Q24.40 P - Chemistry: Molecular Nature Of Matter And Change

Question

If $\mathbf{6}\mathbf{.40}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{ }\mathbf{\text{mol}}$  of  ${}^{\mathbf{176}}\mathbf{ }\mathbf{\text{W}}$ emits$\mathbf{1}\mathbf{.07}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}{\mathit{\beta }}^{\mathbf{+}}$ in  $\mathbf{1}\mathbf{.00}\mathbf{ }\mathbf{\text{h}}$, what is the decay constant?

Step-by-Step Solution

Answer

The decay constant is $0.325 {\text{h}}^{-1}.$

Step 1: 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: Calculation of the decay constant

The initial amount of  ${}^{176} \text{W}$is as follows:

 $\begin{array}{c}{\text{N}}_0=6.40\times {10}^{-9} \text{mol}\times \frac{6.022\times {10}^{23}\text{ atoms }}{1 \text{mol}}\\ =3.854\times {10}^{15}\text{ atoms .}\end{array}$

After  $t=1.00 \text{h},1.07\times {10}^{15}$ atoms decayed, therefore

 $\begin{array}{c}{\text{N}}_{\text{t}}={\text{N}}_0-1.07\times {10}^{15}\\ =3.854\times {10}^{15}\text{ atoms }-1.07\times {10}^{15}\text{ atoms }\\ =2.784\times {10}^{15}\text{ atoms.}\end{array}$

Now, using this equation, compute the decays constant.

 $\begin{array}{c}\ln \left(\frac{{\text{N}}_{\text{t}}}{{\text{N}}_0}\right)=-\text{kt}\\ \text{k}=\frac{-\ln \left(\frac{{\text{N}}_{\text{t}}}{{\text{N}}_0}\right)}{\text{t}}\\ =\frac{-\ln \left(\frac{2.784\times {10}^{15}\text{ atoms }}{3.854\times {10}^{15}\text{ atoms }}\right)}{1 \text{h}}\\ =0.325 {\text{h}}^{-1}.\end{array}$

Therefore, the computed value is  $0.325 {\text{h}}^{-1}.$