Atomic Number (Z): The atomic number of an element is equal to the number of protons present inside the nucleus of its atoms. Since, an isolated atom has no net charge on it, in neutral atoms, the total number of electrons is equal to its atomic number. 
Atomic number (Z) = Number of protons in the nucleus of an atom
= Number of electrons in the neutral atoms
Mass Number (A): The sum of the number of neutrons and protons in the nucleus of an atom is called its mass number. Mass number is denoted by A. Thus, for an atom, Mass number (A) = Number of protons (P) + Number of neutrons (n) A = P + n
Neutron: It is neutral particle. It is present in the nucleus of an atom. Expect hydrogen (which contains only one electron and one proton but no neutron), the atoms of all other elements including isotopes of hydrogen contain all the three fundamental particles called neutron, proton and electron.
The relation between mass number, Atomic no. and no. of neutrons is given by the equation:
A = Z+n I
Where A = Mass number
Z = Atomic number
n = Number of neutrons in the nucleus.

(a) Limitations of Rutherford Model:
(i) When a body is moving in an orbit, it achieves acceleration (even if body is moving with constant speed in an orbit, it achieves acceleration due to change in direction). So an electron moving around nucleus in an orbit is under acceleration. However, according to radiation theory of Maxwell, the charged particles when accelerated must emit energy as electromagnetic radiations. This means that the revolving electron must also lose energy continuously in the form of electromagnetic radiation. The loss of energy in revolution of the electron around the nucleus must bring it closer to the nucleus and the electron must ultimately fall into the nucleus by the spiral path. This means that the atom must collapse. But we all know that atom is quite stable in nature.
(ii) Rutherford's model could not explain the existence of different spectral lines in the hydrogen spectrum.
(b) Based upon the postulates of Bohr's theory, it is possible to calculate the energy of the hydrogen electron and also one electron species. (He+, U2+ etc.) The mathematical expression for the energy in the nth orbit is
\(E_{ n }=-\frac { 2\pi ^{ 2 }m_{ e }e^{ 4 }Z^{ 2 } }{ n^{ 2 }h^{ 2 } } \)
By substituting the values of m_e (mass of electron), e (charge of electron) and h (Planck's constant), the value of energy comes out to be
\(E_{ n }=-\frac { 2.178\times 10^{ -18 }Z^{ 2 } }{ n^{ 2 } } \)J per atom.
\(=-\frac { 1312\times Z^{ 2 } }{ n^{ 2 } } \)KJ mol^-1
For hydrogen electron,
Z = 1
\(E_{ n }=-\frac { 1312 }{ n^{ 2 } } \)KJ mol^-1
The value for n = 1, gives the energy of the hydrogen electron in the ground state.
By assigning values, energy in different excited states can be calculated.

Kinetic energy of electron,
\(\frac { 1 }{ 2 } { mv }^{ 2 }=2.275\times { 10 }^{ -25 }J\)
or v² = \(\frac { 2\times 2.275\times { 10 }^{ -25 } }{ 9.1\times { 10 }^{ -31 } } \)
Now,\(\lambda =\frac { h }{ mv } \)
\(=\frac { 6.6\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 } }{ (9.1\times { 10 }^{ -31 }kg)\times (7.07\times { 10 }^{ 3 }{ ms }^{ -1 }) } \)
\(=1.029\times { 10 }^{ -6 }m=1029 \ nm\)

For 3d-orbital, n = 3, l = 2, m_l = -2, -1, 0, +1, +2

Given, velocity of beam of helium atoms = 2.0\(\times \)10³m sec^-1
Mass of helium atom = \(\frac { 4 }{ 6.022\times { 10 }^{ 23 } } \)
= \(6.64\times { 10 }^{ -24 }g=6.64\times { 10 }^{ -27 }kg\)
According to de-Broglie equation, \(\lambda =\frac { h }{ mv } \)
\(=\frac { 6.626\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 } }{ (6.64\times { 10 }^{ -27 }kg)\times (2.0\times { 10 }^{ 3 }{ ms }^{ -1 }) } \)
\(= 4.99\times { 10 }^{ -11 }m=49.9pm \ [2]\)

Energy of electron in unielectron atomic system,
\({ E }_{ n }=\frac { { -2\pi }^{ 2 }{ mZ }^{ 2 }{ e }^{ 4 } }{ { n }^{ 2 }{ h }^{ 2 } } \) 
For H-atom, ionisation energy (I.E) = \({ E }_{ \infty }-{ E }_{ 1 }\) 
I.E = \(0-\left( -\frac { { 2\pi }^{ 2 }me^{ 4 }{ 1 }^{ 2 } }{ { 1 }^{ 2 }{ h }^{ 2 } } \right) \) 
(where, Z = 1 and n = 1 for H - atom)
Given, I.E for the H - atom in the ground state
\(2.18\times { 10 }^{ -18 }J{ \ atom }^{ -1 }\)     
For He⁺, I.E = \({ E }_{ \infty }-{ E }_{ 1 }\)  =\(0-\left( -\frac { { 2\pi }^{ 2 }me^{ 4 }{ 1 }^{ 2 } }{ { 1 }^{ 2 }{ h }^{ 2 } } \right) \)
\(=4\times \frac { { 2\pi }^{ 2 }me^{ 4 } }{ { h }^{ 2 } }\)    
\(=4\times 2.18\times { 10 }^{ -18 }J{ \ atom }^{ -1 }\)
\(=8.72\times { 10 }^{ -18 }J{ \ atom }^{ -1 }\)
\(\therefore \) The energy required for the process     
\(He^{ + }\left( g \right) \longrightarrow { He }^{ 2+ }\left( g \right) +{ e }^{ - } \ is \ 8.72\times { 10 }^{ -18 }J{ \ atom }^{ -1 }\).

According to Bohr model for H-atom, the angular momentum of an electron in a given stationary state,
mvr = \(\frac { nh }{ 2\pi } \ or \ 2\pi r=\frac { nh }{ mv } ...(i)\)
From de-Broglie equation, wavelength, \(\lambda =\frac { h }{ mv } \ ...(ii)\)
From equation (i) and (ii), we get
\(2\pi r=n\lambda \)
Therefore, the circumference (2\(\pi \)r) of the Bohr orbit for H-atom is an integral multiple of de-Broglie wavelength. (Here n = number of waves in nth orbit).

To obtain an atomic number of an element fill the orbitals in order of their increasing energies up to the given orbital configuration.
a) 1s²2s², 2p⁶, 3s² (Z = 12)
b) 1s²2s², 2p³ (Z = 7)
c) 1s²2s², 2p⁶, 3s²,3p⁵ (Z = 17).