-236. This U-236 then undergoes fission:

U-235 + n --> U-236 --> Ba-141 + Kr-92 + 3n

As far as I understand, the energy released in fission is gained as kinetic energy of the products, and also released as gamma photons/beta particles and neutrinos when the products decay. My confusion lies in calculating the energy released, as I do not think the method in the textbook is correct.

The absorbed neutron loses nuclear potential energy. This causes the binding energy of U-236>U-235, meaning the rest mass of U-236 is less than the rest mass of U-235 + n. This increase in binding energy is then used to deform the nucleus into a double-lobed drop allowing the two fragments to separate due to electrostatic repulsion.

The two fragments formed have a greater binding energy per nucleon than U-236, and hence the binding energy of the fragments is greater than U-236. (In turn causing the mass of the products to decrease). However this increase in binding energy is gained as kinetic energy of the products/released as gamma photons etc.

Mo1 = Mass(U235+n)

Mo2= MassU236

Mo3 = Massfissionproducts

B1=binding energy of U235

B2=binding energy of U236

B3 = binding energy of fission products

The energy released in fission is due to the increase in binding energy B3-B2. The increase in binding energy B2-B1 is used to deform the nucleus - it is not 'released'. Therefore the energy released in the fission process = (Mo2-Mo3)c^2.

However my textbook states that the energy released in the fission process =(Mo1-Mo3)c^2. I don't understand this as they are including the energy to deform the nucleus, (Mo1-Mo2)c^2, when this is not actually released.

Any help is greatly appreciated!

Nuclear physicsAnswered question

oglasnak9h01 2022-05-14

Do we know what causes the release of energy in nuclear fission?

I was trying to put together all the things I've read in the last couple of days and I realized that based on my current knowledge of the standard model and the way nuclear fission works, I'm not able to understand why would any number of nuclei packed together by the strong nuclear force, no matter how tight, release energy when split.

This took me on an interesting rabbit hole and then I found one answer. Leaving aside the rant, it made me wonder, do we really understand why the splitting releases energy, or we just know it does based on observations.

The way I see things now, since the nuclear forces are so strong, their tendency should be to always bring nuclei together, even if bombarded with an extra neutron. The extra neutron should stick, or if the kinetic force is strong enough, then break the binding and split the nucleus, but certainly not releasing energy, losing mass and shooting 1-3 neutrons in the process.

So, why is my expectation wrong, since experimentally it obviously is.

I was trying to put together all the things I've read in the last couple of days and I realized that based on my current knowledge of the standard model and the way nuclear fission works, I'm not able to understand why would any number of nuclei packed together by the strong nuclear force, no matter how tight, release energy when split.

This took me on an interesting rabbit hole and then I found one answer. Leaving aside the rant, it made me wonder, do we really understand why the splitting releases energy, or we just know it does based on observations.

The way I see things now, since the nuclear forces are so strong, their tendency should be to always bring nuclei together, even if bombarded with an extra neutron. The extra neutron should stick, or if the kinetic force is strong enough, then break the binding and split the nucleus, but certainly not releasing energy, losing mass and shooting 1-3 neutrons in the process.

So, why is my expectation wrong, since experimentally it obviously is.

Nuclear physicsAnswered question

ureji1c8r1 2022-05-13

Protons (as opposed to neutrons) to mediate nuclear fission?

I am just wondering why are protons (as opposed to neutrons) not used to mediate nuclear fission? Is it because it is charged, so we will have to input more unnecessary energy to overcome the Coulomb force?

I am just wondering why are protons (as opposed to neutrons) not used to mediate nuclear fission? Is it because it is charged, so we will have to input more unnecessary energy to overcome the Coulomb force?

Nuclear physicsAnswered question

hard12bb30crg 2022-05-13

Is it possible that nuclear fission contributes to climate change?

This is probably a really stupid question, please forgive me. Is it possible that the use of nuclear fission on earth contributes to the increased energy in the Earth's system as according with the Laws of Thermodynamics regarding the Conservation of Energy?

This is probably a really stupid question, please forgive me. Is it possible that the use of nuclear fission on earth contributes to the increased energy in the Earth's system as according with the Laws of Thermodynamics regarding the Conservation of Energy?

Nuclear physicsAnswered question

britesoulusjhq 2022-05-10

Coulomb potential in nuclear fission?

In the case of symmetrical fission we must consider the Coulomb potential between the two nuclei :

$E=k\frac{{q}^{2}}{r}=k\frac{(\frac{Ze}{2}{)}^{2}}{2{r}_{0}(\frac{a}{2}{)}^{1/3}}=k\frac{{2}^{1/3}}{8}\frac{{Z}^{2}{e}^{2}}{{r}_{0}{A}^{1/3}}\frac{1}{1.6\times {10}^{-13}}=0.157\frac{{Z}^{2}}{A}$

The process is spontaneous only if $Q>E$ (where $Q$ is the difference between final and initial binding energy).

Why is the process allowed only if $Q>E$ ? Don't the nuclei repel each other rendering the process "easier"?

In the case of symmetrical fission we must consider the Coulomb potential between the two nuclei :

$E=k\frac{{q}^{2}}{r}=k\frac{(\frac{Ze}{2}{)}^{2}}{2{r}_{0}(\frac{a}{2}{)}^{1/3}}=k\frac{{2}^{1/3}}{8}\frac{{Z}^{2}{e}^{2}}{{r}_{0}{A}^{1/3}}\frac{1}{1.6\times {10}^{-13}}=0.157\frac{{Z}^{2}}{A}$

The process is spontaneous only if $Q>E$ (where $Q$ is the difference between final and initial binding energy).

Why is the process allowed only if $Q>E$ ? Don't the nuclei repel each other rendering the process "easier"?

Nuclear physicsAnswered question

William Santiago 2022-05-09

Fission producing Cs-137

This is I suppose quite a precise question about Nuclear fission. What produces, aside from U-235, via a fission process, Cs-137?

Does any isotope of Actinium, for example, undergo a fission process and break into Cs-137 and something else?

Is there any way for me to easily figure this out for myself?

Many thanks.

This is I suppose quite a precise question about Nuclear fission. What produces, aside from U-235, via a fission process, Cs-137?

Does any isotope of Actinium, for example, undergo a fission process and break into Cs-137 and something else?

Is there any way for me to easily figure this out for myself?

Many thanks.

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When you need an example of nuclear fission, the best way to learn how things work is to turn to the questions and answers that contain explanations and the practical implementation of the concept. If you’re trying to reverse-engineer some equations, it will work as a helpful approach, yet you should focus on the lab experiments where the heavy nucleus is attacked by the neutrons with all the instability involved. Start with the nuclear fission equation to determine the level of energy produced to see how the variables affect the outcome of what you’re aiming to achieve in the result.