# Neutron capture and fission reaction

#### dRic2

Gold Member

My professor said that in order to promote a fission reaction you need to provide a certain amount of energy (like in a chemical reaction you need to overcome the activation energy) and the easiest way to do it is to hit the nucleus of the target atom with a neutron. He then gave us this example:

$$^{235}U + n → (^{236}U^*) → F_1 + F_2 + ... + Energy$$

The Uranium-235 is hit by a neutron, then the neutron gets absorbed by the nucleus and the result is the formation of an excited state of the Uranium-236. This excited state has more energy than the minimum energy required to break the nucleus and so the fission reactor may occur ($F_1$ and $F_2$ are the resulting fragments).

My question is the following: "since the energy levels are quantized, when the neutron does not have exactly the same energy corresponding to the difference in energy between $(^{236}U^*)$ and $^{235}U$ how can the reaction take place?"

The explanation, I think, lies in the relation

$$\Delta E \Delta t ≥ \frac {\hbar} 2$$

When the nucleus is hit by the neutron the excited state is formed, thus $\Delta t ≈ 0$, forcing $\Delta E → \infty$ so the excited state can be formed because the uncertainty on energy is very big. As the time passes, $\Delta t$ gets bigger so the uncertainty on $E$ has to get smaller and smaller. When $\Delta E$ is too small, something has to happen and (possibly) the nucleus splits in two. Thinking about the order of magnitude of the energy of the hitting neutron I suppose the expected life for such an excited state is about $10^{-12} s$.

Is it correct up until now ?

This leads me to an other (stupid) question: "what if the energy of the hitting neutron is slightly higher (by an amount $\delta$) than the "activation energy" of the fission reaction?"

Obviously the reaction takes place and an amount of energy $\Delta E$ is released. Then where did $\delta$ go ? is it transformed into kinetic energy of the fragments $F_1$ and $F_2$ or is it emitted as a photon ? Or did I miss something else ?

Thanks
Ric

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#### mathman

For your first question, extra energy goes into kinetic energy of the $^{236}U$.

#### dRic2

Gold Member
For your first question, extra energy goes into kinetic energy of the $^{236}U$.
I thought of that explanation too, but what if the neutron has slightly less energy than the expected values for $^{236}U$ (but still higher than the "activation energy" required) ?

#### anorlunda

Mentor
Gold Member
Think of the kinetic energies of the original neutron, the U235, and each product particle including the U236. Momentum is conserved, mass energy is conserved.

#### dRic2

Gold Member
I'm not following... I'll try make it clear with an example with random numbers.

Suppose the energy of $^{235}U$ is $E_{^{235}U}=1$ and $E_{^{236}U}=2$. Now suppose $E_{att} = 0.8$ and suppose the neutron hitting the nucleus has energy $E_{n} = 0.9$... How can the reaction take place ? It would imply that $E_{^{236}U}=1.9$ but if energy is quantized it has to be $E_{^{236}U}=2$ and thus the reaction can't happen because, although I'm providing enough "activation energy" (I'm giving 0.9, while only 0.8 are necessary), $^{236}U$ can't exist because I do not have enough energy for it (I have to give 1, but I'm giving only 0.9).

#### rpp

I don't think your explanation is correct for fissile atoms. For fissile atoms (like U-235), their is no additional kinetic energy needed to create the excited state. The absorption can take place with very low energy neutrons, and in fact, the reaction is more likely to take place with very low energy neutrons. I believe your explanation is correct for fertile atoms (like U-238), or for other threshold reactions like (n,2n), etc.

For the second part of your question, there are many, many different energy states in a large atom like uranium. Total energy is conserved, but you have to take into account the initial internal state, the final internal state, any change in mass (binding energy), the kinetic energy of the nucleus, and any gamma rays that are released (if any).

#### Astronuc

Staff Emeritus
For the second part of your question, there are many, many different energy states in a large atom like uranium. Total energy is conserved, but you have to take into account the initial internal state, the final internal state, any change in mass (binding energy), the kinetic energy of the nucleus, and any gamma rays that are released (if any).
This is correct. The difference in mass between the U235 + n and U236 allows for fission, or gamma decay (radiative capture), which occurs for some fraction of neutrons captured, and similarly for other fissile nuclides, e.g., Pu-239 and Pu-241.

The OP should consider that a thermal neutron with an energy of 0.025 eV can cause a fission releasing ~200 MeV. Fast neutrons released promptly at fission have most probable energies of ~ 0.9 MeV and the probability of higher neutron energies falls dramatically as a function of energy from 1 to 10 MeV. Fast fissions in U-235 and U-238 are important in reactors using U-based fuel.

Last edited:
• PeterDonis

#### dRic2

Gold Member
Thanks a lot for all the replies, but I'm still a little bit confused. Here's a quote from the book "Introduction to Nuclear Reactor Theory" by Lamarsh

For example, the binding energy of the last neutron in $^{236}U$ is 6.4 MeV, while the critical energy is only 5.3 MeV thus when a neutron with zero kinetic energy is absorbed by $^{235}U$, the compound nucleus $^{236}U$ is produced at bout 1.1 MeV above the critical energy , and fission may immediately occur.
So basically the nucleus of $^{236}U$ is excited 1.1 MeV above the critical energy with a neutron with ZERO kinetic energy. But it is impossible to have zero kinetic energy, so the neutron will have, for example, $k_e$ MeV of kinetic energy. Now $^{236}U$ should be excited 1.1 + $k_e$ MeV above the critical energy, right?
But how can that happen? Since the neutron is free his energy spectrum is continuous ($k_e$ can assume ANY value), but the energy levels inside a nucleus are quantized! This seems a paradox to me...

#### rpp

I think the confusion you are having is that there isn't a "single" critical energy. There are many, many different energy levels in the nucleus, you can find plots of the energy levels on the internet. Many of these are above the "critical energy". However, as you pointed out, they are discrete values. Therefore, you also have to take into account the other mechanisms to transfer energy. The mass will change (binding energy), the nucleus can emit a gamma ray (discrete), the atom will have a final kinetic energy (non-discrete), and finally, it can fission and release all kinds of particles with different kinetic energy (non-discrete).

#### dRic2

Gold Member
Therefore, you also have to take into account the other mechanisms to transfer energy. The mass will change (binding energy), the nucleus can emit a gamma ray (discrete), the atom will have a final kinetic energy (non-discrete)...
Ok, but here you are assuming that the compound nucleus is formed in a "wrong" energy level and then by emitting $y-rays$ it goes to the "right" level and then fission occurs. I don't understand how is even possible that the compound nucleus is created with "wrong" energy... Maybe I misunderstood your answer.

Here's an other passage from the previously cited book: Look at the part I highlighted in green: if the energy levels are discrete values, why is the probability just "smaller" and not ZERO ?

Sorry if I'm bothering too much, but I really can't understand it...

#### Attachments

• 105.8 KB Views: 288

#### rpp

Just referring to the notes that you have underlined in green, the resonance has a "width" because of the thermal motion of the neutron and target atoms. The relative motion between the neutron and target increase the probability that the reaction will give the needed energy, which effectively increases the width of the resonance. The relative motion is kinetic energy, and is a function of the temperature. You can look up "Doppler Broadening" in most textbooks to get a better understanding of how the width of the resonance is related to the thermal temperature.

But note that resonance absorption only occurs in part of the energy spectrum. For U-235, resonance absorption only occurs at energies greater than 1 eV (approximately). See the following cross section plot of U-235:
https://t2.lanl.gov/nis/data/endf/endfvii-n-pdf/u/235.pdf

In LWR's, most fission reactions actually occur at much lower energies (note the log scale on the magnitude of the cross sections).
You do not need a resonance for a U-235 absorption to occur. Resonances greatly increase the probability of absorption around the resonance energy, but absorption can occur outside of a resonance.

#### dRic2

Gold Member
You do not need a resonance for a U-235 absorption to occur. Resonances greatly increase the probability of absorption around the resonance energy, but absorption can occur outside of a resonance.
This is my problem. I don't understand why resonance is not necessary for fission to occur... If the total energy coming from the neutron ($B + E_c$) is not exactly equal to the energy of the excited state how can the reaction occur ? (I think the answer lies in quantum mechanics)

Since english is not my main language I don't know if maybe I'm not understanding your answers or I'm not making myself very clear... I apologize if I'm annoying you #### Astronuc

Staff Emeritus
Code:
https://physics.nist.gov/cgi-bin/cuu/Value?mnu
https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=U

1 amu c2 = 931.49432 MeV

Mass U235     235.0439301   amu
Mass neutron    1.008664916 amu

Mass U235+n   236.052595016 amu
Mass U236     236.0455682   amu

Excess Mass     0.007026816 amu

6.545439 MeV
If one looks at the mass of U235 and the neutron (no kinetic energy involved), the total mass is greater than that of U-236 (not excited). Subtracting the mass of U236, there is an extra 6.54544 MeV, which greater than the binding energy of the last neutron, so the excited nucleus is unstable. Fission usually occurs, but about 15 to 16% of the time, U236 emits a gamma ray and settles to a ground state without fissioning. The binding energy of the last neutron is simply the energy required to expel a neutron, the last neutron from the nucleus. A 6.4 MeV gamma ray could induce a photoneutron reaction. The actual gamma energy required is slightly higher because the U-236 would recoil when absorbing the gamma ray.

#### snorkack

The answer is that there is no requirement that there should be a compound nucleus in between.

You can have reactions:
235U + n → 236U + γ
or
235U + n → 236U*236U + γ

The second reaction requires that the energy of neutron fit some excited level of 236U, within the precision of its lifetime. The first has somewhat lower but not nonzero cross-section - but does not limit the energy, because free neutron and free photon both have unquantized energy.

#### dRic2

Gold Member
Thanks to all for your replies

Code:
https://physics.nist.gov/cgi-bin/cuu/Value?mnu
https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=U

1 amu c2 = 931.49432 MeV

Mass U235     235.0439301   amu
Mass neutron    1.008664916 amu

Mass U235+n   236.052595016 amu
Mass U236     236.0455682   amu

Excess Mass     0.007026816 amu

6.545439 MeV
If one looks at the mass of U235 and the neutron (no kinetic energy involved), the total mass is greater than that of U-236 (not excited). Subtracting the mass of U236, there is an extra 6.54544 MeV, which greater than the binding energy of the last neutron, so the excited nucleus is unstable. Fission usually occurs, but about 15 to 16% of the time, U236 emits a gamma ray and settles to a ground state without fissioning. The binding energy of the last neutron is simply the energy required to expel a neutron, the last neutron from the nucleus. A 6.4 MeV gamma ray could induce a photoneutron reaction. The actual gamma energy required is slightly higher because the U-236 would recoil when absorbing the gamma ray.
this is pretty clear to me. @snorkack seems to understand my problem.

within the precision of its lifetime.
so are you saying the same thing I was in my #1 post or not?

#### PeterDonis

Mentor
so are you saying the same thing I was in my #1 post or not?
Not quite. You said in post #1:

When the nucleus is hit by the neutron the excited state is formed, thus $\Delta t ≈ 0$
Actually, you will have $\Delta t$ be the lifetime of the excited state of 236U that is formed. So $\Delta E$, i.e., the width of the range of energies the 235U + n system can have and still have the reaction occur, is given by the uncertainty relation with that $\Delta t$. If there are multiple possible 236U excited states that can form, there will be multiple possible corresponding ranges of energies for the 235U + n system. (And the above is still somewhat oversimplified because you have to factor in recoil, etc.)

• dRic2

"Neutron capture and fission reaction"

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