Chapter 1, Problem 27RE

The half-life of palladium-100, ¹⁰⁰Pd, is four days. (So half of any given quantity of ¹⁰⁰Pd will disintegrate in four days.) The initial mass of a sample is one gram.

(a) Find the mass that remains after 16 days.

(b) Find the mass m(t) that remains after t days.

(c) Find the inverse of this function and explain its meaning.

(d) When will the mass be reduced to 0.01g?

To determine

To find: The quantity of mass that remains after 16days.

Answer to Problem 27RE

Solution:

The quantity of mass that remains after 16days is $\frac{1}{16}\text{g}$.

Explanation of Solution

Given:

The initial mass of the sample = 1g.

The half-life of Palladium-100 $\left({}^{100}Pd\right)$ = 4 days.

Calculation:

Let the mass be m.

The initial mass = m and the mass of Palladium-100 in 4days = $\frac{1}{2}m$.

The mass after 8days=$\frac{1}{2}\left(\frac{1}{2}m\right)={\left(\frac{1}{2}\right)}^{2}m=\frac{1}{4}m$.

The mass after 12days= $\frac{1}{2}{\left(\frac{1}{2}\right)}^{2}m={\left(\frac{1}{2}\right)}^{3}m=\frac{1}{8}m$.

The mass after 16days=$\frac{1}{2}{\left(\frac{1}{2}\right)}^{3}m={\left(\frac{1}{2}\right)}^{4}m=\frac{1}{16}m$.

Since the initial mass m = 1g, the quantity of mass after 16 days is, $\frac{1}{16}\left(1\text{g}\right)$.

Thus, the mass that remains after 16days is $\frac{1}{16}\text{g}$.

To determine

To find: The mass that remains after t days.

Answer to Problem 27RE

Solution:

The mass that remains after t days is $2^{\frac{-t}{4}}$.

Explanation of Solution

Given:

The initial mass of the sample=1g.

The half-life of Palladium-100 $\left({}^{100}Pd\right)$ = 4days.

Calculation:

From part (a), the mass after 16 days is, ${\left(\frac{1}{2}\right)}^{\frac{16}{4}}m=\frac{1}{16}\text{g}$.

The general form of the mass after n days is given by, ${\left(\frac{1}{2}\right)}^{\frac{n}{4}}$

Therefore, the mass remains after t days is ${\left(\frac{1}{2}\right)}^{\frac{t}{4}}$.

To determine

To find: The inverse of ${\left(\frac{1}{2}\right)}^{\frac{t}{4}}$ and explanation of its meaning.

Answer to Problem 27RE

Solution:

The inverse of the function is $f^{-1}\left(m\right)=-4{\log }_2\left(m\right)$ and the meaning of the inverse function in which the time is passed when m gram is left.

Explanation of Solution

Laws of logarithms:

If x and y are positive numbers, then

$\begin{array}{l}1.{\log }_b\left(x^r\right)=r{\log }_{b}x\\ 2.{\log }_b\left(b\right)=1\end{array}$

Calculation:

Let $m\left(t\right)=2^{\frac{-t}{4}}$ be the mass that remains after t days.

Then its inverse function is, $t=f^{-1}\left(m\right)$.

Here, $m=2^{\frac{-t}{4}}$.

Taking ${\log }_2$ on both sides and simplify as follows.

$\begin{array}{c}{\log }_2\left(m\right)={\log }_2\left(2^{\frac{-t}{4}}\right)\\ =\frac{-t}{4}{\log }_2\left(2\right)\\ =\frac{-t}{4}\left(1\right)\left[\text{by laws of logarithms}\right]\\ =\frac{-t}{4}\end{array}$

Simplify further as,

$\begin{array}{l}{\log }_2\left(m\right)=\frac{-t}{4}\\ -4{\log }_2\left(m\right)=t\end{array}$ $t=-4{\log }_2\left(m\right)$

Substitute t value in the equation (2)

Thus, $f^{-1}\left(m\right)=-4{\log }_2\left(m\right)$.

Therefore, the inverse of the function is $f^{-1}\left(m\right)=-4{\log }_2\left(m\right)$ it represents the time when is passed when m grams left.

To determine

To find: When the mass is reduced to 0.01g.

Answer to Problem 27RE

Solution:

The mass is reduced to 0.01g approximately in 26.6 days.

Explanation of Solution

Calculation:

From part (a), $m=2^{\frac{-t}{4}}$.

Substitute $m=0.01$ in $m=2^{\frac{-t}{4}}$,

$0.01=2^{\frac{-t}{4}}$

Taking $\log$ on both sides.

$\begin{array}{c}\log \left(0.01\right)=\log \left(2^{\frac{-t}{4}}\right)\\ \log \left({10}^{-2}\right)=\frac{-t}{4}\log \left(2\right)\\ -2\log \left(10\right)=\frac{-t}{4}\left(\log 2\right)\\ -2=\frac{-t}{4}\left(\log 2\right)\end{array}$

On further simplification,

$\begin{array}{l}\frac{-2\times -4}{\log 2}=t\\ t=\frac{8}{\log 2}\\ t\approx 26.6\end{array}$

Therefore, the mass is reduced to 0.01g approximately in 26.6 days.