Chapter 11.6, Problem 60E

a.

To determine

Find the constant $k$ .

Answer to Problem 60E

  $0.000427$ .

Explanation of Solution

Given information:

Chemistry Radium $226$ , which had been used for cancer treatment, decays with a half-life of $1622$ years. Using $1$ gram as the original amount of Radium $226$ .

Calculation:

Consider the formula for half life,

  $\begin{array}{l}{t}_{\frac{1}{2}}=\frac{0.693}{k}\\ k=cons\tan t\\ k=\frac{0.693}{t_{\frac{1}{2}}}\\ put{t}_{\frac{1}{2}}=1622inequationk=\frac{0.693}{t_{\frac{1}{2}}}weget,\\ k=\frac{0.693}{1622}\\ k=0.000427\end{array}$

Hence, the constant $k$ is $0.000427$ .

b.

To determine

Find the time to have $1.7gram$ remaining.

Answer to Problem 60E

  $707years$ .

Explanation of Solution

Given information:

Chemistry Radium $226$ , which had been used for cancer treatment, decays with a half-life of $1622$ years. . Suppose there is a $2.3gram$ sample of Radium $226$ .

Calculation:

Consider the formula,

  $N=N_o{\left(\frac{1}{2}\right)}^t$

Here $N$ is the final value, $N_o$ is the initial value, $t$ is the time period,

  $\begin{array}{l}N=N_o{\left(\frac{1}{2}\right)}^t\\ \frac{N}{N_o}={\left(\frac{1}{2}\right)}^t\\ taking\log onbothsides,\\ \log \frac{N}{N_o}=\log {\left(\frac{1}{2}\right)}^t\\ \log \frac{N}{N_o}=t\log \left(\frac{1}{2}\right)\\ t=\frac{\log \frac{N}{N_o}}{\log \left(\frac{1}{2}\right)}\end{array}$

Substitute $N=1.7,N_o=2.3$ we get,

  $\begin{array}{l}t=\frac{\log \frac{1.7}{2.3}}{\log \left(\frac{1}{2}\right)}\\ t=\frac{\log \left(0.73913\right)}{\log \left(0.5\right)}=\frac{-01313}{-0.3010}\\ t=0.4366\\ so,timetakenis,\\ \left(no.oftimeperiod\right)\times \left(halflife\right)\\ \left(0.4366\right)\times \left(1622\right)=707years\end{array}$

Hence, the time to have $1.7gram$ remaining is $707years$ .