Chapter 30, Problem 65A

(a)

To determine

To calculate:

The percentage of original material left after $6$ days.

Answer to Problem 65A

The original material percentage left after $6$ days will be $25\%$ .

Explanation of Solution

Given:

For an isotope,

Half life cycle $\left(\frac{1}{2}\right)=3.0$ days

Formula used:

Calculatingthe percentage of original material using the formula of half life cycle.

  $A\left(t\right)=A_0\times {\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Where,

  $A\left(t\right)$:Amount of material left after years.

  $A_0$ : Initial amount of substance

  $t_{1/2}$:Half life cycle of the decaying material.

Calculation:

Calculating the percentage of original material using the formula of half life cycle.

In the first step,

  $A\left(t\right)=A_0\times {\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Substitute all the values in above formula,

  $\begin{array}{l}A\left(t\right)={\left(\frac{1}{2}\right)}^{\left(6/3\right)}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}\\ =0.25=25\%\end{array}$

Conclusion:

The percentage of theoriginal material left $6$ days after will be $25\%$ .

(b)

To determine

To calculate:

The percentage of original material left after $9$ days.

Answer to Problem 65A

The percentage of original material left after $9$ days is $12.5\%$ .

Explanation of Solution

Given:

For an isotope,

Half life cycle $\left(\frac{1}{2}\right)=3.0$ days

Formula used:

Calculating the percentage of original material using the formula of half life cycle.

  $A\left(t\right)=A_0\times {\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Where,

  $A\left(t\right)$: Amount of material left after years.

  $A_0$ : Initial amount of substance

  $t_{1/2}$: Half life cycle of the decaying material.

Calculation:

Calculating the percentage of original material using the formula of half life cycle.

In the first step,

  $A\left(t\right)=A_0\times {\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Substitute all the values in above formula,

  $\begin{array}{l}A\left(t\right)={\left(\frac{1}{2}\right)}^{\left(9/3\right)}={\left(\frac{1}{2}\right)}^3=\frac{1}{8}\\ =0.125=12.5\%\end{array}$

Conclusion:

The percentage of original material left after $9$ days is $12.5\%$ .

(c)

To determine

To calculate:

The percentage of original material left after $12$ days.

Answer to Problem 65A

The original material percentage left after $12$ days is $6.25\%$ .

Explanation of Solution

Given:

For an isotope,

Half life cycle $\left(\frac{1}{2}\right)=3.0$ days

Formula used:

Calculating the percentage of original material using the formula of half life cycle.

  $A\left(t\right)=A_0\times {\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Where,

  $A\left(t\right)$: Amount of material left after years.

  $A_0$ : Initial amount of substance

  $t_{1/2}$: Half life cycle of the decaying material.

Calculation:

To calculate the percentage of original material, using the formula of half life cycle,

In the first step,

  $A\left(t\right)=A_0\times {\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Substitute all the values in above formula

  $\begin{array}{l}A\left(t\right)={\left(\frac{1}{2}\right)}^{\left(12/3\right)}={\left(\frac{1}{2}\right)}^4=\frac{1}{16}\\ =0.0624=6.25\%\end{array}$

Conclusion:

The original material percentage left after $12$ days is $6.25\%$ .