Chapter 29, Problem 68P

(a)

To determine

The ratio of the decay rates or activity at $t=0$

Answer to Problem 68P

The ratio of the decay rates or activity at $t=0$ is $4.0$.

Explanation of Solution

Two radioactive sample with equal number of nuclides A and B, $(N_A=N_B=N_0)$  at $t=0$ has half-life of $3.0\text{ h}$ and $12.0\text{ h}$ respectively.

Write the formula for half-life

$T_{1/2}=\tau \ln 2$                                                                                             (I)

Here, $T_{1/2}$ is the half-life and $\tau$ is the time constant.

Write the formula for the decay constant

$\lambda =\frac{1}{\tau }$                                                                                                    (II)

Here, $\lambda$ is the decay constant

Write the formula for the activity

$R=R_0{2}^{-t/T_{1/2}}$                                                                                             (III)

Here, $R$ is the activity after time $t$, $R_0$ is the activity at $t=0$ and $t$ is the time.

Write the formula for the activity

$R=\lambda N$                                                                                           (III)

Here, $N$ is the number of nuclei.

Conclusion:

Equating (IV) in (III) to solve for $R$

$R=R_0{2}^{-t/T_{1/2}}=\lambda N$

The ratio of activities of A and B is

$\frac{R_A}{R_B}=\frac{R_{A0}{2}^0}{R_{B0}{2}^0}=\frac{R_{A0}}{R_{B0}}=\frac{{\lambda }_A{N}_0}{{\lambda }_B{N}_0}=\frac{{\lambda }_A}{{\lambda }_B}$

Substitute equation (II)  and (I) in above equation

$\frac{R_A}{R_B}=\frac{{\tau }_B}{{\tau }_A}=\frac{T_B/\ln 2}{T_A/\ln 2}=\frac{T_B}{T_A}$

Substitute $3.0\text{ h}$ for $T_A$ and $12.0\text{ h}$ for $T_B$ in the above equation to find the ratio of $\frac{R_A}{R_B}$

$\frac{R_A}{R_B}=\frac{12.0\text{ h}}{3.0\text{ h}}=4.0$

Thus, the ratio of the decay rates or activity at $t=0$ is $4.0$.

(b)

To determine

The ratio of the decay rates or activity at $t=12.0\text{ h}$

Answer to Problem 68P

The ratio of the decay rates or activity at $t=12.0\text{ h}$ is $0.50$.

Explanation of Solution

The ratio of $\frac{R_A}{R_B}$ at $t=12.0\text{ h}$ is

$\frac{R_A}{R_B}=\frac{R_{A0}{2}^{-t/T_A}}{R_{B0}{2}^{-t/T_B}}$

Substitute $\frac{R_{A0}}{R_{B0}}=4.0$, $3.0\text{ h}$ for $T_A$ and $12.0\text{ h}$ for $T_B$ in the above equation to find the ratio of $\frac{R_A}{R_B}$ at $t=12.0\text{ h}$

$\begin{array}{l}\frac{R_A}{R_B}=4.0\times 2^{t\left(\frac{1}{T_B}-\frac{1}{T_A}\right)}\\ \frac{R_A}{R_B}=4.0\times 2^{12.0\text{ h}\times \left(\frac{1}{12.0\text{ h}}-\frac{1}{3.0\text{ h}}\right)}\\ \frac{R_A}{R_B}=0.50\end{array}$

Thus, the ratio of the decay rates or activity at $t=12.0\text{ h}$ is $0.50$.

(c)

To determine

The ratio of the decay rates or activity at $t=24.0\text{ h}$

Answer to Problem 68P

The ratio of the decay rates or activity at $t=24.0\text{ h}$ is $0.063$.

Explanation of Solution

The ratio of $\frac{R_A}{R_B}$ at $t=24.0\text{ h}$ is

$\frac{R_A}{R_B}=\frac{R_{A0}{2}^{-t/T_A}}{R_{B0}{2}^{-t/T_B}}$

Substitute $\frac{R_{A0}}{R_{B0}}=4.0$, $3.0\text{ h}$ for $T_A$ and $12.0\text{ h}$ for $T_B$ in the above equation to find the ratio of $\frac{R_A}{R_B}$ at $t=24.0\text{ h}$

$\begin{array}{l}\frac{R_A}{R_B}=4.0\times 2^{t\left(\frac{1}{T_B}-\frac{1}{T_A}\right)}\\ \frac{R_A}{R_B}=4.0\times 2^{24.0\text{ h}\times \left(\frac{1}{12.0\text{ h}}-\frac{1}{3.0\text{ h}}\right)}\\ \frac{R_A}{R_B}=0.063\end{array}$

Thus, the ratio of the decay rates or activity at $t=24.0\text{ h}$ is $0.063$.