Chapter 29, Problem 38P

(a)

To determine

The activity of the radioactive nuclide after $600.00\text{ s}$.

Answer to Problem 38P

The activity of the radioactive nuclide after $600.00\text{ s}$ is $10000{\text{ s}}^{-1}$.

Explanation of Solution

A radioactive nuclide has half-life $T_{1/2}=800.0\text{ s}$ and initial activity $R_0=80000{\text{ s}}^{-1}$ undergoes decay for a period of $t=600.0\text{ s}$.

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 activity

$R=R_0{e}^{-t/\tau }$                                                                                           (II)

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

Conclusion:

Substitute equation (I) in (II) and rewrite to find the $R$

$\begin{array}{l}R=R_0{e}^{-t\ln 2/T_{1/2}}\\ R=R_0{2}^{-t/T_{1/2}}\\ R=R_0{\left(\frac{1}{2}\right)}^{t/T_{1/2}}\end{array}$

Substitute $600.00\text{ s}$ for $t$, $200.0\text{ s}$ for $T_{1/2}$ and $80000.0{\text{ s}}^{-1}$ for $R_0$ in the above equation to find the value of $R$

$\begin{array}{l}R=80000.0{\text{ s}}^{-1}\times {\left(\frac{1}{2}\right)}^{600.0\text{ s}/200.0\text{ s}}\\ R=80000.0{\text{ s}}^{-1}\times {\left(\frac{1}{2}\right)}^3\\ R=80000.0{\text{ s}}^{-1}\times 0.125\\ R=10000{\text{ s}}^{-1}\end{array}$

Thus, the activity of the radioactive nuclide after $600.00\text{ s}$ is $10000{\text{ s}}^{-1}$.

(b)

To determine

The initial number of nuclei in a radioactive sample.

Answer to Problem 38P

The initial number of nuclei in a radioactive sample is $2.308\times {10}^7$.

Explanation of Solution

Write the formula for the activity

$R=\lambda N$                                                                                           (III)

Here,$\lambda$ is the decay constant and $N$ is the number of nuclei after time $t$.

Write the formula for the number of nuclei

$N=N_0{e}^{-t/\tau }$                                                                                           (IV)

Here, $N_0$ is the number of nuclei at $t=0$.

Write the formula for the time constant

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

Conclusion:

Substitute equation (II) and (IV) in (III) and rewrite to find $N_0$

$\begin{array}{c}{R}_0{e}^{-t/\tau }=\lambda N_0{e}^{-t/\tau }\\ R_0=\lambda N_0\\ N_0=\frac{1}{\lambda }{R}_0\end{array}$

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

$\begin{array}{l}{N}_0=\tau R_0\\ N_0=\frac{T_{1/2}{R}_0}{\ln 2}\end{array}$

Substitute $200.0\text{ s}$ for $T_{1/2}$ and $80000.0{\text{ s}}^{-1}$ for $R_0$ in the above equation to find the value of $N_0$

$\begin{array}{l}{N}_0=\frac{200.0\text{ s}\times 80000.0{\text{ s}}^{-1}}{\ln 2}\\ N_0=2.308\times {10}^7\end{array}$

Thus, the initial number of nuclei in a radioactive sample is $2.308\times {10}^7$.

(c)

To determine

The probability per second of any one of the nuclei decay.

Answer to Problem 38P

The probability per second of any one of the nuclei decay is $3.466\times {10}^{-3}{\text{ s}}^{-1}$.

Explanation of Solution

Substitute equation (I) in (V) and solve for $\lambda$

$\begin{array}{l}\lambda =\frac{1}{\tau }\\ \lambda =\frac{\ln 2}{T_{1/2}}\end{array}$

Substitute $200.0\text{ s}$ for $T_{1/2}$ in the above equation to find the value of $\lambda$

$\begin{array}{l}\lambda =\frac{\ln 2}{200.0\text{ s}}\\ \lambda =3.466\times {10}^{-3}{\text{ s}}^{-1}\end{array}$

Thus, the probability per second of any one of the nuclei decay is $3.466\times {10}^{-3}{\text{ s}}^{-1}$.