91Ó°ÊÓ

the relation between the activity R of a sample and the number of radioactive nuclei N in the sample is given by:

$R=\mathrm{λ}N$ ...............(i)

the number N of remaining nuclei after time t is given by:

$N=N_0{e}^{-\mathrm{λ}t}$ ...............(ii)

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

Multiplying equation (ii) by A and substituting equation (i), we get

$R=R_0{e}^{-\mathrm{λ}t}$ ...............(iii)

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

the relation between the half-life $T_{1/2}$and the decay constant $\mathrm{λ}$is given by:

$\mathrm{λ}=\frac{\ln 2}{T_{1/2}}$ ..................(iv)

To take any information from any measured data, we need to fit the data first.

The equation that describes the dependence of the activity on the time is equation (iii); Taking the natural logarithm for equation (iii), we get:

$\ln R=\ln R_0{e}^{-\mathrm{λ}t}$

So, if we graph In R versus t, we should get a straight line, where its slope is $-\mathrm{λ}$and intercept ln R axis at $\ln R_0$. Thus, let us relist the given data but add one more column to is the values of In R, so we get:

Now, we graph ln R versus t, as shown below.

From the graph, the slope is -0.60, thus, the decay constant is:

$$\begin{gathered} \mathrm{λ}=0.60h^{-1} \\ =1.67×{10}^{-4}{s}^{-1} \end{gathered}$$

Now, we substitute his value into equation (iv), so we get the half-life of the isotope:

$$\begin{gathered} T_{1/2}=\frac{\ln 2}{0.60h^{-1}} \\ =1.16h \end{gathered}$$

At t = 0, R= 20000, so we substitute our values for $\mathrm{λ}$and R into equation (i) and evaluate for N, so we get the number of nuclei present in the sample at [color(red)t = 0]

$$\begin{gathered} N=\frac{R}{\mathrm{λ}} \\ =\frac{20,000decay/s}{1.67×{10}^{-4}{s}^{-1}} \\ =1.20×{10}^{8}nuclei \end{gathered}$$

At t= 7.0 h,

R = 300,

So we substitute our values for $\mathrm{λ}$and R into equation (i) and evaluate for N, so we get the number of nuclei present in the sample at [color(red)t = 7]

$$\begin{gathered} N=\frac{R}{\mathrm{λ}} \\ =\frac{300decay/s}{1.67×{10}^{-4}{s}^{-1}} \\ =1.80×{10}^{6}nuclei \end{gathered}$$