Draw a graph showing the variation of decay rate with number of active nuclei.
We know that, rate of decay $=\frac{-d N}{d t}=\lambda N$
where decay constant $(\lambda)$ is constant for a given radioactive material. Therefore, graph between $N$ and $\frac{d N}{d t}$ is a straight line as shown in the diagram.
Which sample A or B shown in figure has shorter mean-life?
From the given figure, we can say that
$$\begin{aligned} & \text { at } t=0,\left(\frac{d N}{d t}\right)_A=\left(\frac{d N}{d t}\right)_B \\ & \Rightarrow \quad\left(N_0\right)_A=\left(N_0\right)_B \end{aligned}$$
Considering any instant $t$ by drawing a line perpendicular to time axis, we find that $\left(\frac{d N}{d t}\right)_A>\left(\frac{d N}{d t}\right)_B$
$$ \begin{array}{l} \Rightarrow & \lambda_A N_A>\lambda_B N_B\quad \text { (rate of decay of } B \text { is slower ) } \\ \because & N_A>N_B \\ \therefore & \lambda_B>\lambda_A \\ \Rightarrow & \tau_A>\tau_B \quad \left[\because \text { Average life } \tau=\frac{1}{\lambda}\right] \end{array}$$
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.