24.133 CP - Chemistry: Molecular Nature Of Matter And Change

Question

A sample of cobalt-60 (t_1/2  = 5.27 yr), a powerful emitter used to treat cancer, was purchased by a hospital on March 1, 2007. The sample must be replaced when its activity reaches 70% of the original value. On what date must it be replaced?

Step-by-Step Solution

Answer

The replaced date is 11 November 2009.

Step 1: Definition of Emitter

A substance that emits electrons as a radioactive material is called an emitter. It is also classified as the “part of a transistor” where the charge-carrying holes or electrons come from.

Step 2: Calculation of the number of years

Calculate the rate constant using the provided half-life.

$$\begin{gathered} t_{1/2}=\frac{\ln 2}{k} \\ k=\frac{\ln 2}{5.27} \\ =0.132year{}^{-1} \end{gathered}$$

 

To calculate the time, use the rate equation.

$$\begin{gathered} \ln \frac{A_o}{A_t}=kt \\ t=\frac{\ln \frac{A_o}{A_t}}{k} \\ t=\frac{\ln \frac{1}{0.70}}{0.132year{s}^{-1}} \\ =2.70years \end{gathered}$$

Step 3: Calculation of the date

To calculate the replacement date, multiply the time by March 1, 2007.

 $2.70years\times \frac{365.25days}{year}=986.175days$

 

Therefore, the replaced date is March 1, 2007 + 986.175 days = 11 November, 2009.