91Ó°ÊÓ

a) Ratio of the resulting number of BB atoms to the resulting number of CC atoms, $\frac{N_{BB}}{N_{CC}}=\frac{2}{1}$

b) Half-life of decay from to AA either BB or CC ,$T_{1/2}=8.00d$

c) Ratio of the resulting number of CC atoms to the resulting number of AA atoms,$\frac{N_{CC}}{N_{AA}}=1.50$

Here, the decay of the pure sample AA can be either into AA, BB, or CC.Thus, we can get the initial present undecayed AA atoms that are before the decay by adding all the remaining undecayed atoms of the three product nuclei. Using the given exponential expression for the decay of the atoms, we can get the time of decay.

Formulas:

The disintegration constant is as follows:

$\mathrm{λ}=\frac{\ln 2}{T_{{\displaystyle \frac{1}{2}}}}$ …… (i)

Here, $T_{\frac{1}{2}}$is the half-life of the substance.

The undecayed nuclei of the sample is as follows:

$N=N_0{e}^{-\mathrm{λ}t}$ …… (ii)

Let,$N_{AA0}$be the number of element AA at t = 0 . At a later time t , due to radioactive decay, we have

$N_{AA0}=N_{AA}+N_{BB}+N_{CC}$

The disintegration constant of the decay can be calculated as:

$$\begin{gathered} \mathrm{λ}=\frac{\ln 2}{8.00d} \\ =\frac{0.0866}{d} \end{gathered}$$

Since,$\frac{N_{BB}}{N_{CC}}=\frac{2}{1}$when$\frac{N_{CC}}{N_{AA}}=1.50$ , then ratio of resultingatoms to resulting AA atoms can be calculated as:

$$\begin{gathered} \frac{N_{BB}}{N_{AA}}=\frac{N_{BB}}{N_{CC}}×\frac{N_{CC}}{N_{AA}} \\ =\frac{2}{1}×\frac{1.50}{1} \\ =\frac{3}{1} \end{gathered}$$

Now, using the above values in equation (ii), determine the time of decay for the required situation as follows:

$$\begin{gathered} \frac{N_{AA0}}{N_{AA}}=e^{-\mathrm{λ}t} \\ \frac{N_{AA}+N_{BB}+N_{CC}}{N_{AA}}=e^{-\left(\frac{0.0866}{d}\right)t} \end{gathered}$$

Substitute the value and solve as:

$$\begin{gathered} 1+3+1.50=e^{-\left(\frac{0.0866}{d}\right)t} \\ t=\frac{\left(\ln 5.50\right)}{\left({\displaystyle \frac{0.866}{d}}\right)} \\ t=19.7d \end{gathered}$$

Hence, the time of decay is 19.7d .