^60Co is an example of a radioactive isotope used in cancer therapy. For a speci

Question

^60Co is an example of a radioactive isotope used in cancer therapy. For a specific manufactured product type, just after a sample of^60Co has been manufactured, it has an activity of 5000 Ci (1 Ci = 1 Curie = 3.70 times 10^10 decays/second). When the sample's activity has dropped below 3500 Ci, the sample is considered too weak to be used. The half-life of^60Co is 5.271 years, (a) If a specific sample was manufactured on October 6, 2013, and it is now April 6, 2016, what is the activity of the sample? (b) Is it still usable?

Explanation / Answer

(a)

half life of Co-60 is t1/2 = 5.271 years

So decay constant is = ln2/t1/2 = (0.693)/(5.271y)

initial activity is A0 = 5000 Ci, then the final activity after time t is given as

A = A0e-t

So activity after t = 2.5 years is,

A = (5000 Ci)e-[(0.693)/(5.271y)]2.5y

or A = 3600 Ci

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(b)

As we saw above Activity A is 3600 Ci which is till above the threshold value of 3500 Ci.

So, the sample is still usable.

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