How Do I Solve This Radioactive Decay Problem?

[Sanosuke Sagara]

I have my doubt,solution and question in the attachment that followed.Thanks for anybody that spend some time on this question.

 

[Astronuc]

The problem is a bit difficult to follow.

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_o exp(-\lambdat), where A_o is the initial activity and \lambda is the decay constant = ln 2/t_1/2. So at the beginning of the text t=30, and at end of test t=90. Also A(t) = \lambda N(t) where N is the number of atoms of the radionuclide.

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

 

[Sanosuke Sagara]

Thanks for your spending time in this question and I will try to understand better for the question before doing calculation.

 

[Astronuc]

Ah, I think I understand part 'd'.

First of all, activity (A) is a product of specific activity (A_s) and mass M. Specific activity is simply activity A divided by the mass M.

Remember that activity is simply \lambda*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 10⁷Bq and a piston mass of 1.6 kg, and assuming that the radionuclide is uniformly distributed, this gives a specific activity of 3.5625 x 10⁷Bq/kg. Assuming the mass of the piston to be constant, the specific activity decreases as the same rate as the activity. However, we know that the piston is losing mass to wear! On the other hand, the wear rate is very small, m(loss) << 1.6 kg. The activity of lost material at anytime is 620 Bq compared to the initial activity of 5.7 x 10⁷Bq.

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_s * M or expressed as a function of time, A = A_s(t) * M(t). In order for A to be constant, M(t) must have a form such that the rate of loss of M (conversely, the rate of accumulation of metal lost) must offset the rate of decrease of the specific activity.

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) = \int_{30}^{90} \dot{m} dt.

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.

 

[Sanosuke Sagara]

Thanks for your help ,Astronuc for your detail explanation that make me understand better with the question and I will try to solve the question by myself after given the 'important clue' to this question.