Question

Figure 42-19 shows the decay of parents in a radioactive sample. The axes are scaled by ${\mathit{N}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{x}{\mathbf{10}}^{\mathbf{6}}$and ${\mathit{t}}_{\mathbf{S}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathit{s}$. What is the activity of the sample at?

Short answer

The activity of the sample at t = 27s is 60 Bq.

Step-by-step solution

Determine the formula for the decay

Write the formula forundecayed sample remaining after a given time,as follows:

$N=N_0{e}^{-\lambda t}$ …… (i)

Write the formula for the rate of decay:

$R=\lambda N$

Here, $\lambda$is the disintegration constant, Nis the number of undecayed nuclei.

Calculate the activity of the sample at 27h

The number of atoms present initially at t = 0 is ${\mathrm{N}}_0=2\times {10}^6$. From Fig. 42-19, determine that the number is halved at t = 2s. Thus, using equation (i), determine the disintegration constant of the sample as follows:

$$\begin{gathered} \lambda =\frac{1}{t}\ln \left(\frac{N_0}{N}\right) \\ =\frac{1}{2s}\ln \left(\frac{N_0}{{\displaystyle \frac{N_0}{2}}}\right) \\ =0.3466s^{-1} \end{gathered}$$

Now, using this value in equation (i), determine value of the remaining nuclei of the radioactive sample at$t=27s$ as follows:

$$\begin{gathered} N=\left(2\times {10}^6\right)e^{-\left(0.3466s^{-1}\right)\left(27s\right)} \\ =173 \end{gathered}$$

Now, the activity of the sample at the given time is given using the above values in equation (ii) as follows:

$$\begin{gathered} R=\left(0.3466s^{-1}\right)\left(173\right) \\ \approx 60s^{-1} \\ =60Bq \end{gathered}$$

Hence, the activity of the sample is 60 Bq.