Chapter 0.3, Problem 23E

a

To determine

Express the amount of substance remaining as function of time $t$ .

Answer to Problem 23E

The required expression is $N\left(t\right)=6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}$ .

Explanation of Solution

Given:

  $\begin{array}{c}{N}_0=6.6grams\\ t_{1/2}=14days\end{array}$

Concept Used:

The formula of half-life is:

  $N\left(t\right)=N_0{b}^t$

Where,

  $N\left(t\right)\to$ Remaining amount of substance

  $N_0\to$ Starting amount of substance

  $t\to$ Time elapsed

Calculation:

To find the formula for half-life.

  $\begin{array}{c}\frac{1}{2}{N}_0=N_0{b}^{14}\\ \frac{1}{2}=b^{14}\\ b={\left(\frac{1}{2}\right)}^{\frac{1}{14}}\end{array}$

Substitute the value of $b$ into the formula and the initial value 6.6 grams:

  $N\left(t\right)=6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}$

b

To determine

Time taken when the amount remain will be 1 gram.

Answer to Problem 23E

Time taken when the amount remain will be 1 gram is 38.1 days.

Explanation of Solution

Given:

  $N\left(t\right)=1gram$

Calculation:

Substitute the values in the formula:

  $\begin{array}{c}N\left(t\right)=6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}\\ 1=6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}\\ \frac{1}{6.6}={\left(\frac{1}{2}\right)}^{\frac{t}{14}}\end{array}$

Take natural logarithm to both sides:

  $\begin{array}{l}\ln \left(\frac{1}{6.6}\right)=\ln \left[{\left(\frac{1}{2}\right)}^{\frac{t}{14}}\right]\\ \ln \left(\frac{1}{6.6}\right)=\frac{t}{14}\ln \left(\frac{1}{2}\right)\\ t=14\left(\frac{\ln \left(\frac{1}{6.6}\right)}{\ln \left(\frac{1}{2}\right)}\right)\\ t\approx 38.1days\end{array}$

Hence, time taken when the amount remain will be 1 gram is 38.1 days.