Energy acquired by the electron (as kinetic energy) after being accelerated by a potential difference of 1 KV
i.e., 1000 volts = 1000 eV
= 1000 x 1.602 x 10^-19 J
= 1.602 x 10^-16 J (1 eV = 1.602 x 10^-19 J)
Energy in Joules = charge on the electron in coulombs x potential difference in volts
i.e., Kinetic Energy (KE) = \(\frac{1}{2}\) mv²
= 1.602 x 10^-16 J or \(\frac{1}{2}\) x 9.1 x 10^-31 v²
v²  = 3.521 x 10¹⁴ or v = 1.88 x 10^-1 ms^-1
\(\therefore \lambda=\frac{h}{mv}=\frac{6.626\times 10^{-34}kgm^2 s{-1}}{9.1\times 10^{-31}kg\times 1.88\times 10^{-1}ms^{-1}}\)
= 3.87 x 10^-11 m.

For hydrogen atom
E_n = \(\frac{-21.8\times 10^{-19}}{n^2}\) = J atom (or)
E_n = \(\frac{-13.6}{n^2}\) eV atom^-1 (1 eV = 1.6 x 10^-9 J)
Energy is emitted when an electron jumps from n = 4 to n = 2 will be given by
\(\Delta\)E = E₄ - E₂ = 21.8 x 10^-19 \([\frac{1}{2^2}-\frac{1}{4^2}]\)
= 21.8 x 10^-19 x \(\frac{3}{16}\) = 10^-19 J
= 4.0875 x 10^-19 J.
The wavelength corresponding to this transition is
E = hv = \(h\frac{c}{\lambda}\) or \(\lambda =\frac{hc}{E}\)
= \(\frac{(6.620\times 10^-34 JS)\times (3\times 10^8 ms^{-1})}{4.863\times10^{-19}J}\)
= 4.863 x 10^-7 m = 4863 \(\overset { o }{ A } \) or 4863.3 nm.

(i) \({ _{ 30 }^{ 64 }{ Zn } }^{ 2+ }\)
No. of protons = 30
No. of electrons = 30 - 2 = 28
No. of neutrons = 64 - 30 = 34
(ii) ³⁵Cl^-
Atomic number = Mass number - No. of neutrons
= 35 - 18 = 17
Atomic number = No. of protons = 17
(iii) Ar
Mass number = No. of protons + No. of neutrons
= 18 + 22 = 40
Atomic number = No. of protons = 18
No. of electrons =  No. of protons = 18.

(i) Isotopes: Atoms of same element having same atomic number and different mass number Eg : _IH^I, _IH², _IH³ (_IT³) .
(ii) Isotones: These species possess same number of neutrons. Their atomic and mass numbers are different. Eg : \(_{ 6}^{ 14 }{C }\) ₇N¹⁵ ₈O¹⁶
(iii) Isobars: Atoms of same element having same mass number and different atomic number Eg : \(_{ 6}^{ 14 }{C }\) , ₇N^I4
(iv) Isoelectronic species: The species (atoms or ions) containing same number of electrons. Eg : .0^2-, F^-, Na⁺, Mg^2+, A1³⁺,Ne
(v) Nucleon: Protons and neutrons present in the nucleus are collectively called nucleons. The total number of nucleons is termed as mass number (A) of an atom.

Given; velocity of beam of helium atoms = 20 10³ m 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,  \(\lambda =\frac { h }{ mv } \)
= \(\frac { 6.626\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 } }{ (6.626\times { 10 }^{ -27 }kg)\times (2.0\times { 10 }^{ 3 }{ ms }^{ -1 }) } \)
= 4.99 \(\times\)10^-11 m = 49.9 pm 

h = 626\(\times\)10^-34 kg m² s^-1
Δv = 5.27\(\times\)10^-24 ms^-1
Δx = 1 nm = 1\(\times\)10^-9 m
Mass of the particle
\(m=\frac { h }{ 4\pi \Delta x } kg\)
= \(\frac { 6.626\times { 10 }^{ -34 }{ kg }{ m }^{ 2 }{ s }^{ -1 } }{ 4\times 3.143\times 5.27\times { 10 }^{ -24 }{ ms }^{ -1 }\times 1\times { 10 }^{ -9 }m } \)
= \(1\times { 10 }^{ -9 }kg\)
\(\boxed{m=1\times{10}^{-2} kg}\)

h = \(6.626\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 }\)
m = 0.1 mg = 1 X 10^-7 kg
Δv =\(5\times { 10 }^{ -19 }{ ms }^{ -1 }\)
Uncertainty in position
\(\Delta x=\frac { h }{ 4\pi m\Delta v } \)
= \(\frac { 6.626\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 } }{ 4\times 3.143\times { 10 }^{ -7 }kg\times 5\times { 10 }^{ -9 }{ ms }^{ -1 } } \)
= \(1.05\times { 10 }^{ -9 }m\)
\(\boxed{\triangle x=1.05\ nm}\)

h = \(6.626\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 }\)
m = \(9.1\times { 10 }^{ -28 }g=9.1\times { 10 }^{ -31 }kg\)
Product of uncertainties,
\(\Delta x\Delta c\ge \frac { h }{ 4\pi m } { m }^{ 2 }{ s }^{ -1 }\)
\(\ge \frac { 6.626\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 } }{ 4\times 3.143\times 9.1\times { 10 }^{ -31 }kg } \)
\(\ge 5.77\times { 10 }^{ -5 }{ m }^{ 2 }{ s }^{ -1 }\)
\(\boxed{\Delta x.\Delta v\ge 5.77\times { 10 }^{ -5 }{ m }^{ 2 }{ s }^{ -1 }}\)

h =  \(6.626\times { 10 }^{ -34 }kg{ \quad m }^{ 2 }{ s }^{ -1 }\)
\(\Delta v =5.27\times { 10 }^{ -24 }{ ms }^{ -1 };\Delta x={ 10 }^{ -2 }m\)
Mass of the particle,
\(m=\frac { h }{ 4\pi \Delta x\Delta v } kg\)
= \(\frac { 6.626\times { 10 }^{ -34 }kg{ m }^{ 2 }{ s }^{ -1 } }{ 4\times 3.143\times { 10 }^{ -2 }m\times 5.27\times { 10 }^{ -24 }{ ms }^{ -1 } } \)
= \(1\times { 10 }^{ -9 }kg\)
\(\boxed{m = 1\times { 10 }^{ -9 }kg}\)

Given m = 3000 kg; Δx = 10 pm 
= \(10\times { 10 }^{ -12 }m={ 10 }^{ -11 }m\)
By uncertainty principle
Δv = \(\frac { h }{ 4\pi \times m\times \Delta x } \)
= \(\frac { 6.626\times { 10 }^{ -34 } }{ 4\times \frac { 22 }{ 7 } \times 3000\times { 10 }^{ -11 } } { ms }^{ -1 }\)
= \(1.76\times { 10 }^{ -27 }{ ms }^{ -1 }\)
\(\boxed{\Delta v=1.76\times { 10 }^{ -27 }{ ms }^{ -1 }}\)

(a) The number of subshells is the same as the number used to designate the shell ie., n = 4 so contains 4 sub shells.
(b) The subshells in the fourth shell are designated as 4s. 4p. 4d and 4f
(c) The number of orbitals in s, p, d and subshell is 1,3,5 and 7-respectively.
(d) Each orbital can contain a maximum of two electrons.
4s - 1 orbital - 2 electrons
4p - 3 orbital - 6 electrons
4d - 5 orbital -10 electrons
4s - 7 orbital - 14 electrons

The electronic configuration of the element with 2K, 8L and 5M electrons will be
1s² 2s² 2p_x² 2p_y² 2p_z² 3s² 3p_x¹ 3p_y¹ 3p_z1
(i) Total number of electrons = 2 + 8 + 5 = 15
∴ Atomic number of the element = 15
(ii) Total number of s-electrons = 2 + 2 + 2 = 6
(iii) Total number of p-electrons = 6 + 3 = 9
(iv) Since, the atom is neutral
∴  Number of protons = Number of electrons = Atomic number = 15
(v) valency of the element = 3

∵ (V) < (ii) = (iv) < (iii) = (vi) < (i)
∴ 3p < 3d = 3d < 4p = 4p <4d

(i) For chromium - 24
 Expected configuration:
1s² 2S²2p⁶ 3s² 3p⁶ 3d⁴ 4s²
Actual configuration:
1s² 2S²2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
For copper - 29
Expected:
1s² 2s²2p⁶ 3s² 3p⁶ 3d⁹ 4s²
Actual configuration:
1s² 2s²2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
(ii)The extraordinary stability of half-filled and completely filled electron configurations is due to symmetrical distribution of electrons .Moreover, in such configuration electron can exchange their positions among themselves to maximum extent. This exchange leads to stabilisation.
(iii) Due to this stability, one of the 4s electrons occupies the 3d orbital in chromium and copper to attain the half filled and the completely filled configurations respectively.

(i) The net charge experienced by the electron is called effective nuclear charge.
(ii) The effective nuclear charge depends on the shape of the orbitals and it decreases with increase in azimuthal quantum number 1.
(iii) The order of the effective nuclear charge felt by a electron in an orbital within the given shell is s > p > d > f.
(iv) Greater the effective nuclear charge, greater is the stability of the orbital.
(v) Hence, within a given energy level, the energy of the orbitals are in the following order s < p < d < f.
(vi) The energies of same orbital decrease with an increase in the atomic number.

(i) The energy of electrons in atoms is quantised
(ii) The existence of quantized electronic energy levels is a direct result of the wave like properties of electrons. The solutions of Schrodinger wave equation gives the allowed energy levels (orbits).
(iii) According to Heisenberg uncertainty principle, the exact position and momentum of an electron can not be determined with absolute accuracy. As a consequence, quantum mechanics introduced the concept of orbital. Orbital is a three dimensional space in which the probability of finding the electron is maximum.
(iv) The solution of  Schrodinger wave equation for the allowed energies of an atom gives the wave function ψ,which represents an atomic orbital. The wave nature of electron present in an orbital can be well defined by the wave function ψ.
(v) The wave function ψ itself has no physical meaning.However, the probability of finding the electron in a small volume dxdydz around a point (x,y,z) is proportional to |ψ(x,y,x)|² dxdydz |ψ(x,y,x)|² is known as probability density and is always positive.

de-Broglie combined the following two equations of energy of which one represents wave character (hu) and the other represents the particle nature(mc²).(i) Planck's quantum hypothesis: E = hv
(ii) Einstein's mass - energy relationship :
E = mc2
From (i) and (ii)
hv = mc²
hc/λ =mc²
λ = h/mc
(a) The equation represents the wavelength of photons whose momentum is given by mv (photons have zero rest mass)
(b) For a particle of matter with mass m and moving with a velocity v, the equation can be written as  λ = h/mv
(c) This is valid only when the particle travels at speeds much less than the speed of light

Limitation of Bohr's atom model:
(a) The Bohr's atom model is applicable only to species having one electron such as hydrogen, Li²⁺ etc ... and not applicable to multi electron atoms.
(b) It was unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
(c) Bohr's theory was unable to explain why the electron is restricted to revolve around the nucleus in a fixed orbit in which the angular momentum of the electron is equal to nh/2π

Applying Bohr's postulates to a hydrogen like atom (one electron species such as H, He⁺ and Li²⁺ etc...)the radius of the n^th orbit and the energy of the electron revolving in the n^th orbit were derived. The results are as follows:
r_n = \(\frac { (0.529){ n }^{ 2 } }{ x } \mathring { A } \)  ...(1)
E_n = \(\frac { -13.6({ z }^{ 2 }) }{ { n }^{ 2 } } ev\quad { atom }^{ -1 }\)   or ...  (2) or
E_n = \(\frac { (-1312.8){ z }^{ 2 } }{ { n }^{ 2 } } kJ\quad { mol }^{ -1 }\) .....(3)

Bohr's atom is based on the following assumptions:
(a) The energies of electrons are quantised
(b) The electron is revolving around the nucleus in a certain fixed circular path called stationary orbit.
(c) Electron can revolve, only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2π i.e mvr =nh/2π
where n = 1,2,3 ...etc...
As long as an electron revolves in the fixed stationary orbit, it doesn't lose its energy.
However, when an electron jumps from higher energy state (E₂)to a lower energy state (E₁)the excess energy is emitted as radiation. The frequency of the emitted radiation is
E2 = E1 = hv and 
\(v=\frac { ({ E }_{ 2 }-{ E }_{ 1 }) }{ h } \)
Conversely, when suitable energy is supplied to an electron, it will jump from lower energy orbit to a higher energy orbit.