Why the K.E. Increases in a 10-Stage Linac Accelerator with 200 kV Supply

[jsmith613]

Lets say we had a 10-stage 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?

 

[jsmith613]

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^-17J (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^-17J *2 = 9.6*10^-17J ?If not then what will it be?

 

[Bob S]

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^-17J (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^-17J *2 = 9.6*10^-17J ?

It is easier to do the calculations using eV (electron-volt) units. The electron mass in eV units is mc² = 511,000 eV. The kinetic energy is
$$KE=\frac{1}{2}mv^2=\frac{1}{2}\beta^2mc^2$$
where v=βc and c = 3 x 10⁸ meters per second. So
$$\beta=\sqrt{\frac{2\cdot KE}{mc^2}}$$
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 $L=\frac{\beta c}{2f}=$ 0.05141, 0.07269, .08903 meters etc.

 

[jsmith613]

It is easier to do the calculations using eV (electron-volt) units. The electron mass in eV units is mc² = 511,000 eV. The kinetic energy is
$$KE=\frac{1}{2}mv^2=\frac{1}{2}\beta^2mc^2$$
where v=βc and c = 3 x 10⁸ meters per second. So
$$\beta=\sqrt{\frac{2\cdot KE}{mc^2}}$$
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 $L=\frac{\beta c}{2f}=$ 0.05141, 0.07269, .08903 meters etc.

OK so a good way to see it is as follows:

tube 1 --> tube 2 - KE doubles
tube 2 --> tube 3 - KE * 1.5

so
Tube 1 = L
tube 2 = L*sqrt(2) = L-new
tube 3 = L-new * sqrt(1.5)

but this has answered by question so thanks a lot :)