Chapter 3, Problem 79RE

Solution

(a)

To calculate: Substance remaining in function.

  $S(t)=S_0\cdot 2^{-t/1.5}$

Given information: 1.5 sec.

Calculation: If the mass halves every 1.5 seconds, then after t seconds, it will have halved $\frac{t}{1.5}$ . Hence, the initial mass $s_0$ is multiplied by $\frac{1}{2}$ a number of times equal to $\frac{t}{1.5}$ after t seconds:

  $S(t)=S_0\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \cdot \cdot \frac{1}{2}=S_0\cdot {\left(\frac{1}{2}\right)}^{t/1.5}$

We can simplify the formula as

  $S_0\cdot \frac{1}{2^{t/1.5}}=S_0\cdot 2^{-t/1.5}$

(b)

To calculate: Number of guppies after 1.5 sec and after 3 sec.

After 1.5 sec: $150\%$

After 3 sec: $25\%$

Given information: 1.5 sec.

Calculation: if the half-life of the substance is 1.5 seconds, then by definition, $50\%$ will remain after 1.5 seconds have passed.

After 3 seconds, two half-lives will have passed, and so $50\%\times 50\%=25\%$ will remain after 3 seconds.

(c)

To calculate: 1g after 1 min .

  $1,099,512$ metric tons.

Given information: 1.5 sec.

Calculation: in 79a we found the formula for radioactive decay as

  $S(t)=S_0\cdot 2^{-t/1.5}$

Where initial mass in grams and mass after t seconds.

If there was 1 gram after 60 seconds, then

  $S(60)=1$

We substitute our formula:

  $S_0\cdot 2^{-60/1.5}=1$

Simplify:

  $S_0\cdot 2^{-40}=1$

Now we will multiply both sides by $2^{40}$ :

  $\begin{array}{c}{S}_0\cdot 2^{-40}\cdot 2^{40}=1\cdot 2^{40}\\ S_0=2^{40}\end{array}$

And so the initial mass was

  $\begin{array}{c}{S}_0\cdot 2^{-40}\cdot 2^{40}=1\cdot 2^{40}\\ S_0=2^{40}\end{array}$