Define the different terms used for the composition of a nucleus.
Free proton is stable, while free neutron is unstable.
Following nuclear equation for decay of neutron
${ }_{0} n^{1} \rightarrow{ }_{1} p^{1}+{ }_{-1} e^{0}+{ }_{0} v^{0}$ where ${ }_{0} v^{0}$ is antineutrino
Hence, one neutron decays into a proton, an electron and a antineutrino and has a mean life of about $1000 \mathrm{~s}$.
Neutron is stable inside the nucleus.
The composition of nucleus :
$\Rightarrow$ Nucleon $(A)$ : The nucleon is the sum of the number of protons and neutrons presents in
the nucleus of the atom.
$\Rightarrow$ Atomic number $(Z)$ : The number of electrons in the atom and hence the number of proton
in the neutral atom is called the atomic number.
$\Rightarrow$ Mass number $(A)$ : The total mass of proton and neutrons in the nucleus of the $\mathrm{Z}+\mathrm{N}$
atom is called the atomic number $(A)$.
$\Rightarrow$ Neutron number $\mathrm{N}:$ The number of neutron present in the nucleus of a atom is called
the neutron number $\mathrm{N}$.
$\therefore \mathrm{N}=\mathrm{A}-\mathrm{Z}$
$\Rightarrow$ Nuclide : The type of nucleus is called nuclide and is represented by symbol ${ }_{\mathrm{Z}} \mathrm{X}$ or $\mathrm{Z}^{\mathrm{A}}$
where $\mathrm{X}$ is the chemical symbol of element.
For example: Nuclide of gold ${ }_{79}^{197}$ Au in which $197$ nucleon, $79$ proton and $197-79=118$ neutron.
The force acting between proton and proton inside the nucleus is
In the nucleus of ${ }_{11} N a^{23}$, the number of protons, neutrons and electrons are
Specify the mass of proton in kilogram and in the $'u'$ unit.
In the following reaction $_{12}M{g^{24}}{ + _2}H{e^4}{ \to _{14}}S{i^X}{ + _0}{n^1},X$ is
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason $R$. Assertion A : The nuclear density of nuclides ${ }_5^{10} B ,{ }_3^6 Li ,{ }_{26}^{56} Fe ,{ }_{10}^{20} Ne$ and ${ }_{83}^{209} Bi$ can be arranged as $\rho_{ Bi }^{ N }>\rho_{ Fe }^{ N }>\rho_{ Ne }^{ N }>\rho_{ B }^{ N }>\rho_{ Li }^{ N }$.
Reason $R$ : The radius $R$ of nucleus is related to its mass number $A$ as $R=R_0 A^{1 / 3}$, where $R_0$ is a constant.
In the light of the above statement, choose the correct answer from the options given below :