Chapter 40, Problem 73P

a.

To determine

To Justify:The equation $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$

Explanation of Solution

Introduction:

Often, the daughter nucleus of a radioactive parent nucleus is itself radioactive. The number of daughter nuclei $N_D$ are given by the solution to the differential equation $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$

The rate of decay of D nuclei subtracted from the rate of change of $N_D$ is the rate of generation of D nuclei. The generation rate is equal to the decay rate if P nuclei, which equals ${\lambda }_P{N}_P$ . The decay rate of D nuclei is ${\lambda }_D{N}_D$ .

Conclusion:

The rate of change of $N_D$ is the rate of generation of D nuclei minus the rate of decay of D nuclei. The generation rate is equal to the decay rate if P nuclei, which equals ${\lambda }_P{N}_P$ . The decay rate of D nuclei is ${\lambda }_D{N}_D$ .

b.

To determine

To Show: The equation $N_D(t)=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$ where $N_{Po}$ is the number of parent nuclei present at $t=0$ when there are no daughter nuclei.

Answer to Problem 73P

  $N_D(t)=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$ is the solution of the equation $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$

Explanation of Solution

Given:

Parent nucleus decay constant ${\lambda }_P$

Daughter nucleus decay constant ${\lambda }_D$

  $N_D(t)=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$

Formula Used:

  $N_D(t)=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$

Calculations:

Differentiate $N_D(t)=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$ with respect to t

  $\frac{d}{dt}\left[N_D(t)\right]=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}\frac{d}{dt}(e^{-}{}^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}\left(-{\lambda }_p{e}^{-{\lambda }_{p}t}+{\lambda }_D{e}^{-{\lambda }_{D}t}\right)$

Substituting in $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$

  $\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}\left(-{\lambda }_p{e}^{{\lambda }_{p}t}+{\lambda }_D{e}^{-{\lambda }_{D}t}\right)=N_{Po}{\lambda }_P{e}^{-}{}^{{\lambda }_{p}t}-{\lambda }_D\left[\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}\left({\lambda }_p{e}^{-{\lambda }_{p}t}-{\lambda }_D{e}^{-{\lambda }_{D}t}\right)\right]$

Multiply both sides by $\frac{{\lambda }_D-{\lambda }_P}{{\lambda }_D{\lambda }_P}$ and simplify

  $\frac{N_{Po}}{{\lambda }_D}\left(-{\lambda }_p{e}^{-{\lambda }_{p}t}+{\lambda }_D{e}^{-{\lambda }_{D}t}\right)=\frac{{\lambda }_D-{\lambda }_P}{{\lambda }_D}{N}_{Po}{e}^{-}{}^{{\lambda }_{p}t}-N_{Po}(e^{-}{}^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$

  $=N_{Po}{e}^{-}{}^{{\lambda }_{p}t}-\frac{N_{Po}{\lambda }_P}{{\lambda }_D}{e}^{-{\lambda }_{p}t}-N_{Po}{e}^{-}{}^{{\lambda }_{p}t}+N_{Po}{e}^{-{\lambda }_{D}t}$

  $=-\frac{N_{Po}{\lambda }_P}{{\lambda }_D}{e}^{-{\lambda }_{p}t}+N_{Po}{e}^{-{\lambda }_{D}t}$

  $=\frac{N_{Po}}{{\lambda }_D}\left[-{\lambda }_p{e}^{-{\lambda }_{p}t}+{\lambda }_D{e}^{-{\lambda }_{D}t}\right]$

Which is an identity and confirms that $N_D(t)=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$ is the solution of the equation $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$

Conclusion:

  $N_D(t)=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$ is the solution of the equation $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$

c.

To determine

Show that the expression for $N_D$ in part b gives $N_D(t)>0$ whether ${\lambda }_P>{\lambda }_D$ or ${\lambda }_D>{\lambda }_P$ .

Explanation of Solution

Introduction:

Frequently, the daughter nucleus of a radioactive parent nucleus is itself radioactive. The number of daughter nuclei $N_D$ are given by the solution to the differential equation $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$

  $N_D(t)=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$ is the solution of the equation $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$

If ${\lambda }_P>{\lambda }_D$ , the denominator and $(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$ are both negative for $t>0$ . If ${\lambda }_P<{\lambda }_D$ the denominator and $(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$ are both positive for $t>0$ .

d.

To determine

Make a plot of $N_P(t)$ and $N_D(t)$ as a function of time when ${\tau }_D=3{\tau }_P$ .

Explanation of Solution

Introduction:

The number of daughter nuclei $N_D$ are given by the solution to the differential equation $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$ . $N_D(t)=\frac{N_{Po}{\lambda }_P}{{\lambda }_D-{\lambda }_P}(e^{{\lambda }_{p}t}-e^{-{\lambda }_{D}t})$ is the solution of the equation $\frac{d{N}_D}{dt}={\lambda }_P{N}_P-{\lambda }_D{N}_D$ .