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Average mass of iron worn off the ring per day

  1. Oct 29, 2016 #1
    1. The problem statement, all variables and given/known data
    When iron is irradiated with neutrons an isotope of iron is formed. This isotope is radioactive with a half-value period (half-life) of 45 days. Give the meanings of the terms printed in italics.

    A steel piston ring of mass 16 g was irradiated with neutrons until its activity due to the formation of this isotope was 10 microcurie. Ten days after the irraditation the ring was installed in an engine and after 80 days continuous use the crankcase oil was found to have a total activity of 1.85 * 103 disintegrations per second. Determine the average mass of iron worn off the ring per day assuming that all the metal removed from the ring accumulated in the oil and that one curie is equivalent to 3.7 * 1010 disintegrations per second.

    Answer: 4.0 mg per day.

    2. The attempt at a solution
    First of all I calculated the decay constant λ:

    A = A0 e- λ t
    1850 = 370 000 e- λ 90 days or 7 776 000 s
    λ = 6.8 * 10-7 s-1

    I then calculated the number of atoms at t = 0 and t = 90 days:
    dN / dt = - λ N
    N0 = 370 000 / 6.8 * 10-7 = 5.4 * 1011 atoms.
    N90 = 1850 / 6.8 * 10-7 = 2 720 588 235 atoms

    Then I found the relative atomic number Ar: N = m NA / Ar → Ar = m NA / N = 16 * 6 * 1023 / 5.4 * 1011 = 1.78 * 1013

    Then I found the mass at t = 90: m = Ar N / NA = 1.78 * 1013 * 2 720 588 235 / 6 * 1023 = 0.08 g.

    And now find the difference 16 g - 0.08 g = 15.92 g mass worn off in total. 15.92 / 90 = 0.177 g per day worn off.

    Not 0.004 g per day, what's wrong?
     
  2. jcsd
  3. Oct 29, 2016 #2

    gneill

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    Staff: Mentor

    The decay constant stems from the half life, not the activities of different quantities of the material. The material in the oil is not all of the remaining material: most of the ring still exists as the ring; only some scrapings end up in the oil. Use the isotope half-life to find the decay constant.

    I think that you would profit from using Specific Activity (curies per gram) in this problem. The SA decays over time just like activity and number of particles.
     
  4. Oct 29, 2016 #3
    λ = ln 2 / 45 * 24 * 3600 = 1.78 * 10-7 s-1.

    What else shall I find?
     
  5. Oct 29, 2016 #4

    gneill

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    Investigate the hint (or suggestion rather) that I gave in post #2.

    The specific activity of a material decays over time and doesn't depend upon the size of the sample. If you cut a piece off a thing with a given SA, that piece will also have the same SA, and the SA of the original and the piece will follow identical decay curves; their SA's will always match.
     
  6. Oct 29, 2016 #5
    I just didn't want to introduce a concept that isn't covered in my textbook (since I assume I'm expected to solve the problem using generic formulas). But the formula doesn't look that difficult, so SA is a = NA ln 2 / m T1 / 2 = 6.685 * 1015 Bq / g (m = 16 g).

    We have t = 0 days dN / dt = 370 000 Bq and t = 10 + 80 = 90 days dN / dt = 1850 Bq.

    370 000 * 6.685 * 1015 = 2.47345 * 1018 g
    1850 * 6.685 * 1015 = 1.236725 * 1019 g

    Doesn't really look like I'm going in the right direction.
     
  7. Oct 29, 2016 #6

    gneill

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    Specific activity is curies per gram. You're given the initial activity in microcuries and the mass of the ring in grams. No need to for anything fancy at all. Just one divided by the other.

    Then all you need to do is "age" the sample by the decay formula. As I said, SA decays in exactly the same way as other radiation values. So what's the material's SA after 90 days?
     
  8. Oct 29, 2016 #7
    a = 6.685 * 1015 Ci / g
    1 * 10-5 * 6.685 * 1015 = 6.685 * 1010 g at t = 0.

    Update
    You mean just 1 * 10-5 Ci divide by 16 g = 6.25 * 10-7 Ci / g? But here it says a = ... the formula in my previous post.

    SA = SA0 e- λ t?

    Update 2
    SA90 = 6.25 * 10-7 * e- (ln 2 / 3 888 000) * 7 776 000 = 1.56 * 10-7 Ci / g.
    m90 = Activity in Ci / SA90 = (1850 / 3.7 * 1010) / 1.56 * 10-7 = 0.32 g.

    16 - 0.32 = 15.68 g.

    15.68 / 90 = 0.174 g per day, same as I found in post 1.
     
    Last edited: Oct 29, 2016
  9. Oct 29, 2016 #8

    gneill

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    Specific activity is curies per gram. So yes, initially it is (10 μCi)/(16 g). The web site you sited goes into alternative ways to find SA when you're dealing with moles of substances rather than mass.

    And your formula above will "age" the SA over time. So that's good.
     
  10. Oct 29, 2016 #9
    I get the same answer as in post # 1.
     
  11. Oct 29, 2016 #10

    gneill

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    1. They asked for the amount of material lost from the ring, not the amount of material left in the ring.
    2. The machine was not in operation for all 90 days.
     
  12. Oct 29, 2016 #11

    gneill

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    Your m90 above is the amount of the ring material in the oil....
     
  13. Oct 29, 2016 #12
    But m90 is the mass on day 90. The one which is left. So 15.68 g is lost and then we divide by days and get how much is lost per day.

    We had 100 apples on day 1, we have 30 apples on day 30.

    100 - 30 = 70 apples are eaten. So 70 apples / 30 days = 2.3 apples were consumed per day.
     
  14. Oct 29, 2016 #13

    gneill

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    No, the activity given to you for day 90 is the activity of the oil. That's where the abraded ring material ended up.
    The ring was already 10 days old when it was put into operation.
     
  15. Oct 30, 2016 #14
    T1 / 2 iron = 3 888 000 s
    mring day 0 = 16 g
    dN / dt = 370 000 Bq
    dN / dt in 90 days = ?
    t = 7 776 000 s

    dN / dt oil = 1850 Bq

    I find final activity of iron at t = 90 days:
    A = A0 e- (ln 2 / 3 888 000) * 7 776 000 = 370 000 e- (ln 2 / 3 888 000) * 7 776 000 = 92 500 Bq of iron at t = 90 day.

    The total activity is 92 500 + 1850 = 94 350 Bq. This is activity of the ring and oil at t = 90.

    dN / dt = - λ N
    N = 5.29 * 1011 atoms at t = 90 day.
    N = 2.08 * 1012 atoms at t = 0 day.

    N = (m / Ar) * NA
    Find Ar first: Ar = m NA / N = 16 * 6 * 1023 / 2.08 * 1012 = 4.63 * 1012.
    Find mass at t = 90: m = N Ar / NA = 5.19 * 1011 * 4.63 * 1012 / 6 * 1023 = 4.08 g.

    So mass on day 0 = 16 g, mass on day 90 = 4.08 g, mass lost 16 - 4.08 = 11.9 g, mass lost in 90 days = 0.13 g per day.
     
  16. Oct 30, 2016 #15

    gneill

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    No, the total activity of all the iron no matter where it is located is 92 500. The 1850 is the part of it that was abraded and ended up in the oil. That is the part you want to find the mass of.

    The ring itself will still have a mass close to 16 g after 80 days of operation. It only loses a few milligrams per day to abrasion.
     
  17. Oct 30, 2016 #16
    How to find it?

    I don't think that changing 94 350 to 92 500 will lead to a large difference + I don't see a different way to calculate the answer from what I have in # 14.

    Maybe something like 1850 / 92 500 = 0.02. And, if you say that 16 g will remain the same, then 16 * 0.02 = 0.32 g / 80 = 4 * 10-3 g.

    This looks like the answer but the problem in my opinion is too complicated in its phrasing. In my texbook I have a similar problem (after this one) but regarding volume and I solved it relatively fast. I guess that one was better since I could imagine the process and see the logic of the process, while here I only somehow understand what happened (some iron material was placed in oil and some of it was worn off over time and was left in the oil so the total mass didn't change but the activity did).

    Thank you!
     
    Last edited: Oct 30, 2016
  18. Oct 30, 2016 #17

    gneill

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    Yes, that is correct.
    The problem may seem complicated in its presentation, but it represents a real-life scenario. This is one way that part wear can be monitored over time without dismantling the machine to make direct observation and measurements of the part.
     
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