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Q&A What happens if an electron collides with a proton?

posted 1y ago by dmckee‭  ·  edited 1y ago by dmckee‭

Answer
#3: Post edited by user avatar dmckee‭ · 2021-08-20T04:45:29Z (about 1 year ago)
  • Answering from the point of view of a particle physicist. Meaning we consider an event where and electron and proton approach each other in an initially free condition (not already bound).
  • **They can scatter elastically or inelastically.**
  • ## Elastic scattering.
  • In this case the final state will consist only of one electron and one protons, and their combined kinetic-energy after the event is the same as before the event though their final momenta may be different from their initial momenta. (Indeed the case in which the vector momenta remains the same might well be treated as not "colliding" at all.)
  • At some level this is a unrealizable ideal: you can't build a detector system that can absolutely rule out the presence of very low energy photons in the final state. That said, you can often engineer an experiment which can reduce undetected deviation of elasticity to experimentally uninteresting levels.
  • ## Inelastic scattering
  • This represent all other outcomes, and can be usefully subdivided into several cases:
  • 1. Final states including and electron and a proton (possible bound) and one or more additional particles
  • 2. Final states with a proton and some non-electron lepton (and possibly other particles)
  • 3. Final states with an electron and some non-proton baryon (and possibly some other particles)
  • 4. Final states with neither a proton nor an electron but still with lepton number 1 and baryon number 1
  • The additional particles might be any of photons, mesons, leptons, or even baryon pairs, depending on the available energy.
  • There are conservation rules that require the final state lepton and baryon numbers to be the same as the initial state values. This controls how the initial state particles may be replaced and prevents the system from simply disintegrating into photons.
  • ### Replacing the electron
  • A charged-current weak interaction can convert the electron to a electron-neutrino, and will necessarily introduce a change in the charge of the remainder of the system (from +1 to 0). These events are rare.
  • It is allowed by conservation rules to replace the electron with a heavier charged lepton (muon or tau), though this must be accomplished by an unspecified interaction with the baryonic part of the system.
  • ### Replacing (or not) the proton
  • Being a compound object the proton can be transformed in several ways. It can be excited by an electromagnetic interaction (to form a delta baryon), it may have one of its constituent quarks' flavor changed by a weak interaction to form some other heavier baryon, it can have mesons or lepton pairs boosted out of it's sea, or get blasted into a quark jet that then condenses into multiple hadrons.
  • When pairs are boosted out of the proton, it can retain it's identity (be detected as a proton in the final state), but excitation, changing a constituent quark's flavor and blasting it apart all lead to a final state in which some other baryon is detected (at least if you look soon enough).
  • ### Energy concerns
  • Note that the total energy (including mass) of the system is conserved even in an inelastic event. So the production of massive final-state particles can only occur if the center-of-mass energy is sufficient.
  • Even the simple production of photons (relatively common) steals kinetic energy from the stars of this show, and if they lose enough kinetic energy they may emerge from the event in a bound state which will inevitably decay towards the ground-state of hydrogen.
  • Answering from the point of view of a particle physicist. Meaning we consider an event where and electron and proton approach each other in an initially free condition (not already bound).
  • # Electron-Proton scattering
  • Believe it or not, the whole story is
  • >They can scatter elastically or inelastically.
  • ## Elastic scattering.
  • In this case the final state will consist of one electron, one proton, and nothing else; their combined kinetic-energy after the event is the same as before the event though their final momenta may be different from their initial momenta. (Indeed the case in which the vector momenta remains the same might well be treated as not "colliding" at all.)
  • At some level this is a unrealizable ideal: you can't build a detector system that can absolutely rule out the presence of very low energy photons in the final state. That said, you can often engineer an experiment which can reduce undetected deviation of elasticity to experimentally uninteresting levels.
  • ## Inelastic scattering
  • This represent all other outcomes, and can be usefully subdivided into several cases:
  • 1. Final states including and electron and a proton (possible bound) and one or more additional particles
  • 2. Final states with a proton and some non-electron lepton (and possibly other particles)
  • 3. Final states with an electron and some non-proton baryon (and possibly some other particles)
  • 4. Final states with neither a proton nor an electron but still with lepton number 1 and baryon number 1
  • The additional particles might be any of photons, mesons, leptons, or even baryon pairs, depending on the available energy.
  • There are conservation rules that require the final state lepton and baryon numbers to be the same as the initial state values. This controls how the initial state particles may be replaced and prevents the system from simply disintegrating into photons.
  • ### Replacing the electron
  • A charged-current weak interaction can convert the electron to a electron-neutrino, and will necessarily introduce a change in the charge of the remainder of the system (from +1 to 0). These events are rare.
  • It is allowed by conservation rules to replace the electron with a heavier charged lepton (muon or tau), though this must be accomplished by an unspecified interaction with the baryonic part of the system.
  • ### Replacing (or not) the proton
  • Being a compound object the proton can be transformed in several ways. It can be excited by an electromagnetic interaction (to form a delta baryon), it may have one of its constituent quarks' flavor changed by a weak interaction to form some other heavier baryon, it can have mesons or lepton pairs boosted out of it's sea, or get blasted into a quark jet that then condenses into multiple hadrons.
  • When pairs are boosted out of the proton, it can retain it's identity (be detected as a proton in the final state), but excitation, changing a constituent quark's flavor and blasting it apart all lead to a final state in which some other baryon is detected (at least if you look soon enough).
  • ### Energy concerns
  • Note that the total energy (including mass) of the system is conserved even in an inelastic event. So the production of massive final-state particles can only occur if the center-of-mass energy is sufficient.
  • Even the simple production of photons (relatively common) steals kinetic energy from the stars of this show, and if they lose enough kinetic energy they may emerge from the event in a bound state which will inevitably decay towards the ground-state of hydrogen.
#2: Post edited by user avatar dmckee‭ · 2021-08-20T04:24:05Z (about 1 year ago)
  • Answering from the point of view of a particle physicist. Meaning we consider an event where and electron and proton approach each other in an initially free condition (not already bound).
  • **They can scatter elastically or inelastically.**
  • ## Elastic scattering.
  • In this case the final state will consist only of one electron and one protons, and their combined kinetic-energy after the event is the same as before the event though their final momenta may be different from their initial momenta. (Indeed the case in which the vector momenta remains the same might well be treated as not "colliding" at all.)
  • At some level this is a unrealizable ideal: you can't build a detector system that can absolutely rule out the presence of very low energy photons in the final state. That said, you can often engineer an experiment which can reduce undetected deviation of elasticity to experimentally uninteresting levels.
  • ## Inelastic scattering
  • This represent all other outcomes, and can be usefully subdivided into several cases:
  • 1. Final states including and electron and a proton (possible bound) and one or more additional particles
  • 2. Final states with a proton and some non-electron lepton (and possibly other particles)
  • 3. Final states with an electron and some non-proton baryon (and possibly some other particles)
  • 4. Final states with neither a proton nor an electron but still with lepton number 1 and baryon number 1
  • The additional particles might be any of photons, mesons, leptons, or even baryon pairs, depending on the available energy.
  • There are conservation rules that require the final state lepton and baryon numbers to be the same as the initial state values. This controls how the initial state particles may be replaced.
  • ### Replacing the electron
  • A charged-current weak interaction can convert the electron to a electron-neutrino, and will necessarily introduce a change in the charge of the remainder of the system (from +1 to 0). These events are rare.
  • It is allowed by conservation rules to replace the electron with a heavier charged lepton (muon or tau), though this must be accomplished by an unspecified interaction with the baryonic part of the system.
  • ### Replacing (or not) the proton
  • Being a compound object the proton can be transformed in several ways. It can be excited by an electromagnetic interaction (to form a delta baryon), it may have one of its constituent quarks' flavor changed by a weak interaction to form some other heavier baryon, it can have mesons or lepton pairs boosted out of it's sea, or get blasted into a quark jet that then condenses into multiple hadrons.
  • When pairs are boosted out of the proton, it can retain it's identity (be detected as a proton in the final state), but excitation, changing a constituent quark's flavor and blasting it apart all lead to a final state in which some other baryon is detected (at least if you look soon enough).
  • ### Energy concerns
  • Note that the total energy (including mass) of the system is conserved even in an inelastic event. So the production of massive final-state particles can only occur if the center-of-mass energy is sufficient.
  • Even the simple production of photons (relatively common) steals kinetic energy from the stars of this show, and if they lose enough kinetic energy they may emerge from the event in a bound state which will inevitably decay towards the ground-state of hydrogen.
  • Answering from the point of view of a particle physicist. Meaning we consider an event where and electron and proton approach each other in an initially free condition (not already bound).
  • **They can scatter elastically or inelastically.**
  • ## Elastic scattering.
  • In this case the final state will consist only of one electron and one protons, and their combined kinetic-energy after the event is the same as before the event though their final momenta may be different from their initial momenta. (Indeed the case in which the vector momenta remains the same might well be treated as not "colliding" at all.)
  • At some level this is a unrealizable ideal: you can't build a detector system that can absolutely rule out the presence of very low energy photons in the final state. That said, you can often engineer an experiment which can reduce undetected deviation of elasticity to experimentally uninteresting levels.
  • ## Inelastic scattering
  • This represent all other outcomes, and can be usefully subdivided into several cases:
  • 1. Final states including and electron and a proton (possible bound) and one or more additional particles
  • 2. Final states with a proton and some non-electron lepton (and possibly other particles)
  • 3. Final states with an electron and some non-proton baryon (and possibly some other particles)
  • 4. Final states with neither a proton nor an electron but still with lepton number 1 and baryon number 1
  • The additional particles might be any of photons, mesons, leptons, or even baryon pairs, depending on the available energy.
  • There are conservation rules that require the final state lepton and baryon numbers to be the same as the initial state values. This controls how the initial state particles may be replaced and prevents the system from simply disintegrating into photons.
  • ### Replacing the electron
  • A charged-current weak interaction can convert the electron to a electron-neutrino, and will necessarily introduce a change in the charge of the remainder of the system (from +1 to 0). These events are rare.
  • It is allowed by conservation rules to replace the electron with a heavier charged lepton (muon or tau), though this must be accomplished by an unspecified interaction with the baryonic part of the system.
  • ### Replacing (or not) the proton
  • Being a compound object the proton can be transformed in several ways. It can be excited by an electromagnetic interaction (to form a delta baryon), it may have one of its constituent quarks' flavor changed by a weak interaction to form some other heavier baryon, it can have mesons or lepton pairs boosted out of it's sea, or get blasted into a quark jet that then condenses into multiple hadrons.
  • When pairs are boosted out of the proton, it can retain it's identity (be detected as a proton in the final state), but excitation, changing a constituent quark's flavor and blasting it apart all lead to a final state in which some other baryon is detected (at least if you look soon enough).
  • ### Energy concerns
  • Note that the total energy (including mass) of the system is conserved even in an inelastic event. So the production of massive final-state particles can only occur if the center-of-mass energy is sufficient.
  • Even the simple production of photons (relatively common) steals kinetic energy from the stars of this show, and if they lose enough kinetic energy they may emerge from the event in a bound state which will inevitably decay towards the ground-state of hydrogen.
#1: Initial revision by user avatar dmckee‭ · 2021-08-20T04:23:05Z (about 1 year ago)
Answering from the point of view of a particle physicist. Meaning we consider an event where and electron and proton approach each other in an initially free condition (not already bound).

**They can scatter elastically or inelastically.**

## Elastic scattering.  
  
  In this case the final state will consist only of one electron and one protons, and their combined kinetic-energy after the event is the same as before the event though their final momenta may be different from their initial momenta. (Indeed the case in which the vector momenta remains the same might well be treated as not "colliding" at all.)
  
  At some level this is a unrealizable ideal: you can't build a detector system that can absolutely rule out the presence of very low energy photons in the final state. That said, you can often engineer an experiment which can reduce undetected deviation of elasticity to experimentally uninteresting levels.
 
## Inelastic scattering
  
  This represent all other outcomes, and can be usefully subdivided into several cases:

  1. Final states including and electron and a proton (possible bound) and one or more additional particles
  2. Final states with a proton and some non-electron lepton (and possibly other particles)
  3. Final states with an electron and some non-proton baryon (and possibly some other particles)
  4. Final states with neither a proton nor an electron but still with lepton number 1 and baryon number 1
  
  The additional particles might be any of photons, mesons, leptons, or even baryon pairs, depending on the available energy.

  There are conservation rules that require the final state lepton and baryon numbers to be the same as the initial state values. This controls how the initial state particles may be replaced.

### Replacing the electron

A charged-current weak interaction can convert the electron to a electron-neutrino, and will necessarily introduce a change in the charge of the remainder of the system (from +1 to 0). These events are rare.

It is allowed by conservation rules to replace the electron with a heavier charged lepton (muon or tau), though this must be accomplished by an unspecified interaction with the baryonic part of the system.

### Replacing (or not) the proton

Being a compound object the proton can be transformed in several ways. It can be excited by an electromagnetic interaction (to form a delta baryon), it may have one of its constituent quarks' flavor changed by a weak interaction to form some other heavier baryon, it can have mesons or lepton pairs boosted out of it's sea, or get blasted into a quark jet that then condenses into multiple hadrons.

When pairs are boosted out of the proton, it can retain it's identity (be detected as a proton in the final state), but excitation, changing a constituent quark's flavor and blasting it apart all lead to a final state in which some other baryon is detected (at least if you look soon enough).

### Energy concerns

Note that the total energy (including mass) of the system is conserved even in an inelastic event. So the production of massive final-state particles can only occur if the center-of-mass energy is sufficient.

Even the simple production of photons (relatively common) steals kinetic energy from the stars of this show, and if they lose enough kinetic energy they may emerge from the event in a bound state which will inevitably decay towards the ground-state of hydrogen.