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Measuring the mass of tritium

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data

    An isotope of hydrogen passes, without deflection, through a velocity selector that has an electric field of 2.40 × 10^5 N/C and a magnetic field of 0.400 T. It then enters a mass spectrometer that has an applied magnetic field of 0.494 T and consequently describes a circular path with a radius of 3.80 cm.
    (a) What is the mass of the particle?

    2. Relevant equations

    v (velocity) = E/B

    where E = electric field intensity
    B = magnetic field intensity

    .5mv^2 = qV

    V = kq/r


    3. The attempt at a solution

    step 1. find the velocity

    v = E/B

    (2.4 * 10^5)/.4 = 6 * 10^5

    step 2. find voltage

    I tried to find q and V, then find the mass but I couldn't so I'm working backwards, starting with the answer and trying to find V

    .5mv^2 = qV

    .5* (5.01*10^-27)*(6*10^5)^2 = 1.6 * 10^-19 * V

    step 3 get V by itself

    (.5* (5.01*10^-27)*(6*10^5)^2)/(1.6 * 10^-19) = V

    which is

    1.44 * 10^20

    step 4 look for an equation which will give me V (electric potential difference)

    V = kq/r

    ((9*10^9)(1.6*10^-19)/.238 =

    I forgot what that number is, but it did not add up to 1.44 * 10^20, so there's something I'm doing wrong. Let's try a different equation.

    E (electric field intensity) = ΔV/Δd

    step 5 E * d = V

    The E with a magnetic field intensity of .494 is 2.9*10^5 so

    2.9*10^5 * .238 =

    I also forget what that number is but it also does not add up to around 10^20
     
  2. jcsd
  3. Jan 16, 2012 #2
    Hold on, I redid my calculation and found the volts to be 5625
     
  4. Jan 16, 2012 #3
    Well, I'm also there

    step 5 E * d = V

    The E with a magnetic field intensity of .494 is 2.9*10^5 so

    2.9*10^5 * .238 = 69020, not that far away from 5625
     
  5. Jan 16, 2012 #4
    Any ideas on this one?
     
  6. Jan 16, 2012 #5
    you have the velocity of the isotope = 6 x 10^5 m/s as it enters the magnetic field.
    The force on the isotope due to the magnetic field is Bqv where q will be 1.6 x 10^-19 C
    this force equals the centripetal force = (mv^2)/r
    equating Bqv = (mv^2)/r gives Bq =mv/r
    if you substitute into this equation you should get m = 5 x 10^-27kg
     
  7. Jan 16, 2012 #6
    The force on the isotope due to the magnetic field is Bqv where q will be 1.6 x 10^-19 C

    But given the information in the question, how do I know that the charge of the isotope will be as you say? If it's because a hydrogen isotope's by definition has one charge then that will be my answer but i'm not sure that's true.
     
  8. Jan 16, 2012 #7
    You are correct. Hydrogen has only 1 electron to lose and therefore must be singly + charged.
     
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