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Drift tube lengths 
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#1
May412, 10:06 AM

P: 614

Lets say we had a 10stage linac accelerator and we were using a 200 kV supply.
Apparently the K.E at stage 2 (in the second tube) is twice that of the first tube (stage 1). why is this? 


#2
May412, 10:16 AM

P: 614

In fact a better question is:
If an electron, starting at 0 m/s is accelerated across a p.d of 300V its new KE is 4.8 * 10^{17}J (E = QV). My question is, if I accelerate this electron (with this KE) through another p.d of 300V then, disregarding relativistic effects, will the new KE be 4.8 * 10^{17}J *2 = 9.6*10^{17}J ? If not then what will it be? 


#3
May412, 01:30 PM

P: 4,663

[tex] KE=\frac{1}{2}mv^2=\frac{1}{2}\beta^2mc^2 [/tex] where v=βc and c = 3 x 10^{8} meters per second. So [tex] \beta=\sqrt{\frac{2\cdot KE}{mc^2}} [/tex] So β = 0.03427, 0.04846, 0.05935, etc. for n=1, 2, 3, etc. and KE =n·300 eV. When the electron is inside a drift tube, it is completely shielded from the applied voltage. You have to use an ac voltage and a drift tube so that when the applied voltage is the wrong polarity, the drift tube shields the electron, and the electron only "sees" the correct polarity voltage. [added] If you use an ac frequency of f = 100 MHz, the drift tube lengths are [itex] L=\frac{\beta c}{2f}= [/itex] 0.05141, 0.07269, .08903 meters etc. 


#4
May512, 03:34 PM

P: 614

Drift tube lengths
tube 1 > tube 2  KE doubles tube 2 > tube 3  KE * 1.5 so Tube 1 = L tube 2 = L*sqrt(2) = Lnew tube 3 = Lnew * sqrt(1.5) but this has answered by question so thanks a lot :) 


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