Consider a radioactive nucleus $A$ which decays to a stable nucleus $C$ through the following sequence
$$A \rightarrow B \rightarrow C$$
Here $B$ is an intermediate nuclei which is also radioactive. Considering that there are $N_0$ atoms of $A$ initially, plot the graph showing the variation of number of atoms of $A$ and $B$ versus time.
Consider the situation shown in the graph.
At $t=0, N_A=N_0$ (maximum) while $N_B=0$. As time increases, $N_A$ decreases exponentially and the number of atoms of $B$ increases. They becomes $\left(N_B\right)$ maximum and finally drop to zero exponentially by radioactive decay law.
A piece of wood from the ruins of an ancient building was found to have a ${ }^{14} \mathrm{C}$ activity of 12 disintegrations per minute per gram of its carbon content. The ${ }^{14} \mathrm{C}$ activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of ${ }^{14} \mathrm{C}$ is 5760 yr .
Given, $R=12 \mathrm{dis} / \mathrm{min}$ per $\mathrm{g}, R_0=16 \mathrm{dis} / \mathrm{min}$ per $\mathrm{g}, T_{1 / 2}=5760 \mathrm{yr}$ Let $t$ be the span of the tree.
According to radioactive decay law,
$$R=R_0 e^{-\lambda t} \text { or } \frac{R}{R_0}=e^{-\lambda t} \text { or } e^{\lambda t}=\frac{R_0}{R}$$
Taking log on both the sides
$$ \begin{aligned} &\begin{aligned} \lambda t \log _e e & =\log _e \frac{R_0}{R} \Rightarrow \lambda t=\left(\log _{10} \frac{16}{12}\right) \times 2.303 \\ t & =\frac{2.303(\log 4-\log 3)}{\lambda} \\ & =\frac{2.303(0.6020-4.771) \times 5760}{0.6931} \quad \left(\because \lambda=\frac{0.6931}{T_{1 / 2}}\right)\\ & =2391.20 \mathrm{~yr} \end{aligned}\\ \end{aligned}$$
Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately $10^{-15} \mathrm{~m}$.
Each particle (neutron and proton) present inside the nucleus is called a nucleon.
Let $\lambda$ be the wavelength $\lambda=10^{-15} \mathrm{~m}$
To detect separate parts inside a nucleon, the electron must have wavelength less than $10^{-15} \mathrm{~m}$.
We know that
$$\begin{aligned} \lambda & =\frac{h}{p} \text { and } K E=P E \quad\text{.... (i)}\\ \text{Energy}\quad & =\frac{h c}{\lambda}\quad\text{.... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { From Eq. (i) and Eq. (ii), }\\ &\begin{aligned} \text { kinetic energy of electron } & =\mathrm{PE}=\frac{h \mathrm{c}}{\lambda}-\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10^{-15} \times 1.6 \times 10^{-19}} \mathrm{eV} \\ \mathrm{KE} & =10^9 \mathrm{eV} \end{aligned} \end{aligned}$$
A nuclide 1 is said to be the mirror isobar of nuclide 2 if $Z_1=N_2$ and $Z_2=N_1$.(a) What nuclide is a mirror isobar of ${ }_{11}^{23} \mathrm{Na}$ ? (b) Which nuclide out of the two mirror isobars have greater binding energy and why?
(a) According to question, a nuclide 1 is said to be mirror isobar of nuclide 2 , if $Z_1=N_2$ and $Z_2=N_1$.
Now in ${ }_{11} \mathrm{Na}^{23}, Z_1=11, N_1=23-11=12$
$\therefore$ Mirror isobar of ${ }_{11} \mathrm{Na}^{23}$ is ${ }_{12} \mathrm{Mg}^{23}$, for which $Z_2=12=N_1$ and $N_2=23-12=11=Z_1$
(b) $\mathrm{As}_{12}^{23} \mathrm{Mg}$ contains even number of protons (12) against ${ }_{11}^{23} \mathrm{Na}$ which has odd number of protons (11), therefore ${ }_{11}^{23} \mathrm{Mg}$ has greater binding energy than ${ }_{11} \mathrm{Na}^{23}$.
Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is
$${ }^{38} \text { Sulphur } \xrightarrow[=2.48 \mathrm{~h}]{\text { half-life }}{ }^{38} \mathrm{Cl} \xrightarrow[=0.62 \mathrm{~h}]{\text { half-life }}{ }^{38} \mathrm{Ar} \text { (stable) }$$
Assume that we start with $1000{ }^{38} \mathrm{~S}$ nuclei at time $t=0$. The number of ${ }^{38} \mathrm{Cl}$ is of count zero at $t=0$ and will again be zero at $t=\infty$. At what value of $t$, would the number of counts be a maximum?
Consider the chain of two decays
$${ }^{38} \mathrm{~S} \xrightarrow[2.48 \mathrm{~h}]{ }{ }^{38} \mathrm{Cl} \xrightarrow[0.62 \mathrm{~h}]{ }{ }^{38} \mathrm{Ar}$$
At time $t$, Let ${ }^{38} \mathrm{~S}$ have $N_1(t)$ active nuclei and ${ }^{38} \mathrm{Cl}$ have $N_2(t)$ active nuclei.
$$\begin{aligned} \frac{d N_1}{d t} & =-\lambda_1 N_1=\text { rate of formation of } \mathrm{Cl}^{38} \\ \text{Also,}\quad\frac{d N_2}{d t} & =-\lambda_1 N_2+\lambda_1 N_1 \\ \text{But}\quad N_1 & =N_0 e^{-\lambda_1 t} \\ \frac{d N_2}{d t} & =\lambda_1 N_0 e^{-\lambda_1 t}-\lambda_2 N_2\quad\text{.... (i)} \end{aligned}$$
Multiplying by $e^{\lambda z^t} d t$ and rearranging
$$e^{\lambda_2 t} d N_2+\lambda_2 N_2 e^{\lambda_2 t} d t=\lambda_1 N_0 e^{\left(\lambda_2-\lambda_1\right) t} d t$$
Integrating both sides
$$N_2 e^{\lambda_2 t}=\frac{N_0 \lambda_1}{\lambda_2-\lambda_1} e^{\left(\lambda_2-\lambda_1\right) t}+C$$
$$\begin{aligned} \therefore \quad N_2 e^{\lambda_2 t} & =\frac{N_0 \lambda_1}{\lambda_2-\lambda_1}\left(e^{\left(\lambda_2-\lambda_1\right) t}-1\right) \quad\text{.... (ii)}\\ N_2 & =\frac{N_0 \lambda_1}{\lambda_2-\lambda_1}\left(e^{-\lambda_1 t}-e^{-\lambda_2 t}\right) \end{aligned}$$
For maximum count, $\quad \frac{d N_2}{d t}=0$
$$\lambda_1 N_0 e^{-\lambda_1 t}-\lambda_2 N_2=0\quad$$ [From Eq. (i)]
$\Rightarrow \quad \frac{N_0}{N_2}=\frac{\lambda_2}{\lambda_1} e^{\lambda_1 t}\quad$ [From Eq. (ii)]
$$\begin{gathered} e^{\lambda_2 t}-\frac{\lambda_2}{\lambda_1} \cdot \frac{\lambda_1}{\left(\lambda_1-\lambda_1\right)} e^{\lambda_1 t}\left[e^{\left(\lambda_2-\lambda_1\right) t}-1\right]=0 \\ \text{or}\quad e^{\lambda_2 t}-\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\lambda_2 t}+\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\lambda_1 t}=0 \end{gathered}$$
$$\begin{aligned} 1-\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)}+\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\left(\lambda_1-\lambda_2\right) t} & =0 \\ \frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\left(\lambda_1-\lambda_2\right) t} & =\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)}-1 \\ e^{\left(\lambda_1-\lambda_2\right) t} & =\frac{\lambda_1}{\lambda_2} \\ t & =\left(\log _e \frac{\lambda_1}{\lambda_2}\right) /\left(\lambda_1-\lambda_2\right) \\ & =\frac{\log _e\left(\frac{2.48}{0.62}\right)}{2.48-0.62} \\ & =\frac{\log _e 4}{1.86}=\frac{2.303 \times 2 \times 0.3010}{1.86} \quad\left(\because \lambda=\frac{0.693}{T_{1 / 2}}\right) \\ & =0.745 \mathrm{~S} \end{aligned}$$