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- Thread starter Sanosuke Sagara
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Astronuc

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One is given an initial activity, at t=0. The piston is installed on day 30, and the engine tested for 60 days (to day 90).

The equation for activity A(t) is A(t) = A

I find part 'd' a bit confusing, but I need to look at it again.

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Astronuc

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First of all, activity (A) is a product of specific activity (A

Remember that activity is simply [itex]\lambda[/itex]*N, where N is the number of radionuclide atoms. Well the total N has a mass M = N*m where m is the mass of the radionuclide atom. This relationship (M = N * m) would work for a pure (100%) mass of the radionuclide, but in this case, the radioactive nuclide is disolved into the piston head.

However, on is given an initial activity 5.7 x 10

Now the problem states "During testing, any worn-off metal is found to have an activity of 620 Bq," i.e. the activity of the lost metal is constant during the test. But how can this be if the activity (and specific activity) is decreasing with time? Remember A = A

The total metal loss, is simply the integral over time (with appropriate limits 30-90 days) of the metal loss rate, i.e.

m(loss) = [itex]\int_{30}^{90} \dot{m} dt[/itex].

So to solve part 'd', one must determine the expression for the rate of mass loss.

Thanks for the intereseting problem. This represents a practical application of a tracer isotope, although the constant activity during the test is not realistic - hopefully the wear of a real piston does not increase exponentially.

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