Why do stable nuclei never have more protons than neutrons?
Because protons are positively charged and repel one another electrically. This repulsion becomes so great in nuclei with more than 10 protons or so, that an excess of neutrons which produce only attractive forces, is required for stability.
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 ${ }_{11} \mathrm{Na}^{23}$.