Chapter 43, Problem 18P

(a)

To determine

The differential equation that defines the number of daughter nucleus.

Answer to Problem 18P

The daughter nucleus satisfies the differential equation is $\frac{d{N}_2}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$.

Explanation of Solution

Let $N_{10}$ be the number of parent nuclei at time $t=0$, $N_1(t)$ be the number of parent nuclei at any instant $t$ and ${\lambda }_1$ is the decay constant of the parent nuclei. The number of daughter nucleus at time $t=0$ is zero, $N_2(t)$ be the number of daughter nuclei at any instant $t$ and ${\lambda }_2$ is the decay constant of the daughter nuclei.

Write the expression for decay rate

  $\frac{dN}{dt}=\lambda N$                                                                                                  (I)

Here, $\frac{dN}{dt}$ is the rate of decay, $\lambda$ is the decay constant and $N$ is the number of nuclei.

Write the expression for rate of change of the daughter nuclei

  $\frac{d{N}_2}{dt}=\frac{d{N}_{2+}}{dt}-\frac{d{N}_{2-}}{dt}$                                                                                                  (II)

Here, $\frac{d{N}_2}{dt}$ is the rate of change of the daughter nuclei, $\frac{d{N}_{2+}}{dt}$ is the rate of production of the daughter nuclei and $\frac{d{N}_{2-}}{dt}$ is the rate of decay of the daughter nuclei.

The rate of decay of the parent nucleus is same as the rate of production of the daughter nuclei.

Substitute $\frac{d{N}_{1-}}{dt}$ for $\frac{d{N}_{2+}}{dt}$ in (II)

  $\frac{d{N}_2}{dt}=\frac{d{N}_{1-}}{dt}-\frac{d{N}_{2-}}{dt}$                                                                                                 (III)

Conclusion:

Substitute ${\lambda }_1{N}_1$ for $\frac{d{N}_{1-}}{dt}$  and ${\lambda }_2{N}_2$ for $\frac{d{N}_{2-}}{dt}$ in (III) to find $\frac{d{N}_2}{dt}$.

  $\frac{d{N}_2}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$                                                                                                (IV)

Thus, the daughter nucleus satisfies the differential equation is $\frac{d{N}_2}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$.

(b)

To determine

The solution to the above differential equation using the verification by substitution method.

Answer to Problem 18P

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

Explanation of Solution

Write the expression for trail solution

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

Differentiate the above equation with respect to $t$

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

Multiply equation (V) by ${\lambda }_2$ and add it with (V)

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

Write the expression for $N_1$

  $N_1=N_{10}{e}^{-{\lambda }_{1}t}$                                                                                                 (VIII)

Conclusion:

Solve (VII) further

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

Substitute (VIII) in the above equation and rearrange.

  $\frac{d{N}_2}{dt}={\lambda }_1{N}_1-{\lambda }_2{N}_2$

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

(c)

To determine

The number of $\text{P}\text{o}$ and $\text{P}\text{b}$ nuclei versus time graph.

Answer to Problem 18P

The number of $\text{P}\text{o}$ and $\text{P}\text{b}$ nuclei versus time graph is given below:

Explanation of Solution

Write the expression for decay constant

  $\lambda =\frac{\ln 2}{T_{1/2}}$                                                                                                             (IX)

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

Substitute $3.10\text{min}$ for $T_{1/2}$ to find ${\lambda }_1$. Here, ${\lambda }_1$ is the decay constant of $\text{P}\text{o}$

  $\begin{array}{c}{\lambda }_1=\frac{\ln 2}{3.10\text{ min}}\\ =0.2236{\text{min}}^{-1}\end{array}$

Substitute $26.8\min$ for $T_{1/2}$ to find ${\lambda }_2$. Here, ${\lambda }_2$ is the decay constant of $\text{P}\text{b}$

  $\begin{array}{c}{\lambda }_2=\frac{\ln 2}{26.8\text{ min}}\\ =0.0259{\text{min}}^{-1}\end{array}$

Substitute $0.2236{\text{min}}^{-1}$ for ${\lambda }_1$ and $1000$ for $N_{10}$ in (VIII) to find $N_1$

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

Substitute $0.0259{\text{min}}^{-1}$ for ${\lambda }_2$, $0.2236{\text{min}}^{-1}$ for ${\lambda }_1$ and $1000$ for $N_{10}$ in (V) to find $N_2$

  $\begin{array}{c}{N}_2=\frac{\left(1000\right)\left(0.2236{\text{min}}^{-1}\right)}{\left(0.2236{\text{min}}^{-1}-0.0259{\text{min}}^{-1}\right)}\left(e^{-\left(0.0259{\text{min}}^{-1}\right)t}-e^{-\left(0.2236{\text{min}}^{-1}\right)t}\right)\\ =-1131\left(e^{-\left(0.0259{\text{min}}^{-1}\right)t}-e^{-\left(0.2236{\text{min}}^{-1}\right)t}\right)\end{array}$                   (XI)

Conclusion:

Using expression (X) and (XI), table the number of $\text{P}\text{o}$ and $\text{P}\text{b}$ nuclei for various $t$ values

$t\text{ (min)}$	$N_1$	$N_2$
0	1000	0
2	445	350
4	408	557
6	261	673
8	167	730
10	107	752
12	68.3	751
14	43.7	737
16	27.9	715
18	17.9	689
20	11.4	660
22	7.30	631
24	4.67	602
26	2.99	573
28	1.91	545
30	1.22	519
32	0.781	493
34	0.499	468
36	0.319	445

Using the above data in the table construct the graph

Thus, the above graph shows the number of nuclei as a function of time.

(d)

To determine

The instant when the number of $\text{P}\text{b}$ is maximum inferred from the above graph.

Answer to Problem 18P

The number of $\text{P}\text{b}$ is maximum in the time interval $t=10\text{min}$ to $t=12\text{min}$.

Explanation of Solution

From the above graph, the $\text{P}\text{b}$ is maximum in the time interval $t=10\text{min}$ to $t=12\text{min}$.

(e)

To determine

An expression for maximum number of $\text{P}\text{b}$ in terms of ${\lambda }_1$ and ${\lambda }_2$.

Answer to Problem 18P

The time for maximum $\text{P}\text{b}$ is given by the expression, $t_{\text{m}}=\frac{\ln \left({\lambda }_1/{\lambda }_2\right)}{{\lambda }_1-{\lambda }_2}$.

Explanation of Solution

Equate (VI) to zero to find the maximum

  $\frac{d{N}_2}{dt}=\frac{N_{10}{\lambda }_1}{{\lambda }_1-{\lambda }_2}\left(-{\lambda }_2{e}^{-{\lambda }_{2}t}+{\lambda }_1{e}^{-{\lambda }_{1}t}\right)=0$                                                                 (XII)

Conclusion:

Simplify and rearrange for $t$

  $\begin{array}{c}{\lambda }_1{e}^{-{\lambda }_{1}t}={\lambda }_2{e}^{-{\lambda }_{2}t}\\ t=t_{\text{m}}=\frac{\ln \left({\lambda }_1/{\lambda }_2\right)}{{\lambda }_1-{\lambda }_2}\end{array}$                                                                                       (XIII)

Here, $t_{\text{m}}$ is the time when the number of $\text{P}\text{b}$ is maximum.

Thus, time for maximum $\text{P}\text{b}$ is given by the expression, $t_{\text{m}}=\frac{\ln \left({\lambda }_1/{\lambda }_2\right)}{{\lambda }_1-{\lambda }_2}$.

(f)

To determine

The instant when the number of $\text{P}\text{b}$ is maximum calculated using the above expression.

Answer to Problem 18P

The time when the number of $\text{P}\text{b}$ is maximum is $10.9\text{min}$.

Explanation of Solution

The time for maximum $\text{P}\text{b}$ is given in the above part.

Conclusion:

Substitute $0.0259{\text{min}}^{-1}$ for ${\lambda }_2$ and $0.2236{\text{min}}^{-1}$ for ${\lambda }_1$ in (XIII) to find $t_{\text{m}}$

  $\begin{array}{c}{t}_{\text{m}}=\frac{\ln \left({\lambda }_1/{\lambda }_2\right)}{{\lambda }_1-{\lambda }_2}\\ =\frac{\ln \left(0.2236{\text{min}}^{-1}/0.0259{\text{min}}^{-1}\right)}{0.2236{\text{min}}^{-1}-0.0259{\text{min}}^{-1}}\\ =10.9\text{min}\end{array}$

Thus, the time when the number of $\text{P}\text{b}$ is maximum is $10.9\text{min}$ and it agrees with the answer in part (c).