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Speed, force, magnetic flux of an ion?

  1. Oct 27, 2016 #1
    1. The problem statement, all variables and given/known data
    The diagram shows a mass spectrometer used for measuring the masses of isotopes. It consists of an ion generator and accelerator, a velocity selector and an ion separator, all in a vacuum.

    af11bacf44df.jpg

    In one experiment, tin ions, each of which carries a charge of +1.6 * 10-19 C, are produced in the ion generator and are then accelerated by a p. d. of 20 000 V. Tin has a number of isotopes, two of which are tin-118 (118Sn) and tin-120 (120Sn).

    (a) (i) Assuming that an ion of tin-120 is at rest before being accelerated, show that the final speed after acceleration is 177 km s-1. Mass of nucleon = 1.7 * 10-27.
    (a) (ii) What will be the final speed of an ion of tin-118?

    (b) In practice all ions produced by the ion generator have a range of speeds. A velocity selector is used to isolate ions with a single speed. In the velocity selector the force produced by the electric field is balanced by that due to the magnetic field which is perpendicular to the plane of the paper.
    (b) (i) The plates producing the electric field have a separation of 2.0 cm. The potentials of the plates are marked on the diagram. What is the magnitude of the force on an ion due to this electric field in the velocity selector?
    (b) (ii) Write down the equation which must be satisfied if the ions are to emerge from the exit hole of the velocity selector. Define the terms in the equation.
    (b) (iii) What magnetic flux density is required if ions travelling with a speed of 177 km s-1 are to be selected?

    (c) After selection the ions are separated using a magnetic filed on its own, as shown in the diagram.
    (c) (i) Explain why the ions move in circular paths in this region.
    (c) (ii) Show that the radius of the path is directly proportional to the mass of the ion.
    (c) (iii) The ions are detected using the photographic plate P. Determine the distance between the points of impact on the photographic plate of the two isotopes of tin when a magnetic flux density of 0.75 T is used in the ion separator.

    Answers: (a) (ii) 179 m s-1, (b) (i) 3.2 * 10-15 N, (iii) 0.11 T, (c) (iii) 0.50 cm.

    2. The attempt at a solution
    (a) (i) (1 / 2) M v2 = Q V, where Q = 1.6 * 10-19 C, V = 20 000 V, v = ? and M = ?.

    We find M first: Relative atomic mass Ar = Mass of atom M / (1 / 12( the mass of a 126C atom u → Ar = M / u → M = Ar * u = 120 * 1.7 * 10-27 = 2.04 * 10-25 kg.
    So v = √ 2 Q V / M = √ 2 * 1.6 * 10-19 * 20 000 / 2.04 * 10-25 = 177 123 m s-1 or 177 km s-1.

    (a) (ii) Doing the same thing, find M first: M = Ar * u = 118 * 1.7 * 10-27 = 2.006 * 10-25 kg.

    Plug in: v = √ 2 Q V / M = √ 2 * 1.6 * 10-19 * 20 000 / 2.006 * 10-25 = 178 617.7 m s-1.

    I didn't continue since it's either a typo in my book answer, or I did something wrong. I get 179 km s-1, not 179 m s-1.
     
  2. jcsd
  3. Oct 27, 2016 #2

    cnh1995

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    I think that is indeed a typo. Final speed of tin-120 is 177km/s. So the answer should be in km/s and not m/s. Also, 179m/s practically seems too small (slower than a bullet shot from a gun).
     
  4. Oct 27, 2016 #3
    Good : ).

    I continued the work. For (b) (i) I need to find the force.
    So I used F = e E, where E = V / d so F = e V / d = 1.6 * 10-19 * 200 / 0.02 = 1.6 * 10-15 N. But the answer is two times larger. What's the reason?

    Update
    Maybe since the potentials are +200 and -200 we can assume that the electric field is being created by a scale of between +200, 0 and -200 or a 400 V in total. F = e V / d = 1.6 * 10-19 * 400 / 0.02 = 3.2 * 10-15 N
     
  5. Oct 27, 2016 #4

    cnh1995

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    Right.
     
  6. Oct 27, 2016 #5
    Great.

    Though I'm now struggling with the (b) (iii) and (c) (iii) answers.
    I can only think of formulas involving B like r = M v / Q B, where we don't have radius.

    As I understand we need to find the radius. I used the formula above: r = M v / Q B. So we have: B = 0.75, I took 120Sn so M = 2.04 * 10-25 kg, Q = 1.6 * 10-19 C and v is 177 000 m s-1. So r = 2.04 * 10-25 * 177 000 / 1.6 * 10-19 * 0.75 = 0.3009 m or 30.09 cm. Doesn't stand even close to 0.5 cm in the answer.

    What am I missing?
     
  7. Oct 27, 2016 #6

    cnh1995

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    That is when the ions enter the ion separator. I believe the question is asking about the selection of ions in the velocity selector. In the velocity selector, electric and magnetic forces balance each other. What is the expression describing this? You have solved a similar problem recently.
     
  8. Oct 27, 2016 #7

    cnh1995

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    This only gives the radius of tin-120 ion. The question is asking you the distance between two impact points. What is that distance in terms of the radii of the ions?
     
  9. Oct 27, 2016 #8
    e E = e v B?
    B = E / v = (400 / 0.02) / 177 000 = 0.11299435 T = 0.11 T?
     
  10. Oct 27, 2016 #9

    cnh1995

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    Yes.
     
  11. Oct 27, 2016 #10
    Hm, maybe I'm calculating wrong. Not sure.

    r1 = 0.3009 m or 0.3011091 m when I use 177 123 m s-1 instead of 177 000 m s-1.

    r2 = 2.006 * 10-25 * 179 000 (or 178 617.7) / 1.6 * 10-19 * 0.75 = 0.299 m (or 0.298 589 255 m).

    If I use rounded values like 177k and 179k I get the difference around 0.1 cm, while if I use 177 123 m / s and 178 617.7 m / s I get 0.25 cm. But still not 0.50 cm.

    What am I missing?
     
  12. Oct 27, 2016 #11

    cnh1995

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    You are off by a factor of two. This is a big clue. See the diagram carefully. Do you want to calculate the difference between the radii?
     
    Last edited: Oct 27, 2016
  13. Oct 27, 2016 #12
    Yes, isn't it the answer?
     
  14. Oct 27, 2016 #13

    cnh1995

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    No. Its the difference between the diameters. See the diagram carefully.
     
  15. Oct 27, 2016 #14
    Aaah, yes indeed! It's the difference between the diameters and not the radii. In that case we get 0.5 cm.

    Thank you, didn't expect to solve such a large problem so fast.
     
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