Which processes are posible and why?

  • Thread starter Thread starter Ene Dene
  • Start date Start date
Ene Dene
Messages
47
Reaction score
0
1. Photon runs into motionless electron and gives the electron all of its energy
2. Fast positron interacts with motionless electron producing one photon

1. I know that if that electron were in atom, photon of some energy could give its energy, and electron would jump into a higher energy state. But what about free electron, as in this case? And why not if it's not possible?

2. positron and electron should anihilate producing photon, but I'm not sure is it 1 or two photons, and I don't know the argument why.
 
Physics news on Phys.org
If you calculate the momenta and energy using Conservation laws, you should be able to see whether they are possible or not.
 
1. Are p and E conserved?
2. Calculate E^2-p^2 for the initial and final state.
 
clem said:
1. Are p and E conserved?
p=(E(photon)+E(electron), p1+p2=0) before, after I should have E of electron that exceeds rest mass of electron, and no momentum, so the answer should be that process is not possible.

clem said:
2. Calculate E^2-p^2 for the initial and final state.
It should not be possible because photon should have some momentum (three component momentum) and since it doesn't have any mass, it can't be possible.

Are these correct explanations?
 
Ene Dene said:
p=(E(photon)+E(electron), p1+p2=0) before, after I should have E of electron that exceeds rest mass of electron, and no momentum, so the answer should be that process is not possible.


It should not be possible because photon should have some momentum (three component momentum) and since it doesn't have any mass, it can't be possible.

Are these correct explanations?
Neither is possible, but your reasoning is flawed.
For #1, look at the barycentric (or "center of mass") system where the total momentum is zero. The final state would have total energy mc^2, while the initial system would have a higher total energy.
For #2, one photon cannot be produced because there is no center of mass system for a single photon.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top