Q24.42 P - Chemistry: Molecular Nature Of Matter And Change

Question

The half-life of radium- $\text{226}$ is  $\mathbf{\text{1.60}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{3}}}$yr. How many hours will it take for a  $\mathbf{\text{2.50}}$g sample to decay to the point where $\mathbf{\text{0.185}}$ g of the isotope remains?

Step-by-Step Solution

Answer

The time taken is $5.267\times {10}^7 \text{hrs}$ _.

Step 1: Half-life

Half-life is the time taken by the radioactive substance to decay to half of its amount. The radioactive decay follows the first order kinetics.

 

Step 2: Calculation

The half-life of radium226 is${\text{t}}_{\text{1/2}}{\text{=1.60×10}}^{\text{3}}\text{yr}$  

The initial amount of radium-226  is ${\text{N}}_{\text{0}}\text{=2.50 g}$ 

The amount of radium-226  after time (t) is${\text{N}}_{\text{t}}\text{=0.185 g}$  

We have to calculate the elapsed time in hours (t).

First, let us calculate the decay constant of radium226

$\begin{array}{c}\text{k =}\frac{\text{ln(2)}}{{\text{t}}_{\text{1/2}}}\\ \text{ =}\frac{\text{0.693}}{{\text{1.60×10}}^{\text{3}}\text{yr}}\\ {\text{=4.33×10}}^{\text{-4}}{\text{yr}}^{\text{-1}}\end{array}$

Therefore, the time elapsed is

 $\begin{array}{c}\text{ln}\left(\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\right)\text{ =-kt}\\ \text{t =}\frac{\text{-ln}\left(\frac{{\text{N}}_t}{{\text{ N}}_{\text{0}}}\right)}{\text{k}}\end{array}$

$\begin{array}{c}\text{t=}\frac{\text{-ln}\left(\frac{\text{0.185 g}}{\text{2.50 g}}\right)}{{\text{4.33×10}}^{\text{-4}}\text{yr}}\\ {\text{ =6.013×10}}^{\text{3}}\text{yr}\\ {\text{ =6.013×10}}^{\text{3}}\text{yr×}\frac{\text{365 d}}{\text{1yr}}\text{×}\frac{\text{24 hrs}}{\text{1 d}}\\ {\text{ =5.267×10}}^{\text{7}}\text{ hrs}\end{array}$

 

Hence,It would take   $5.267\times {10}^7 \text{hrs}$