Chapter 30, Problem 58GP

To determine

To find: Radioactive isotope of lead.

Answer to Problem 58GP

  $\text{P}\text{b}$

Explanation of Solution

Given info:

Measured activity, $\mathrm{R}\left(\mathrm{T}\right)=1.05\mathrm{\%}\text{ of }{\mathrm{R}}_0=0.0105{\mathrm{R}}_0$

Time interval, $\mathrm{t}=4.00\text{ h}$

Formula used:

Half-life can be given as:

  $\mathrm{\lambda }=\frac{0.693}{{\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-\frac{1}{\mathrm{t}}\ln \left(\frac{\mathrm{R}\left(\mathrm{T}\right)}{{\mathrm{R}}_0}\right)$

Here, ${\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is the half-life and ${\mathrm{R}}_{\mathrm{o}}$ is the initial activity.

Calculation:

Substituting all the values in the above expression to find the half-life:

  $\begin{array}{l}\frac{0.693}{{\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-\frac{1}{\mathrm{t}}\ln \left(\frac{\mathrm{R}\left(\mathrm{T}\right)}{{\mathrm{R}}_0}\right)\\ \frac{0.693}{{\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-\frac{1}{4.00\text{ h}}\ln \left(\frac{0.0105{\mathrm{R}}_0}{{\mathrm{R}}_0}\right)\\ {\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=0.6084\times \text{60 min}\\ {\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=36.50\text{ min}\end{array}$

Conclusion:

Thus, the radioactive isotope of lead is $\text{P}\text{b}$ .