Mixing
How fast do nuclei need to be traveling when colliding for them to break apart?
Clash Royale CLAN TAG#URR8PPP
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I don't mean fission or fusion reactions with uranium or such. I mean gold or lead or even hydrogen collisions, such as in these particle colliders or outer space, wherein the nuclei seem to touch, and possibly break apart.
homework-and-exercises nuclear-physics collision estimation binding-energy
edited 4 hours ago
Qmechanic♦
98.2k121721066
asked 6 hours ago
Kurt Hikes
1705
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up vote
1
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I don't mean fission or fusion reactions with uranium or such. I mean gold or lead or even hydrogen collisions, such as in these particle colliders or outer space, wherein the nuclei seem to touch, and possibly break apart.
homework-and-exercises nuclear-physics collision estimation binding-energy
edited 4 hours ago
Qmechanic♦
98.2k121721066
asked 6 hours ago
Kurt Hikes
1705
1
You seem confused about energy scales and the definitions of processes. Fusion has nothing to do with uranium, and touching is not in contradistinction to fusion. The typical outcome of touching is fusion. The electron shells are irrelevant because the energies we're talking about are so high. If you want to know about reactions where the nuclei basically get blown to bits, then the energy scale you want is much higher than the energy scale where they just touch. Please edit the question to clarify what you're asking about.
– Ben Crowell
4 hours ago
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
I don't mean fission or fusion reactions with uranium or such. I mean gold or lead or even hydrogen collisions, such as in these particle colliders or outer space, wherein the nuclei seem to touch, and possibly break apart.
homework-and-exercises nuclear-physics collision estimation binding-energy
edited 4 hours ago
Qmechanic♦
98.2k121721066
asked 6 hours ago
Kurt Hikes
1705
I don't mean fission or fusion reactions with uranium or such. I mean gold or lead or even hydrogen collisions, such as in these particle colliders or outer space, wherein the nuclei seem to touch, and possibly break apart.
homework-and-exercises nuclear-physics collision estimation binding-energy
homework-and-exercises nuclear-physics collision estimation binding-energy
edited 4 hours ago
Qmechanic♦
98.2k121721066
asked 6 hours ago
Kurt Hikes
1705
edited 4 hours ago
Qmechanic♦
98.2k121721066
asked 6 hours ago
Kurt Hikes
1705
edited 4 hours ago
Qmechanic♦
98.2k121721066
edited 4 hours ago
Qmechanic♦
98.2k121721066
edited 4 hours ago
Qmechanic♦
98.2k121721066
98.2k121721066
asked 6 hours ago
Kurt Hikes
1705
asked 6 hours ago
Kurt Hikes
1705
asked 6 hours ago
Kurt Hikes
1705
1705
1
You seem confused about energy scales and the definitions of processes. Fusion has nothing to do with uranium, and touching is not in contradistinction to fusion. The typical outcome of touching is fusion. The electron shells are irrelevant because the energies we're talking about are so high. If you want to know about reactions where the nuclei basically get blown to bits, then the energy scale you want is much higher than the energy scale where they just touch. Please edit the question to clarify what you're asking about.
– Ben Crowell
4 hours ago
add a comment |Â
1
You seem confused about energy scales and the definitions of processes. Fusion has nothing to do with uranium, and touching is not in contradistinction to fusion. The typical outcome of touching is fusion. The electron shells are irrelevant because the energies we're talking about are so high. If you want to know about reactions where the nuclei basically get blown to bits, then the energy scale you want is much higher than the energy scale where they just touch. Please edit the question to clarify what you're asking about.
– Ben Crowell
4 hours ago
1
1
You seem confused about energy scales and the definitions of processes. Fusion has nothing to do with uranium, and touching is not in contradistinction to fusion. The typical outcome of touching is fusion. The electron shells are irrelevant because the energies we're talking about are so high. If you want to know about reactions where the nuclei basically get blown to bits, then the energy scale you want is much higher than the energy scale where they just touch. Please edit the question to clarify what you're asking about.
– Ben Crowell
4 hours ago
You seem confused about energy scales and the definitions of processes. Fusion has nothing to do with uranium, and touching is not in contradistinction to fusion. The typical outcome of touching is fusion. The electron shells are irrelevant because the energies we're talking about are so high. If you want to know about reactions where the nuclei basically get blown to bits, then the energy scale you want is much higher than the energy scale where they just touch. Please edit the question to clarify what you're asking about.
– Ben Crowell
4 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
A back-of-the-envelope estimate for the energy at which this might happen is the total binding energy of the nuclei. This binding energy $E_B$ can be approximated for heavier nuclei using the well-known semi-empirical mass formula:
$$E_B=a_V A - a_S A^2/3 - a_C fracZ(Z-1)A^1/3 - a_A frac(A-2Z)^2A - delta(A,Z)$$
for atomic number $Z$ and mass number $A$. The coefficients are empirically measured, though there is substantial uncertainty on their exact values, depending on the particular way in which data is fitted and the parametrization chosen for $delta(A,Z)$; two such parametrizations are given here. We can say, for least-squares fits, that $a_Vapprox 15$ MeV, $a_S approx 17$ MeV, $a_Capprox 0.7$ MeV, $a_Aapprox 23$ MeV, and:
$delta=0$ if $A$ is odd- If $A$ is even and $Z$ is odd, $deltaapproxfrac12A^1/2$ or $deltaapproxfrac34A^3/4$
- If $A$ is even and $Z$ is even, $deltaapproxfrac-12A^1/2$ or $deltaapproxfrac-34A^3/4$
Note that this formula is not accurate for very light nuclei, as it comes from Gamow's "liquid-drop model", and as such ignores the shell structure of nuclear energy levels that is very important in light nuclei. If there's any doubt, there's a table of nuclear binding energies here: http://dbserv.pnpi.spb.ru/elbib/tablisot/toi98/www/astro/table2.pdf
In any case, if the total energy in the collision is somewhere close to or above the binding energy of the nuclei, it is possible that one or more of the nuclei will break apart. However, smaller amounts of energy can cause the ejection of one or a few nucleons while the rest of the nucleus itself stays together - for such a process, the energy requirement is (again, very roughly) the difference of the binding energies of the final and initial nuclei.
Given this energy, the speed at which this will happen is given by
$$fracvc=sqrt1-left(fracmc^2Eright)^2$$
In our situation, we want our nuclei to be traveling with kinetic energies around $E_B$, so that $Eapprox mc^2+E_B$. Looking at the table, you can see that the binding energies of nuclei are usually in the tens or hundreds of MeV, whereas $mc^2$ (the mass of the nucleus) is usually in the tens or hundreds of GeV, so the ratio $fracmc^2E$ will differ nontrivially from 1, and they will be traveling an appreciable fraction of the speed of light. Exactly how fast depends on the particular nuclei involved in the collision, as well as the energy we're assuming they're traveling at. For two lead-208 nuclei, $E_B=1636$ MeV, $mc^2=194$ GeV, so $v=0.129c$, if we assume that each has enough kinetic energy to dissociate the other.
answered 6 hours ago
probably_someone
14.5k12451
Thanks for the answer, really, but I wonder if someone could give me the actual speed at which this happens.....
– Kurt Hikes
5 hours ago
@KurtHikes edited
– probably_someone
5 hours ago
add a comment |Â
up vote
1
down vote
Physicists generally think in terms of energy rather than speed. This is because many phenomena vary linearly in energy, while the relationship between energy and speed is nonlinear. For slow speeds, where you can ignore relativity, the relationship between speed and kinetic energy $T$ is
$$
T = frac12 mv^2,
$$
while if the kinetic energy $T$ is a non-negligible fraction of the rest energy $mc^2$ one must instead use
$$
T=(gamma-1)mc^2
text, with
frac1gamma = sqrt1-v^2/c^2.
$$
In general, collisions where the center-of-mass energy is a few mega-eV can cause a large nucleus to vibrate or rotate, and it will eventually release this extra internal energy by emitting photons. For most nuclei, the energy required to knock free a single proton or neutron is about 8 MeV.
For energies above 140 MeV, you must also take into account the probability of creating pions (and heavier mesons at higher energies), but creating a spray of nuclear fragments has a larger phase space in collisions between heavy nuclei.
A practical use case that may interest you is neutron spallation, in which a hard proton beam stops in heavy metal target, like tungsten or mercury. The metal nuclei are totally disrupted, breaking into fragments which cool by boiling off twenty to forty neutrons for each incoming proton. The spallation neutron sources at Low Alamos and Oak Ridge have proton beams with $Tapprox m_textprotonc^2$, which is a speed around $0.85c$. Much less speed would be required in mercury-mercury collision; somewhere in my answer history is an explanation of why spallation sources use proton accelerators anyway, whose gist is "it's the energy that matters."
answered 4 hours ago
rob♦
38.2k971159
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
A back-of-the-envelope estimate for the energy at which this might happen is the total binding energy of the nuclei. This binding energy $E_B$ can be approximated for heavier nuclei using the well-known semi-empirical mass formula:
$$E_B=a_V A - a_S A^2/3 - a_C fracZ(Z-1)A^1/3 - a_A frac(A-2Z)^2A - delta(A,Z)$$
for atomic number $Z$ and mass number $A$. The coefficients are empirically measured, though there is substantial uncertainty on their exact values, depending on the particular way in which data is fitted and the parametrization chosen for $delta(A,Z)$; two such parametrizations are given here. We can say, for least-squares fits, that $a_Vapprox 15$ MeV, $a_S approx 17$ MeV, $a_Capprox 0.7$ MeV, $a_Aapprox 23$ MeV, and:
$delta=0$ if $A$ is odd- If $A$ is even and $Z$ is odd, $deltaapproxfrac12A^1/2$ or $deltaapproxfrac34A^3/4$
- If $A$ is even and $Z$ is even, $deltaapproxfrac-12A^1/2$ or $deltaapproxfrac-34A^3/4$
Note that this formula is not accurate for very light nuclei, as it comes from Gamow's "liquid-drop model", and as such ignores the shell structure of nuclear energy levels that is very important in light nuclei. If there's any doubt, there's a table of nuclear binding energies here: http://dbserv.pnpi.spb.ru/elbib/tablisot/toi98/www/astro/table2.pdf
In any case, if the total energy in the collision is somewhere close to or above the binding energy of the nuclei, it is possible that one or more of the nuclei will break apart. However, smaller amounts of energy can cause the ejection of one or a few nucleons while the rest of the nucleus itself stays together - for such a process, the energy requirement is (again, very roughly) the difference of the binding energies of the final and initial nuclei.
Given this energy, the speed at which this will happen is given by
$$fracvc=sqrt1-left(fracmc^2Eright)^2$$
In our situation, we want our nuclei to be traveling with kinetic energies around $E_B$, so that $Eapprox mc^2+E_B$. Looking at the table, you can see that the binding energies of nuclei are usually in the tens or hundreds of MeV, whereas $mc^2$ (the mass of the nucleus) is usually in the tens or hundreds of GeV, so the ratio $fracmc^2E$ will differ nontrivially from 1, and they will be traveling an appreciable fraction of the speed of light. Exactly how fast depends on the particular nuclei involved in the collision, as well as the energy we're assuming they're traveling at. For two lead-208 nuclei, $E_B=1636$ MeV, $mc^2=194$ GeV, so $v=0.129c$, if we assume that each has enough kinetic energy to dissociate the other.
answered 6 hours ago
probably_someone
14.5k12451
Thanks for the answer, really, but I wonder if someone could give me the actual speed at which this happens.....
– Kurt Hikes
5 hours ago
@KurtHikes edited
– probably_someone
5 hours ago
add a comment |Â
up vote
3
down vote
A back-of-the-envelope estimate for the energy at which this might happen is the total binding energy of the nuclei. This binding energy $E_B$ can be approximated for heavier nuclei using the well-known semi-empirical mass formula:
$$E_B=a_V A - a_S A^2/3 - a_C fracZ(Z-1)A^1/3 - a_A frac(A-2Z)^2A - delta(A,Z)$$
for atomic number $Z$ and mass number $A$. The coefficients are empirically measured, though there is substantial uncertainty on their exact values, depending on the particular way in which data is fitted and the parametrization chosen for $delta(A,Z)$; two such parametrizations are given here. We can say, for least-squares fits, that $a_Vapprox 15$ MeV, $a_S approx 17$ MeV, $a_Capprox 0.7$ MeV, $a_Aapprox 23$ MeV, and:
$delta=0$ if $A$ is odd- If $A$ is even and $Z$ is odd, $deltaapproxfrac12A^1/2$ or $deltaapproxfrac34A^3/4$
- If $A$ is even and $Z$ is even, $deltaapproxfrac-12A^1/2$ or $deltaapproxfrac-34A^3/4$
Note that this formula is not accurate for very light nuclei, as it comes from Gamow's "liquid-drop model", and as such ignores the shell structure of nuclear energy levels that is very important in light nuclei. If there's any doubt, there's a table of nuclear binding energies here: http://dbserv.pnpi.spb.ru/elbib/tablisot/toi98/www/astro/table2.pdf
In any case, if the total energy in the collision is somewhere close to or above the binding energy of the nuclei, it is possible that one or more of the nuclei will break apart. However, smaller amounts of energy can cause the ejection of one or a few nucleons while the rest of the nucleus itself stays together - for such a process, the energy requirement is (again, very roughly) the difference of the binding energies of the final and initial nuclei.
Given this energy, the speed at which this will happen is given by
$$fracvc=sqrt1-left(fracmc^2Eright)^2$$
In our situation, we want our nuclei to be traveling with kinetic energies around $E_B$, so that $Eapprox mc^2+E_B$. Looking at the table, you can see that the binding energies of nuclei are usually in the tens or hundreds of MeV, whereas $mc^2$ (the mass of the nucleus) is usually in the tens or hundreds of GeV, so the ratio $fracmc^2E$ will differ nontrivially from 1, and they will be traveling an appreciable fraction of the speed of light. Exactly how fast depends on the particular nuclei involved in the collision, as well as the energy we're assuming they're traveling at. For two lead-208 nuclei, $E_B=1636$ MeV, $mc^2=194$ GeV, so $v=0.129c$, if we assume that each has enough kinetic energy to dissociate the other.
answered 6 hours ago
probably_someone
14.5k12451
Thanks for the answer, really, but I wonder if someone could give me the actual speed at which this happens.....
– Kurt Hikes
5 hours ago
@KurtHikes edited
– probably_someone
5 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A back-of-the-envelope estimate for the energy at which this might happen is the total binding energy of the nuclei. This binding energy $E_B$ can be approximated for heavier nuclei using the well-known semi-empirical mass formula:
$$E_B=a_V A - a_S A^2/3 - a_C fracZ(Z-1)A^1/3 - a_A frac(A-2Z)^2A - delta(A,Z)$$
for atomic number $Z$ and mass number $A$. The coefficients are empirically measured, though there is substantial uncertainty on their exact values, depending on the particular way in which data is fitted and the parametrization chosen for $delta(A,Z)$; two such parametrizations are given here. We can say, for least-squares fits, that $a_Vapprox 15$ MeV, $a_S approx 17$ MeV, $a_Capprox 0.7$ MeV, $a_Aapprox 23$ MeV, and:
$delta=0$ if $A$ is odd- If $A$ is even and $Z$ is odd, $deltaapproxfrac12A^1/2$ or $deltaapproxfrac34A^3/4$
- If $A$ is even and $Z$ is even, $deltaapproxfrac-12A^1/2$ or $deltaapproxfrac-34A^3/4$
Note that this formula is not accurate for very light nuclei, as it comes from Gamow's "liquid-drop model", and as such ignores the shell structure of nuclear energy levels that is very important in light nuclei. If there's any doubt, there's a table of nuclear binding energies here: http://dbserv.pnpi.spb.ru/elbib/tablisot/toi98/www/astro/table2.pdf
In any case, if the total energy in the collision is somewhere close to or above the binding energy of the nuclei, it is possible that one or more of the nuclei will break apart. However, smaller amounts of energy can cause the ejection of one or a few nucleons while the rest of the nucleus itself stays together - for such a process, the energy requirement is (again, very roughly) the difference of the binding energies of the final and initial nuclei.
Given this energy, the speed at which this will happen is given by
$$fracvc=sqrt1-left(fracmc^2Eright)^2$$
In our situation, we want our nuclei to be traveling with kinetic energies around $E_B$, so that $Eapprox mc^2+E_B$. Looking at the table, you can see that the binding energies of nuclei are usually in the tens or hundreds of MeV, whereas $mc^2$ (the mass of the nucleus) is usually in the tens or hundreds of GeV, so the ratio $fracmc^2E$ will differ nontrivially from 1, and they will be traveling an appreciable fraction of the speed of light. Exactly how fast depends on the particular nuclei involved in the collision, as well as the energy we're assuming they're traveling at. For two lead-208 nuclei, $E_B=1636$ MeV, $mc^2=194$ GeV, so $v=0.129c$, if we assume that each has enough kinetic energy to dissociate the other.
answered 6 hours ago
probably_someone
14.5k12451
A back-of-the-envelope estimate for the energy at which this might happen is the total binding energy of the nuclei. This binding energy $E_B$ can be approximated for heavier nuclei using the well-known semi-empirical mass formula:
$$E_B=a_V A - a_S A^2/3 - a_C fracZ(Z-1)A^1/3 - a_A frac(A-2Z)^2A - delta(A,Z)$$
for atomic number $Z$ and mass number $A$. The coefficients are empirically measured, though there is substantial uncertainty on their exact values, depending on the particular way in which data is fitted and the parametrization chosen for $delta(A,Z)$; two such parametrizations are given here. We can say, for least-squares fits, that $a_Vapprox 15$ MeV, $a_S approx 17$ MeV, $a_Capprox 0.7$ MeV, $a_Aapprox 23$ MeV, and:
$delta=0$ if $A$ is odd- If $A$ is even and $Z$ is odd, $deltaapproxfrac12A^1/2$ or $deltaapproxfrac34A^3/4$
- If $A$ is even and $Z$ is even, $deltaapproxfrac-12A^1/2$ or $deltaapproxfrac-34A^3/4$
Note that this formula is not accurate for very light nuclei, as it comes from Gamow's "liquid-drop model", and as such ignores the shell structure of nuclear energy levels that is very important in light nuclei. If there's any doubt, there's a table of nuclear binding energies here: http://dbserv.pnpi.spb.ru/elbib/tablisot/toi98/www/astro/table2.pdf
In any case, if the total energy in the collision is somewhere close to or above the binding energy of the nuclei, it is possible that one or more of the nuclei will break apart. However, smaller amounts of energy can cause the ejection of one or a few nucleons while the rest of the nucleus itself stays together - for such a process, the energy requirement is (again, very roughly) the difference of the binding energies of the final and initial nuclei.
Given this energy, the speed at which this will happen is given by
$$fracvc=sqrt1-left(fracmc^2Eright)^2$$
In our situation, we want our nuclei to be traveling with kinetic energies around $E_B$, so that $Eapprox mc^2+E_B$. Looking at the table, you can see that the binding energies of nuclei are usually in the tens or hundreds of MeV, whereas $mc^2$ (the mass of the nucleus) is usually in the tens or hundreds of GeV, so the ratio $fracmc^2E$ will differ nontrivially from 1, and they will be traveling an appreciable fraction of the speed of light. Exactly how fast depends on the particular nuclei involved in the collision, as well as the energy we're assuming they're traveling at. For two lead-208 nuclei, $E_B=1636$ MeV, $mc^2=194$ GeV, so $v=0.129c$, if we assume that each has enough kinetic energy to dissociate the other.
answered 6 hours ago
probably_someone
14.5k12451
edited 5 hours ago
answered 6 hours ago
probably_someone
14.5k12451
answered 6 hours ago
probably_someone
14.5k12451
answered 6 hours ago
probably_someone
14.5k12451
14.5k12451
Thanks for the answer, really, but I wonder if someone could give me the actual speed at which this happens.....
– Kurt Hikes
5 hours ago
@KurtHikes edited
– probably_someone
5 hours ago
add a comment |Â
Thanks for the answer, really, but I wonder if someone could give me the actual speed at which this happens.....
– Kurt Hikes
5 hours ago
@KurtHikes edited
– probably_someone
5 hours ago
Thanks for the answer, really, but I wonder if someone could give me the actual speed at which this happens.....
– Kurt Hikes
5 hours ago
Thanks for the answer, really, but I wonder if someone could give me the actual speed at which this happens.....
– Kurt Hikes
5 hours ago
@KurtHikes edited
– probably_someone
5 hours ago
@KurtHikes edited
– probably_someone
5 hours ago
add a comment |Â
up vote
1
down vote
Physicists generally think in terms of energy rather than speed. This is because many phenomena vary linearly in energy, while the relationship between energy and speed is nonlinear. For slow speeds, where you can ignore relativity, the relationship between speed and kinetic energy $T$ is
$$
T = frac12 mv^2,
$$
while if the kinetic energy $T$ is a non-negligible fraction of the rest energy $mc^2$ one must instead use
$$
T=(gamma-1)mc^2
text, with
frac1gamma = sqrt1-v^2/c^2.
$$
In general, collisions where the center-of-mass energy is a few mega-eV can cause a large nucleus to vibrate or rotate, and it will eventually release this extra internal energy by emitting photons. For most nuclei, the energy required to knock free a single proton or neutron is about 8 MeV.
For energies above 140 MeV, you must also take into account the probability of creating pions (and heavier mesons at higher energies), but creating a spray of nuclear fragments has a larger phase space in collisions between heavy nuclei.
A practical use case that may interest you is neutron spallation, in which a hard proton beam stops in heavy metal target, like tungsten or mercury. The metal nuclei are totally disrupted, breaking into fragments which cool by boiling off twenty to forty neutrons for each incoming proton. The spallation neutron sources at Low Alamos and Oak Ridge have proton beams with $Tapprox m_textprotonc^2$, which is a speed around $0.85c$. Much less speed would be required in mercury-mercury collision; somewhere in my answer history is an explanation of why spallation sources use proton accelerators anyway, whose gist is "it's the energy that matters."
answered 4 hours ago
rob♦
38.2k971159
add a comment |Â
up vote
1
down vote
Physicists generally think in terms of energy rather than speed. This is because many phenomena vary linearly in energy, while the relationship between energy and speed is nonlinear. For slow speeds, where you can ignore relativity, the relationship between speed and kinetic energy $T$ is
$$
T = frac12 mv^2,
$$
while if the kinetic energy $T$ is a non-negligible fraction of the rest energy $mc^2$ one must instead use
$$
T=(gamma-1)mc^2
text, with
frac1gamma = sqrt1-v^2/c^2.
$$
In general, collisions where the center-of-mass energy is a few mega-eV can cause a large nucleus to vibrate or rotate, and it will eventually release this extra internal energy by emitting photons. For most nuclei, the energy required to knock free a single proton or neutron is about 8 MeV.
For energies above 140 MeV, you must also take into account the probability of creating pions (and heavier mesons at higher energies), but creating a spray of nuclear fragments has a larger phase space in collisions between heavy nuclei.
A practical use case that may interest you is neutron spallation, in which a hard proton beam stops in heavy metal target, like tungsten or mercury. The metal nuclei are totally disrupted, breaking into fragments which cool by boiling off twenty to forty neutrons for each incoming proton. The spallation neutron sources at Low Alamos and Oak Ridge have proton beams with $Tapprox m_textprotonc^2$, which is a speed around $0.85c$. Much less speed would be required in mercury-mercury collision; somewhere in my answer history is an explanation of why spallation sources use proton accelerators anyway, whose gist is "it's the energy that matters."
answered 4 hours ago
rob♦
38.2k971159
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Physicists generally think in terms of energy rather than speed. This is because many phenomena vary linearly in energy, while the relationship between energy and speed is nonlinear. For slow speeds, where you can ignore relativity, the relationship between speed and kinetic energy $T$ is
$$
T = frac12 mv^2,
$$
while if the kinetic energy $T$ is a non-negligible fraction of the rest energy $mc^2$ one must instead use
$$
T=(gamma-1)mc^2
text, with
frac1gamma = sqrt1-v^2/c^2.
$$
In general, collisions where the center-of-mass energy is a few mega-eV can cause a large nucleus to vibrate or rotate, and it will eventually release this extra internal energy by emitting photons. For most nuclei, the energy required to knock free a single proton or neutron is about 8 MeV.
For energies above 140 MeV, you must also take into account the probability of creating pions (and heavier mesons at higher energies), but creating a spray of nuclear fragments has a larger phase space in collisions between heavy nuclei.
A practical use case that may interest you is neutron spallation, in which a hard proton beam stops in heavy metal target, like tungsten or mercury. The metal nuclei are totally disrupted, breaking into fragments which cool by boiling off twenty to forty neutrons for each incoming proton. The spallation neutron sources at Low Alamos and Oak Ridge have proton beams with $Tapprox m_textprotonc^2$, which is a speed around $0.85c$. Much less speed would be required in mercury-mercury collision; somewhere in my answer history is an explanation of why spallation sources use proton accelerators anyway, whose gist is "it's the energy that matters."
answered 4 hours ago
rob♦
38.2k971159
Physicists generally think in terms of energy rather than speed. This is because many phenomena vary linearly in energy, while the relationship between energy and speed is nonlinear. For slow speeds, where you can ignore relativity, the relationship between speed and kinetic energy $T$ is
$$
T = frac12 mv^2,
$$
while if the kinetic energy $T$ is a non-negligible fraction of the rest energy $mc^2$ one must instead use
$$
T=(gamma-1)mc^2
text, with
frac1gamma = sqrt1-v^2/c^2.
$$
In general, collisions where the center-of-mass energy is a few mega-eV can cause a large nucleus to vibrate or rotate, and it will eventually release this extra internal energy by emitting photons. For most nuclei, the energy required to knock free a single proton or neutron is about 8 MeV.
For energies above 140 MeV, you must also take into account the probability of creating pions (and heavier mesons at higher energies), but creating a spray of nuclear fragments has a larger phase space in collisions between heavy nuclei.
A practical use case that may interest you is neutron spallation, in which a hard proton beam stops in heavy metal target, like tungsten or mercury. The metal nuclei are totally disrupted, breaking into fragments which cool by boiling off twenty to forty neutrons for each incoming proton. The spallation neutron sources at Low Alamos and Oak Ridge have proton beams with $Tapprox m_textprotonc^2$, which is a speed around $0.85c$. Much less speed would be required in mercury-mercury collision; somewhere in my answer history is an explanation of why spallation sources use proton accelerators anyway, whose gist is "it's the energy that matters."
answered 4 hours ago
rob♦
38.2k971159
answered 4 hours ago
rob♦
38.2k971159
answered 4 hours ago
rob♦
38.2k971159
answered 4 hours ago
rob♦
38.2k971159
38.2k971159
add a comment |Â
add a comment |Â
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1
You seem confused about energy scales and the definitions of processes. Fusion has nothing to do with uranium, and touching is not in contradistinction to fusion. The typical outcome of touching is fusion. The electron shells are irrelevant because the energies we're talking about are so high. If you want to know about reactions where the nuclei basically get blown to bits, then the energy scale you want is much higher than the energy scale where they just touch. Please edit the question to clarify what you're asking about.
– Ben Crowell
4 hours ago