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?

Short answer

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

Step-by-step solution

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.

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 ${\text{N}}_{\text{0}}\text{=2.50g}$

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

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

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

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

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