Solving Radioactive Decay Homework: Activity at 2pm

[retupmoc]

Homework Statement

I am looking for some guidance on how to tackle the following problem. Maybe I am a little confused as it is rather wordy.

It is known that an isotope, X, spontaneously decays by positron emission into
another isotope, Y, with a half life of 100 minutes. The isotope X is injected into a
patient and after a certain amount of time a sample of blood (containing some of the
radioactive isotope) is withdrawn. The blood sample is withdrawn at 2:27pm. The
sample is then left overnight and placed in a well counter at 10:24 the next day. The
sample is left in the well counter for 20 minutes, during which time 20,000 gamma
rays (each with an energy of 511 keV) are detected. Assuming that the well counter is
100% efficient (i.e. all of the gamma rays emitted by the sample are detected and
recorded as separate events) what was the activity of the sample (in terms of positron
decays per second) at 2pm the previous day?
NOTE: since the counting time is significant compared with the half life, the activity
at 10:24 am cannot be obtained by simply dividing the number of gamma rays
detected by the sampling time


The way I think I should tackle the problem is as follows:

As there are 20,000 decays in 20 minutes and the radioisotope is a positron emitter the activity at 10:24 am should be 10000/20 as 2 photons are emitted per decay with 100% detection efficiency.

I then would take into account the 100 min half life of the radionuclide using the equation A(t) = A(0)exp(-lambda*t)

The activity at 2:27 pm, A(0) = A(t)/exp{ln2*t/T_{half}}

where t is the time in minutes between 2:27pm and 10:24.

Once I know A(0) I can then find the activity at time 2pm by using t=-7mins in the initial equation although this should change the activity by a relatively small amount

Does my reasoning seem sound?

 

[jbsteele]

Homework Equations A(t) = A(0)exp(-lambda*t) The Attempt at a SolutionYes, your reasoning seems sound. You should be able to use the equation A(t) = A(0)exp(-lambda*t) to calculate the activity at 2pm. To do this, you need to first calculate the activity at 10:24am. Since there are 20,000 decays in 20 minutes and the radioisotope is a positron emitter, the activity at 10:24am should be 10000/20 as 2 photons are emitted per decay with 100% detection efficiency. Once you have this value, you can then take into account the 100 min half life of the radionuclide using the equation A(t) = A(0)exp(-lambda*t). Here, lambda is the decay constant and t is the time in minutes between 2:27pm and 10:24. Then, the activity at 2:27pm, A(0) can be calculated as A(t)/exp{ln2*t/T_{half}}. Finally, you can find the activity at time 2pm by using t=-7mins in the initial equation.