Q24.41 P - Chemistry: Molecular Nature Of Matter And Change

Question

The isotope  ${{}^{212}}_{83}\mathbf{Bi}$ has a half-life of $\mathbf{\text{1.01}}$ yr. What mass (in mg) of a $\mathbf{\text{2.00}}$ -mg sample will remain after ${\mathbf{\text{3.75×10}}}^{\mathbf{\text{3}}}\mathbf{\text{ h}}$ ?

Step-by-Step Solution

Answer

The amount of ${{}^{212}}_{83}\mathbf{Bi}$  that remains is $1.49\text{mg}$ 

 

Step 1: Radioactive decay

The radioactive decay follows the first order kinetics. The relation between the rate constant and the half-life of the reaction is as shown below:

 $\mathbf{k}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{693}}{{\mathbf{t}}_{1/2}}$

Step 2: Calculation

- The half-life of ${}_{\text{83}}^{\text{212}}{\text{Bi is t}}_{\text{1/2}}\text{=1.01yr}$ 

- The initial amount of ${}_{\text{83}}^{\text{212}}{\text{Bi was N}}_{\text{0}}\text{=2.00mg}$ 

We have to find the mass of   ${}_{83}^{212}\text{Bi}$after time  _{$\text{t}=3.75\cdot {10}^3 \text{h}$}

First, let us convert the time elapsed from hours to years

 _{$\text{t}=3.75\times {10}^3 \text{h}\times \frac{1 \text{d}}{24 \text{h}}\times \frac{1\text{yr}}{365 \text{d}}=0.428\text{yr}$}

Now, let us calculate the decay constant of  _{${}_{83}^{212}\text{Bi}$}

 _{$\text{k =}\frac{\text{ln(2)}}{{\text{t}}_{\text{1/2}}}\text{=}\frac{\text{0.693}}{\text{1.01yr}}{\text{=0.686yr}}^{\text{-1}}$}

Therefore, the amount of  ${{}^{212}}_{83}\mathbf{Bi}$ that remains after time t is

_{$\begin{array}{c}\text{ln}\left(\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\right)\text{ =-kt}\\ \text{ln}\left(\frac{{\text{N}}_{\text{t}}}{\text{2.00mg}}\right){\text{ =-0.686yr}}^{\text{-1}}\text{×0.428yr}\\ \text{=-0.2936}\end{array}$} 

_{\$\begin{array}{c}\frac{{\text{N}}_{\text{t}}}{\text{2.00mg}}{\text{ =e}}^{\text{-0.2936}}\\ \frac{{\text{N}}_{\text{t}}}{\text{2.00mg}}\text{ =0.7456 }\\ {\text{N}}_{\text{t}}\text{ =1.49mg}\end{array}$}

 

Hence,the amount of  ${{}^{212}}_{83}\mathbf{Bi}$ that remains is$1.49\text{\hspace{0.17em}mg}$  _.