Q24.39 P - Chemistry: Molecular Nature Of Matter And Change

Question

If  $\mathbf{1}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{ }\mathbf{\text{mol}}$of  ${}^{\mathbf{135}}\mathbf{\text{Cs}}$emits$\mathbf{1}\mathbf{.39}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}{\mathit{\beta }}^{\mathbf{-}}$  particles in $\mathbf{1}\mathbf{.00}\mathbf{\text{yr}}$ , what is the decay constant?

Step-by-Step Solution

Answer

The decay constant is$2.31\times {10}^7{\text{yr}}^{-1}$  .

Step 1: Definition of decay constant

The size of a population of radioactive atoms and the pace at which the population diminishes due to radioactive decay are proportionate.

 

Step 2: Calculate the decay constant

The initial amount of ${}^{135}\text{C}$ is:

 $\begin{array}{c}{\text{N}}_0=1.00\times {10}^{-12} \text{mol}\cdot \frac{6.022\times {10}^{23}\text{ atoms }}{1 \text{mol}}\\ =6.022\times {10}^{11}\text{ atoms .}\end{array}$

After  $t=1.00\text{yr},1.39\times {10}^5{}^{135}\text{Cs}$atoms decayed.

Therefore,

 $\begin{array}{c}{\text{N}}_{\text{t}}={\text{N}}_0-1.39\times {10}^5\\ =6.023\times {10}^{11}\text{atoms }-1.39\times {10}^5\text{atoms}\\ =6.02199861\times {10}^{11}\text{ atoms.}\end{array}$

Now, let us calculate the decay constant

  $\begin{array}{c}\ln \left(\frac{{\text{N}}_{\text{t}}}{{\text{N}}_0}\right)=-\text{kt}\\ \text{k}=\frac{-\ln \left(\frac{{\text{N}}_{\text{t}}}{{\text{N}}_0}\right)}{\text{t}}\\ =\frac{-\ln \left(\frac{6.02199861\times {10}^{11}\text{ atoms }}{6.022\times {10}^{11}\text{ atoms }}\right)}{1\text{yr}}\\ =2.31\times {10}^7{\text{yr}}^{-1}.\end{array}$

Therefore, the computed value is  $2.31\times {10}^7{\text{yr}}^{-1}.$