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 :)

 

[pmacias]

The increase in kinetic energy (K.E.) in a 10-stage linac accelerator with a 200 kV supply can be attributed to the principle of energy conservation. In a linac accelerator, charged particles are accelerated by a series of electric fields created by alternating voltage sources. Each stage of the accelerator provides an additional electric field, allowing the particles to gain more kinetic energy as they pass through each stage.

In this scenario, the first stage of the accelerator is receiving a 200 kV supply. This means that each particle passing through this stage will gain 200 kV of potential energy. As the particles move through the first stage, this potential energy is converted into kinetic energy, resulting in a certain amount of K.E. at the end of the first stage.

When the particles move on to the second stage, they already have a certain amount of kinetic energy from the first stage. This K.E. is then added to by the second stage, which also receives a 200 kV supply. This results in a doubling of the K.E. compared to the first stage.

This process continues for each subsequent stage, with the particles gaining more and more kinetic energy as they pass through each stage. Therefore, at stage 2, the K.E. is twice that of stage 1, and this trend continues for each subsequent stage.

In summary, the increase in K.E. in a 10-stage linac accelerator with a 200 kV supply is due to the accumulation of energy from each stage as the particles progress through the accelerator. This demonstrates the efficient and effective nature of linac accelerators in providing high levels of kinetic energy to charged particles.