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Acceleration of Ion Seperation

  1. Sep 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Ok, so this question is not really a homework problem but just something I'm trying to solve for fun and to improve my problem solving skills.

    You must 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 copper plates. Each plate is 5.0 cm long, 4.0 cm wide, and the two plates are seperated by 3.0 cm. A high voltage is applied to the plates causing the ions passing between them to have a constant acceleration directly toward one of the plates and away from the other plate. You boss asks you to calculate the magnitude of acceleration between the plates necessary to seperate ions with a velocity of 100 m/s from those in the beam going 1000 m/s by 2.0 cm when they hit the ion detector 50 cm away. Before the ions enter the gap between the plates, they are going directly toward the center of the gap parallel to the surfaces of the plates. After the ions leave the gap between the plates, they are no longer accelerated during the remaining 50 cm to the ion detector. You look at the ion detector and find its face be circle with a radius of 7 cm.

    2. Relevant equations
    ##X_f = X_0 + V_0t + (1/2)at^2##

    3. The attempt at a solution
    IMG_20150922_212801965.jpg

    Ignore the 48.04 cm. I realize that's wrong. Same thing with the 2.3. That shouldn't be there.

    So I figured that a lot of the information they gave was irrelevent, such as the dimensions of the plates and the radius of the detector. I used Pythagorean theorem to find the velocity of ##X_{2c}## after it left the plates,

    ##X_{2c} = \sqrt(50^2 + 2^2)##

    I just don't know where to go from here. I've been staring at this problem for hours.
     
  2. jcsd
  3. Sep 25, 2015 #2

    mfb

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    If something is moving with 1000 m/s to the left, how much time does it take to travel 50 cm? To get deflected by some distance (let's say 5 cm as example, you don't know the value you need yet), what is its vertical velocity?
     
  4. Sep 25, 2015 #3
    0.0005 seconds. But the velocity is 1000 m/s before going through the plates.
     
  5. Sep 25, 2015 #4

    mfb

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    Do you expect the horizontal velocity component to change?
     
  6. Sep 25, 2015 #5
    Good point.

    ## X_{f_y} = X_{0_y} + V_{0_y}t + (1/2)at^2 ##
    0.05 m = 0 + 1000 m/s(0.0005s) + (1/2)a(0.0005)^2
    a = 3600000 m/s^2

    ## V_{f_y} = V_{0_y} + at##
    = 1000 m/s + (-3600000 m/s^2)(0.0005 s)
    = -800 m/s
     
  7. Sep 25, 2015 #6

    mfb

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    There is no acceleration over the 50 cm distance.
    There is acceleration over the 5 cm of plates before, however.

    This acceleration does not happen in the direction of the 1000 m/s velocity, however. You are mixing the two components.
     
  8. Sep 25, 2015 #7
    I just realized that. a = 400000 m/s^2
    Vy = 200 m/s

    Edit: never mind, that's wrong too.
     
  9. Sep 25, 2015 #8

    mfb

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    Now you can include the displacement that happens within the plates if you like. It is a small correction (~5%).
     
  10. Sep 25, 2015 #9
    You mean the values in my last post were not wrong?
     
  11. Sep 25, 2015 #10

    mfb

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    Oh right, 200 m/s is wrong.
    There is no acceleration in the 50cm drift distance. Acceleration gets relevant in the step afterwards.
     
  12. Sep 25, 2015 #11
    0.05 m/0.0005 s = 100 m/s. That's vertical velocity, right?
     
  13. Sep 25, 2015 #12
    0.05 + 0.02 = 0.07 m
     
  14. Sep 25, 2015 #13
    Using Pythagorearn theorem, ##\sqrt(1000^2 + 100^2)## = 1005 m/s

    Now I have to find the time between the plates.
     
  15. Sep 25, 2015 #14
    (1005 m/s - 1000 m/s)/0.00005 s

    I think I did it! a = 100000 m/s^2 Is that right?
     
    Last edited: Sep 25, 2015
  16. Sep 25, 2015 #15

    mfb

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    You can edit your posts instead of making a new one if you want to add something.
    Correct if you neglect the deflection that happens between the plates. Which is fine if you just want to find a rough estimate for yourself, it would be wrong in a test.
    What does that mean?

    The total velocity does not matter.
    That calculation doesn't make sense.

    Also keep in mind that I introduced the 5 cm deflection purely as an example. You'll have to find the actual deflection later.
     
  17. Sep 25, 2015 #16
    Alright so I found time for the 50cm drift which is 0.0005 s. The vertical velocity is 100 m/s. I'm not sure where to go from here.

    Edit: Time during plates is 0.00005 s.
     
    Last edited: Sep 25, 2015
  18. Sep 25, 2015 #17

    mfb

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    Well, let's keep that approximation of no deflection between the plates.
    During that timescale, the vertical velocity changes from 0 m/s to 100 m/s. What is the acceleration?
     
  19. Sep 25, 2015 #18
    a = (0 - 100 m/s)/0.00005 s = -2000000 m/s^2
     
  20. Sep 25, 2015 #19

    mfb

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    Writing it as 2*106 m/s^2 is probably easier.

    What is the next step?
     
  21. Sep 25, 2015 #20
    Find deflection which would be change in vertical position.

    (1000)(0.00005)-.5(2*10^6)(0.00005)^2 = 0.0475 m

    So velocity is 950 m/s?
     
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