Chapter 30, Problem 70A

To determine

To calculate:

Time taken by the isotope to decay $100\text{g}$ so that $3.1\text{g}$ of $Po-209$ was left.

Answer to Problem 70A

It will take $515$ years to decay sample of $100g$ so that $3.1\text{g}$ of $Po-209$ was left.

Explanation of Solution

Given:

For a isotope $Po$ ,

Initial mass $=100\text{g}$

Final mass $=3.1\text{g}$

Half life period $=103$ years.

Formula Used:

The formula to calculate the number of half life years is given by,

  $m_{final}=m_{initial}\times {\left(\frac{1}{2}\right)}^t$

Where,

  $m_{final}:$ Final mass of sample

  $m_{initial}:$ Initial mass of sample

  $t:$ Number of half life years

Calculation:

The formula used ,

  $m_{final}=m_{initial}\times {\left(\frac{1}{2}\right)}^t$

Substitute all the values in the given formula,

  $\begin{array}{l}\text{3}\text{.1}=100\times {\left(\frac{1}{2}\right)}^{\left(t\right)}={\left(\frac{1}{2}\right)}^{\left(t\right)}=\frac{3.1}{100}\\ {\left(\frac{1}{2}\right)}^{\left(t\right)}=0.031\end{array}$

Taking Log on both sides,

  $\begin{array}{l}Ln{\left(\frac{1}{2}\right)}^{\left(t\right)}=Ln\left(0.031\right)\\ t\times Ln\left(\frac{1}{2}\right)=Ln\left(0.031\right)\\ t=\frac{Ln\left(0.031\right)}{Ln\left(\frac{1}{2}\right)}=t=5\end{array}$

Thus it takes $5$ years

Isotope $Po$ , has half life of $103\text{years}$ , thus

  $5\times \left(103\right)=515\text{years}$

Conclusion:

Hence, it will take $5$ years to decay sample $100\text{g}$ of so that $\text{3}\text{.1}\text{g}$ of $Po-209$ was left.