Chapter 44, Problem 75AP

To determine

The half-life of isotope X.

Answer to Problem 75AP

The half-life of isotope X is $2.66\text{d}$.

Explanation of Solution

Write the relation connecting the initial amount of the isotopes.

    $N_{\text{X}}\left(0\right)=2.50N_{\text{Y}}\left(0\right)$                                                                                           (I)

Here, $N_{\text{X}}\left(0\right)$ is the initial amount of isotope X and $N_{\text{Y}}\left(0\right)$ is the initial amount of isotope Y.

Write the relation connecting the amount of the isotopes after 3 days.

    $N_{\text{X}}\left(3\text{d}\right)=4.20N_{\text{Y}}\left(3\text{d}\right)$                                                                                      (II)

Here, $N_{\text{X}}\left(3\text{d}\right)$ is the amount of isotope X after 3 days and $N_{\text{Y}}\left(3\text{d}\right)$ is the amount of isotope Y after 3 days.

Write the equation for the amount of isotope left after 3 days.

    $N\left(3\text{d}\right)=N\left(0\right)e^{-\lambda t}$                                                                                          (III)

Here, $\lambda$ is the decay constant and $t$ is the time taken.

Write the equation for decay constant.

    $\lambda =\frac{0.693}{T_{1/2}}$                                                                                                        (IV)

Here, $T_{1/2}$ is the half-life.

Conclusion:

Rewrite equation (II) using equation (III).

    $N_{\text{X}}\left(0\right)e^{-{\lambda }_{\text{X}}\left(3\text{d}\right)}=4.20N_{\text{Y}}\left(0\right)e^{-{\lambda }_{\text{Y}}\left(3\text{d}\right)}$

Here, ${\lambda }_{\text{X}}$ is the decay constant for isotope X and ${\lambda }_{\text{Y}}$ is the decay constant for isotope Y.

Substitute equation (I) in the above equation and rearrange to find $e^{{\lambda }_{\text{X}}\left(3\text{d}\right)}$.

    $\begin{array}{c}{N}_{\text{X}}\left(0\right)e^{-{\lambda }_{\text{X}}\left(3\text{d}\right)}=4.20\frac{N_{\text{X}}\left(0\right)}{2.50}{e}^{-{\lambda }_{\text{Y}}\left(3\text{d}\right)}\\ e^{-{\lambda }_{\text{X}}\left(3\text{d}\right)}=\frac{4.20}{2.50}{e}^{-{\lambda }_{\text{Y}}\left(3\text{d}\right)}\\ e^{{\lambda }_{\text{X}}\left(3\text{d}\right)}=\frac{2.50}{4.20}{e}^{{\lambda }_{\text{Y}}\left(3\text{d}\right)}\end{array}$

Take natural logarithm on both sides and substitute equation (IV) to find $T_{1/2}$ of isotope X.

    $\begin{array}{c}\ln \left(e^{{\lambda }_{\text{X}}\left(3\text{d}\right)}\right)=\ln \left(\frac{2.50}{4.20}{e}^{{\lambda }_{\text{Y}}\left(3\text{d}\right)}\right)\\ \left(\frac{0.693}{T_{1/2\text{X}}}\right)\left(3\text{d}\right)=\ln \left(\frac{2.50}{4.20}\right)+\left(\frac{0.693}{T_{1/2\text{Y}}}\right)\left(3\text{d}\right)\end{array}$

Here, $T_{1/2\text{X}}$ is the half-life of X and $T_{1/2\text{Y}}$ is the half-life of Y.

Substitute $1.60\text{d}$ for $T_{1/2\text{Y}}$ to find $T_{1/2\text{X}}$.

    $\begin{array}{c}\left(\frac{0.693}{T_{1/2\text{X}}}\right)\left(3\text{d}\right)=\ln \left(\frac{2.50}{4.20}\right)+\left(\frac{0.693}{1.60\text{d}}\right)\left(3\text{d}\right)\\ \left(\frac{0.693}{T_{1/2\text{X}}}\right)\left(3\text{d}\right)=-0.5188+1.299\\ \left(\frac{0.693}{T_{1/2\text{X}}}\right)\left(3\text{d}\right)=0.7802\\ T_{1/2\text{X}}=2.66\text{d}\end{array}$

Thus, the half-life of isotope X is $2.66\text{d}$.