Nuclei $\mathrm{He}_2^3$ and $\mathrm{He}_1^3$ have the same mass number. $\mathrm{He}_2^3$ has two proton and one neutron. $\mathrm{He}_1^3$ has one proton and two neutron. The repulsive force between protons is missing in ${ }_1 \mathrm{He}^3$ so the binding energy of ${ }_1 \mathrm{He}^3$ is greater than that of ${ }_2 \mathrm{He}^3$.

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.

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 $\left(\frac{d N}{d t}\right)_A>\left(\frac{d N}{d t}\right)_B$

$$ \begin{array}{l} \Rightarrow & \lambda_A N_A>\lambda_B N_B\quad \text { (rate of decay of } B \text { is slower ) } \\ \because & N_A>N_B \\ \therefore & \lambda_B>\lambda_A \\ \Rightarrow & \tau_A>\tau_B \quad \left[\because \text { Average life } \tau=\frac{1}{\lambda}\right] \end{array}$$

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.