Cobalt-60 is used in radiotherapy; a 100 GBq source is delivered to a hospital on 1/1/1999;
i. How many gamma rays will the source produce in a 15 second treatment period?
ii. if the half-life of cobalt-60 is 5.27 years, what is the activity of the source on 1/9/2004?
iii. If a treatment took 15 seconds on 1/1/1999 how long would it take on 1/9/2004?
Please see the attached file.
i) The activity (rate of disintegration) of the sample is 100 G Bq (1 Bq = 1 dis./sec) means that 100*109 atoms disintegrates per second and the same is the rate of emission of the gamma rays. Hence the number of gamma ray photons emitted is 15 second is given by
n = 100*109*15 = ...
Radioactivity in radiotherapy with Cobalt-60 is examined. The half-life of cobalt-6 and the activity source are determined.