Chapter 13, Problem 58P

(a)

To determine

Prove that $\frac{d{N}_2\left(t\right)}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$.

Answer to Problem 58P

The relation $\frac{d{N}_2\left(t\right)}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$ proved.

Explanation of Solution

$\frac{d{N}_2\left(t\right)}{dt}$ the time rate of change of daughter nuclei. It is equal to the difference  of production rate and decay rate of $N_2$. It is equivalent to difference in decay rate of $N_1$ $N_2$.

    $\begin{array}{c}\frac{d{N}_2}{dt}=P_{N_2}-R_{N_2}\\ =R_{N_1}-R_{N_2}\\ ={\lambda }_1{N}_1-{\lambda }_2{N}_2\end{array}$

Here, $P_{N_2}$ the production rate of $N_2$, $R_{N_2}$ the decay rate of $N_2$ $R_{N_1}$ the decay rate of $N_1$.

Conclusion:

Thus, the relation $\frac{d{N}_2\left(t\right)}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$ proved.

(b)

To determine

Prove that $N_2\left(t\right)=\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left(e^{-{\lambda }_{2}t}-e^{-{\lambda }_{1}t}\right)$ is the solution of differential equation in part(a).

Answer to Problem 58P

It is proved that $N_2\left(t\right)=\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left(e^{-{\lambda }_{2}t}-e^{-{\lambda }_{1}t}\right)$ is the solution.

Explanation of Solution

Write the given trial solution.

    $N_2\left(t\right)=\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left(e^{-{\lambda }_{2}t}-e^{-{\lambda }_{1}t}\right)$

Differentiate the above equation with respect to t.

    $\begin{array}{c}\therefore \frac{d{N}_2}{dt}=\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left(-{\lambda }_2{e}^{-{\lambda }_{2}t}+{\lambda }_1{e}^{-{\lambda }_{1}t}\right)\\ \frac{d{N}_2}{dt}+{\lambda }_2{N}_2=\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left(-{\lambda }_2{e}^{-{\lambda }_{2}t}+{\lambda }_1{e}^{-{\lambda }_{1}t}+{\lambda }_2{e}^{-{\lambda }_{2}t}-{\lambda }_2{e}^{-{\lambda }_{1}t}\right)\\ =\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left({\lambda }_1-{\lambda }_2\right)e^{-{\lambda }_{1}t}\\ ={\lambda }_1{N}_1\end{array}$

For the relation to be valid, $\frac{d{N}_2\left(t\right)}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$ needed.

Conclusion:

Thus, it is proved that $N_2\left(t\right)=\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left(e^{-{\lambda }_{2}t}-e^{-{\lambda }_{1}t}\right)$ is the solution.

(c)

To determine

Plot the variations of $N_1\left(t\right)$ for ${}^{218}\text{Po}$ $N_2\left(t\right)$ for ${}^{214}\text{Pb}$ and find the time at which ${}^{214}\text{Pb}$ maximum number of nuclei.

Answer to Problem 58P

Graph is plotted and the nuclei number is maximum at 10.9 minute.

Explanation of Solution

Write the function for $N_1\left(t\right)$.

    $N_1\left(t\right)=1000e^{-\left(0.2236{\min }^{-1}\right)t}$

Write the function for $N_2\left(t\right)$.

    $N_2\left(t\right)=1130.8\left[e^{-\left(0.2236{\min }^{-1}\right)t}-e^{-\left(0.0259{\min }^{-1}\right)t}\right]$

Plot the above two functions.

From the graph, approximate time which nuclei number is maximum is 10.9 minute.

Conclusion:

Thus, the graph is plotted and the nuclei number is maximum at 10.9 minute.

(d)

To determine

Obtain the symbolic formula for $t_m$ using the condition $\frac{d{N}_2}{dt}=0$ in terms of ${\lambda }_1\text{and}{\lambda }_2$ check whether the value in part (c) matches with this equation or not.

Answer to Problem 58P

Equation is $t_m=\frac{\ln \left({\lambda }_1/{\lambda }_2\right)}{{\lambda }_1-{\lambda }_2}$ the value given by the equation matches with result of part (c).

Explanation of Solution

Write the trial solution given in part (b).

    $N_2\left(t\right)=\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left(e^{-{\lambda }_{2}t}-e^{-{\lambda }_{1}t}\right)$

Differentiate the above equation with respect to t.

    $\therefore \frac{d{N}_2}{dt}=\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left(-{\lambda }_2{e}^{-{\lambda }_{2}t}+{\lambda }_1{e}^{-{\lambda }_{1}t}\right)$

Apply the condition $\frac{d{N}_2}{dt}=0$.

$\begin{array}{c}-{\lambda }_2{e}^{-{\lambda }_{2}t}+{\lambda }_1{e}^{-{\lambda }_{1}t}=0\\ {\lambda }_2{e}^{-{\lambda }_{2}t}={\lambda }_1{e}^{-{\lambda }_{1}t}\\ e^{\left({\lambda }_1-{\lambda }_2\right)t}=\frac{{\lambda }_1}{{\lambda }_2}\end{array}$

Apply ln function on both sides.

  $\begin{array}{c}\ln e^{\left({\lambda }_1-{\lambda }_2\right)t}=\ln \frac{{\lambda }_1}{{\lambda }_2}\\ t=\frac{\ln \left({\lambda }_1/{\lambda }_2\right)}{{\lambda }_1-{\lambda }_2}\end{array}$

Conclusion:

Substitute $0.2236{\min }^{-1}$ ${\lambda }_1$ and $0.0259{\min }^{-1}$ ${\lambda }_2$ the above equation.

    $\begin{array}{c}t=\frac{\ln \left(\frac{0.2236{\min }^{-1}}{0.0259{\min }^{-1}}\right)}{0.2236{\min }^{-1}-0.0259{\min }^{-1}}\\ =10.9\min \end{array}$

The calculated value agrees with result of part (c).

Thus, the equation is $t_m=\frac{\ln \left({\lambda }_1/{\lambda }_2\right)}{{\lambda }_1-{\lambda }_2}$ the value given by the equation matches with result of part (c).