If a certain number of neutrons and protons are brought together to form a nucleus of a certain charge and mass, an energy $E_b$ will be released in the process. The energy $E_b$ is called the binding energy of the nucleus.
$E_b=\Delta M{c}^2$
The ratio of the binding energy $E_b$ of a nucleus to the number of the nucleons, A, in that nucleus is called the binding energy per nucleon, $E_{bn}$.
$E_{bn}=\frac{E_b}{A}$

We have $\frac{dN}{dt}=-\lambda N$
$\frac{dN}{N}=-\lambda dt$
Integrating both sides,
${\int }_{N_0}^N\frac{dN}{N}=-\lambda {\int }_{t_0}^{t}dt$
$ln(\frac{N}{N_0})=-\lambda t$
$N=N_0{e}^{-\lambda t}$
This is the law of radioactive decay

Mean life ($\tau$) = sum of life of all atom / total no of atoms present
$\tau =\frac{1}{\lambda }$
where $\lambda$ is the decay constant.

The half-life $T_{1/2}$ of a radio nuclide is the time at which both N and R have been reduced to one-half their initial values.
Putting $R=(1/2)R_0$ and $t=T_{1/2}$ in the law of radioactive decay, we get:
$T_{1/2}=\frac{ln2}{\lambda }=\frac{0.693}{\lambda }$

When a nucleus undergoes alpha-decay, it transforms to a different nucleus by emitting an alpha-particle (a helium nucleus,${}_2^{4}He$).
For example, when ${}_{92}^{238}U$undergoes alpha-decay, it transforms to ${}_{90}^{234}T$
${}_{92}^{238}U{\to }_{90}^{234}Th{+}_2^{4}He$
In this process, it is observed that since ${}_2^{4}He$ contains two protons and two neutrons, the mass number and the atomic number of the daughter nucleus decreases by four and two, respectively. Thus, the transformation of a nucleus ${}_Z^{A}X$ into a nucleus ${}_{Z-2}^{A-4}Y$ can be expressed as
${}_Z^{A}X{\to }_{Z-2}^{A-4}Y{+}_2^{4}He$
where ${}_Z^{A}X$ is the parent nucleus and ${}_{Z-2}^{A-4}Y$ is the daughter nucleus.

A nucleus that decays spontaneously by emitting an electron or a positron is said to undergo beta decay.In beta minus (${\beta }^{-}$ ) decay, an electron is emitted by the nucleus and a neutron transforms into a proton within the nucleus according to
$n\to p+e^{-}+\bar{\nu }$
In beta plus (${\beta }^{+}$ ) decay, a positron is emitted by the nucleus and a proton transforms into neutron (inside the nucleus) via
$p\to n+e^{+}+\nu$

Example: The $\frac{B.E.}{A}$ ratio for deuteron and $\alpha$- particle are $X_1$ and $X_2$respectively. Find the energy released in the fusion of deuteron into $\alpha$-particle.

Solution:
Energy released is given as per the reaction
${}_1{H}^2{+}_1{H}^2{=}_{2}H{e}^4+Q$
In the two deuterons ${(}_1{H}^2)$ there are total 4 nucleons and in the alpha particle ${(}_{2}H{e}^4)$ there are $4$ nucleons.
Thus we get
$Q=4(X_2-X_1)$

Example: In nuclear fission, $0.1$% mass is converted into energy. Find the energy released in the fission of $1kg$ mass in $kWh$.

Solution:
$0.1$% mass is converted.
$E=(\frac{1}{1000}\times 1kg)(3\times 10^8{)}^{2}J$
$=9\times 10^{13}J$
Let this energy is liberated per second then,
E per hour $=\frac{9\times 10^{13}}{60\times 60}$
$=0.25\times 10^{11}Wh$
$=2.5\times 10^{7}kWh$