Engineering problem (acceleration of ions)

[catalina]

1. Acceleration between plates

You have a summer job as an assistant in a University research group that is designing a devise to sample atmospheric pollution. In this device, it is useful to separate fast moving ions from slow moving ones. To do this the ions are brought into the device in a narrow beam so that all of the ions are going in the same direction. The ion beam then passes between two parallel metal plates. Each plate is 5.0 cm long, 4.0 cm wide and the two plates are separated by 3.0 cm. A high voltage is applied to the plates causing the ions between them to have a constant acceleration directly toward one of the plates and away from the other plate. Before the ions enter the gap between the plates, they are no longer accelerated during the 50 cm journey to the ion detector. Your boss asks you to calculate the magnitude of acceleration between the plates necessary to separate ions with a velocity of 100 m/s from those in the beam going 1000 m/s by 2.0 cm?

2. constant acceleration formulas, parabolic path:

y= y0 + vy0 *t + 1/2(a)t^2
x= x0 + vx0 *t




3. Time during which ions accelerate (calculated with the equation: x=x0 + V0xt , for each ion):
t1=5*10^-5 t2=5*10^-4

 

[LowlyPion]

Welcome to PF.

The 2 cm distance is at 50 cm.

What component of velocity is required to get the slow ion to deflect farther by 2 cm by the time it reaches 50 cm than the fast one?

You know V⊥ = a*t where t is the time the particle is subjected to acceleration between the plates.

That velocity times the time T to get to 50 cm at the original speed then would be the deflection I'd think for each particle. The additional constraint will be that 1 is 2 cm further along when it gets there. One distance will be deflected 2 cm more than the other.

 

[nlightner]

4. Acceleration between plates:

To calculate the acceleration between the plates, we can use the formula a= (Vf-Vi)/t, where Vf is the final velocity, Vi is the initial velocity, and t is the time during which the ions accelerate between the plates.

For the ions with a velocity of 100 m/s, the final velocity (Vf) is 0 m/s since they are stopped by the plates. The initial velocity (Vi) can be calculated using the formula x=x0 + V0xt, where x is the distance between the plates (2 cm), x0 is the initial distance (0 cm), and t is the time during which the ions accelerate (t1=5*10^-5). This gives us an initial velocity of 4000 m/s for the ions with a velocity of 100 m/s.

For the ions with a velocity of 1000 m/s, the final velocity (Vf) is also 0 m/s. The initial velocity (Vi) can be calculated in the same way, giving us an initial velocity of 40000 m/s.

Plugging these values into the acceleration formula, we get a= (0-4000)/5*10^-5 = -80,000,000 m/s^2 for the ions with a velocity of 100 m/s and a= (0-40000)/5*10^-4 = -800,000,000 m/s^2 for the ions with a velocity of 1000 m/s.

Therefore, the magnitude of acceleration between the plates necessary to separate these two ions is 800,000,000 m/s^2. This high acceleration will cause the slower ions to be deflected more than the faster ions, resulting in a separation between the two groups as they exit the plates.