antimatter and Beta

[YT2095]

2 questions really, beta emiters give off Electrons and Positrons so naturaly these must at some time have the occasional collision, and give off 2 photons at Gamma, is this correct?

 

also, would that explain the Blue glow that pure Radium has in the dark?

(possibly Cherenkov radiation).

[[Tycho?]]

Beta emitters give off electrons and positrons in two seperate reactions; I dont know if you get the same reactions happening at the same time.

 

But regardless, positrons are not going to last long. Every atom has electrons whirling around it, so a positron is not going to make it far at all before it finds some electron and is annihilated.

 

Cherenkov radiation is a different beast. It is caused when particles (I dont know what kind) move faster than light in that medium. An example in a nuclear reactor: you have radioactive stuff in water. Light moves more slowly in water than it does in vaccuum. So particles given off can end up beating a beam of light emitted at the same time. For some reason, this gives you the radiation. Wikipedia has more info on it, I obviously do not understand the process.

[insane_alien]

I dont know what kind

 

the charged kind.

[[Tycho?]]

the charged kind.

 

Neutrinos cause it though, they aren't charged.

[swansont]

;318567']Beta emitters give off electrons and positrons in two seperate reactions; I dont know if you get the same reactions happening at the same time.

 

Only in rare cases could you get opposite reactions from the same nuclide. Generally you have either too many or too few neutrons, which means you get one or the other. But if you had stable nuclides that had both one more and one fewer proton, you could. It would most likely be a nuclide with an odd number of both neutrons and protons.

 

Cu-64, Ga-70 and Rb-86 are examples that both beta-plus (or possibly electron capture) and beta-minus decay. Cu-64 lists a positron energy, so it definitely emits that, and has branching ratios of 61%/39%, so you get a good mix.

[[Tycho?]]

Only in rare cases could you get opposite reactions from the same nuclide. Generally you have either too many or too few neutrons, which means you get one or the other. But if you had stable nuclides that had both one more and one fewer proton, you could. It would most likely be a nuclide with an odd number of both neutrons and protons.

 

Cu-64, Ga-70 and Rb-86 are examples that both beta-plus (or possibly electron capture) and beta-minus decay. Cu-64 lists a positron energy, so it definitely emits that, and has branching ratios of 61%/39%, so you get a good mix.

 

Hmm, neat.

[YT2095]

so anyway, does this collision give off 2 Gamma photons?

 

and is that responsible for the blue glow Radium gives off?

[swansont]

so anyway, does this collision give off 2 Gamma photons?

 

and is that responsible for the blue glow Radium gives off?

 

electron-positron annihilation does give off 2 (or sometimes 3) gammas, but these wouldn't be blue. The blue glow is likely from ionization and recombination of the air near the source, caused by the emitted particles. N₂⁺ gives off blue light

[YT2095]

Great, thnx

 

now that this is confirmed (the book I`ve been looking at it quite old!).

 

I`m curious why it (and also you) said 2 and "Sometimes" 3 gammas?

what is it dependant upon?

also are the 2 gammas at a higher energy than when you get the 3 gammas?

and lastly (the trickier part), since the Electron has mass, and so does a Positron (both the same mass just different charges) and the 2 or 3 photons have no mass, does this energy released equal that amount of lost mass?

using the E-Mc^2 equasion.

[swansont]

Great, thnx

 

now that this is confirmed (the book I`ve been looking at it quite old!).

 

I`m curious why it (and also you) said 2 and "Sometimes" 3 gammas?

what is it dependant upon?

also are the 2 gammas at a higher energy than when you get the 3 gammas?

and lastly (the trickier part), since the Electron has mass, and so does a Positron (both the same mass just different charges) and the 2 or 3 photons have no mass, does this energy released equal that amount of lost mass?

using the E-Mc^2 equasion.

2 vs 3 gammas depends on the spin orientation of the electron and positron. If they are both spin up, total spin is 1, and you can't get that with only two gammas.

 

The energy is the mass energy of the particles, or 0.511 MeV each if you get two gammas. (since momentum must also be conserved the two gammas are the same energy, in the rest frame of the e-/e+ pair)

[YT2095]

so it`s a bit like 2 cars for instance, driving in the same direction at 100mph along side each other, the total speed is 100, but if they doing 100mph each and heading towards each other the collision is 200mph and you get 3 gammas so to speak.

[swansont]

It's not energy, it's conservation of angular momentum. Two photons can't give you one unit of angular momentum, since they have one unit each.

[YT2095]

I was considering the cars as being the Electron and the Positron.

 

if both are "Up" (traveling in the same direction) you get 2 Gammas.

if one is "Up" and the other "Down" (traveling in opposite directions) you get 3 gammas as there`s more energy in the collision.

 

would that be correct?

[swansont]

I was considering the cars as being the Electron and the Positron.

 

if both are "Up" (traveling in the same direction) you get 2 Gammas.

if one is "Up" and the other "Down" (traveling in opposite directions) you get 3 gammas as there`s more energy in the collision.

 

would that be correct?

 

There isn't more energy in the collision. It's strictly about being unable to conserve angular momentum with two gammas.

[Norman Albers]

When you collide things to smithereens (where's that word from?) the total spinning, yeah angular momentum of the pieces must equal what went in. So must total energy and momentum, and several other things. This works in accelerator collision kinematics as well as demolition derby.