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Pair Production 
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#1
Jul1603, 01:02 PM

P: 91

Problem:
A photon of energy E strikes an electron at rest and undergoes pair production, producing a positron and another electron. photon + e^{} > e^{+} + e^{} + e^{} The two electrons and the positron move off with identical linear momentum in the direction of the initial photon. All the electrons are relativistic. a) Find the kinetic energy of the final three particles. b) Find the initial energy of the photon. Express your answers in terms of the rest mass of the electron m_{o} and the speed of light c. I set the problem up this way: For part a) E is conserved before and after the collision. E_{photon} = 3E_{e} + 3KE_{e} hc/&lambda = 3m_{o}c^{2} + 3KE_{e} I then solved for KE_{e} and found: KE_{e} = 1/3hc/&lambda  m_{o}c^{2} Is this the correct method to find the kinetic energy of one of the electrons after the collision? The problem says "state your answer in the rest mass of the electron and the speed of light", however if I don't know the wave length of the photon, how do I eliminate that energy term. Since they say the electrons are moving relativistic, do I just assume their speed is equal to 1/10c and then work backward to find the photons wavelength? I may be way off, any help is appreciated! Thanks Edit: I need to use mc^{2}(&gamma  1) for KE! The biggest question I have is linear momentum conserved. I would assume it is, and the problem hints at linear momentum. 


#2
Jul1603, 03:27 PM

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PF Gold
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#3
Jul1603, 04:06 PM

P: 91

Is this okay?
E_{photon} + m_{o}c^{2}= 3m_{o}c^{2} + 3K_{e} E_{photon} = energy of the photon: hc/&lambda m_{o}c^{2} = rest mass energy of electron K_{e} = kinetic energy of the electrons in motion (all three with equal linear momentum and kinetic energy) after the photon collides with the initially motionless electron. 


#4
Jul1603, 04:17 PM

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PF Gold
P: 5,533

Pair Production
That's a good start, but you have to take it further. The problem said to express the answer in terms of m_{0} and c. That means you can't have λ in the answer. So now you need to conserve momentum and get a second equation.
Also, rather than write the total energy of each final particle as m_{0}c^{2}+K, I would write each one as γm_{0}c^{2}. That will make it easier to simultaneously solve the 2 equations. 


#5
Jul1603, 04:38 PM

P: 91

Is the algebra soposed to be really messy on this problem.[?]
I just want to make sure I'm not going through all this for nothing. 


#6
Jul1603, 04:43 PM

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PF Gold
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First, the 2 equations. hc/λ+mc^{2}=3γmc^{2} (Conservation of Energy) h/λ=3γmv (Conservation of Momentum) The clean way is to multiply the second equation by 'c', and move the mc^{2} to the right side in the first equation. Then, the left hand sides of the two equations are equal, so you can set the right hand sides equal to each other. Solve for v, then get the KE from that. Then use either equation (I'd use the second one because it's shorter) to find the initial energy of the photon. 


#7
Jul1603, 04:53 PM

P: 91

I ended up with a &gamma^{2}/[(&gamma +1)(&gamma1)] = 4/9mc^{2}, solving for v would have been fun. I bet I made some mistakes along the way, two pages of algebra and I write small. Plus that I tried to solve for K directly, then subsituted that and tried to go for v. It's ugly. BTW: I found v to be 2/3c 


#8
Jul1603, 05:17 PM

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PF Gold
P: 5,533

It seems you did make a mistake.
Here are your two equations, after the manipulations I described: hc/λ=(3γ1)mc^{2} (Conservation of Energy) hc/λ=3γmcv (Conservation of Momentum) You can straighforwardly solve for v by equating the right hand sides of each equation. For convenience, I would define β=v/c, and I get the following: (3γ1)=3γβ Also note that γ=1/(1β^{2})^{1/2}. You should be able to solve for β without much algebra. BTW, I got β=0.8, so v=0.8c. Edit: fixed math symbol 


#9
Jul1603, 05:20 PM

P: 91

Yeah I found my error. Reworking it right now.
I canceled &gamma, which you can't do. Since E of initial electron is only mc^{2} 


#12
Jul1603, 07:02 PM

P: 91

I get the K of electron after collision:
2/3mc^2 and for initial photon energy of: 4mc^2 If I didn't get those right, I'm quiting for the day. 


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