Which one of the following cannot emit radiation and why? Excited nucleus, excited electron.
Excited electron cannot emit radiation because energy of electronic energy levels is in the range of eV and not MeV ( mega electron volt).
$\gamma$-radiations have energy of the order of MeV.
In pair annihilation, an electron and a positron destroy each other to produce gamma radiations. How is the momentum conserved?
In pair annihilation, an electron and a positron destroy each other to produce $2 \gamma$ photons which move in opposite directions to conserve linear momentum. The annihilation is shown below ${ }_0 \mathrm{e}^{-1}+{ }_0 \mathrm{e}^{+1} \rightarrow 2 \gamma$ ray photons.
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
$$ \begin{aligned} &\begin{aligned} \lambda t \log _e e & =\log _e \frac{R_0}{R} \Rightarrow \lambda t=\left(\log _{10} \frac{16}{12}\right) \times 2.303 \\ t & =\frac{2.303(\log 4-\log 3)}{\lambda} \\ & =\frac{2.303(0.6020-4.771) \times 5760}{0.6931} \quad \left(\because \lambda=\frac{0.6931}{T_{1 / 2}}\right)\\ & =2391.20 \mathrm{~yr} \end{aligned}\\ \end{aligned}$$