Q31.5-56PE
Question
A 5000-Ci \({}^{60}Co\) source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci. How long after its manufacture does this happen?
Step-by-Step Solution
VerifiedThe Manufacture happens in 2.71 years.
Surgery, chemotherapy, radiation therapy, hormonal therapy, targeted therapy, and synthetic lethality can all be used to treat cancer.
The sample's initial activity is determined by
\({R_i} = \frac{{(0.693){N_i}}}{{{t_{1/2}}}}\) …(1)
Finally, the last activity is provided by
\({R_f} = \frac{{(0.693){N_f}}}{{{t_{1/2}}}}\) …(2)
When we divide equation (2) by equation (1), we get
\(\frac{{{R_f}}}{{{R_i}}} = \frac{{{N_f}}}{{{N_i}}}\) …(3)
We know from the radioactive decay equation that
\(\frac{{{N_f}}}{{{N_i}}} = {e^{ - \lambda t}}\) …(4)
As a result of equations (3) and (4), we can write (5)
\(\begin{array}{c}\frac{{{R_f}}}{{{R_i}}} = \frac{{{N_f}}}{{{N_i}}} = {e^{ - \lambda t}}\\\lambda t = \ln \left( {\frac{{{R_i}}}{{{R_f}}}} \right)\\t = \frac{{\ln \left( {\frac{{{R_i}}}{{{R_f}}}} \right)}}{\lambda }\\ = \frac{{\ln \left( {\frac{{{R_i}}}{{{R_f}}}} \right)}}{{0.693}}{t_{1/2}}\end{array}\)
Since we know that
\(\lambda = \frac{{0.693}}{{{t_{1/2}}}}\)
We now have a half-life of \(^{60}{\rm{Co }}is{t_{1/2}} = 5.271\,{\rm{y}}\)
Initial activity \({R_i} = 5000\,{\rm{Ci}}\)
Final activity \({R_f} = 3500\,{\rm{Ci}}\)
Now, if we replace these numbers in equation (5), we get
\(\begin{array}{c}t = \frac{{\ln \left( {\frac{{5000\,{\rm{Ci}}}}{{3500\,{\rm{Ci}}}}} \right)}}{{(0.693)}}(5.271\,{\rm{y}})\\ = 2.71\,{\rm{y}}\end{array}\)
Therefore, the Manufacture happen in 2.71 years.