Electron has wave nature.

Ground state means the lowest energy state. When the electrons absorb energy and jump to outer orbits, this state is called excited state.

Nucleus of Li atom has 3 protons and 4 neutrons.
Charge on one proton =1.60 x 10^-19 coulombs
Charge on 3 protons (i.e. charge on nucleus)
= \(3\times 1.60\times 10^{ -19 }C=4.80\times 10^{ -19 }C\)
Mass of proton mass of neutron \(\simeq 1.67\times 10^{ -27 }kg\) 
Mass of nucleus = \(7\times 1.67\times 10^{ -27 }kg\)
= \(11.69\times 10^{ -27 }kg\)

Heavy atoms such as gold, platinum have nucleus. Heavy nucleus contains large amount of positive charge. When a beam of \(\alpha \)-particles is shot at a thin gold foil, most of them pass through without much effect.
Some however, are deflected back or by small angles due to enormous repulsive force of heavy nucleus. If light aluminium foil is used, the number of \(\alpha \) -particles deflected back or those deflected by small angles will be negligible.

Nuclear charge is defined as the net positive charge experienced by an electron in a multielectron atom. The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.
\({ 13 }^{ Al }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 1 }\)
\({ 14 }^{ Si }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 2 }\)

In a hydrogen atom, the energy of an electron is determined by the value of n and in a multielectron atom, it is determined by n + 1. hence, for a given principal quantum, electrons of s,p,d and f-orbitals have different energy (for,p,d and f = 0,1,2 and 3 respectively).

\(0.7244\times { 10 }^{ 5 }{ ms }^{ -1 }\)

a) ₁H = 1s¹, H^- = 1s²
b) ₁₁Na = 1s²2s², 2p⁶, 3s¹, Na⁺ = 1s²2s²2p⁶
c) ₈O = 1s²2s², 2p², 2p⁴, O^2- = 1s²2s²2p⁶
d) ₉F = 1s²2s²,2p⁵, F^- = 1s²2s²2p⁶.

(a) ₁₅P = 1s², 2s², 2p⁶, 3s², 3p³. 3 unpaired electrons.
(b) ₁₄Si = 1s², 2s², 2p⁶, 3s², 3p². 2 unpaired electrons.
(c) ₁₄Cr = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵, 4s¹. 6 unpaired electrons.
(d) ₂₆Fe = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁶, 4s². 4 unpaired electrons
(e) ₃₆Kr = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶. No unpaired electrons.

Given velocity of electron =1.6 x 10⁶ m sec^-1
mass of elecron = 9.11 x 10^-31 kg
Using de-Broglie wavelength,
\(\lambda =\frac { h }{ mv } =3.5\times 10^{ -11 } \ m\)
\(= \frac { 6.626\times 10^{ -34 } }{ 9.11\times 10^{ -31 }\times 1.6\times 10^{ 6 } }\)
\(=0.455\times 10^{ -9 } \ m=0.455\times 10^{ -12 } \ m\)
\( \lambda =455\ pm\)

The d_xy orbital is exactly like  dx²−y² orbital except that its lobes are at an angle of 45^o to the lobes of dx²−y² orbital.

Nuclear charge experienced by an electron (present in a multi-electron atom) is dependant upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge will decreases. Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atom with (+35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the 2p and 3p orbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge.

If n photons are emitted by a laser, the total energy of the photons emitted is equal to the power of the laser
Energy of 1photon, \(E=\frac { hc }{ \lambda } \)
\(=\frac { 6.626\times10^{ -34 }Js\times3.0\times10^{ 8 }ms^{ -1 } }{ 337.1\times10^{ -9 }m } (1nm=10^{ -9 }m)\)
= 0.05896 x 10^-17J
Energy of 5.6 x 10²⁴ photons
= 0.05896 x 10^-17 x 5.6 x 10²⁴ J
= 0.3302 x 10⁷ J
= 3.302 x 10⁶ J

(i) E₃ nergy in nth orbit, E_n = \(\frac { -2.18 \ \times \ 10^{ -18 } }{ n^{ 2 } } \)
Energy in fifth orbit, 
\(E_{ 5 }=\frac { -2.18 \ \times \ 10^{ -18 } }{ 5^{ 2 } } \)J = -8.72 x 10^-20J
(ii) For H-atom, radius of nth orbit,
r_n = 0.529 x n²A
Radius of 5th Bohr orbit, 
r₅ = 0.529 x 5² = 13.225 A
= 1.3225 nm

The energy of photon is sufficient to disturb a sub-automic particle so that there  is uncertainty in the measurement of position and momentum of the sub-atomic particle. However the energy is insufficient to disturb a macroscopic object.

(A) \(\lambda =300nm=300\times { 10 }^{ -9 }m\)
(B) \(\lambda =300\mu m=300\times { 10 }^{ -6 }m\)
(C) \(\lambda =3nm=3\times { 10 }^{ -9 }m\)
(D) \(\lambda =30\overset { \circ }{ A } =30\times { 10 }^{ -9 }m=3\times { 10 }^{ -9 }m\)
\(\because \) Energy, \(E=\frac { hc }{ \lambda } orE\propto \frac { 1 }{ \lambda } \) 
\(\therefore \) Increasing order of energy is B

Helium nucleus contains 2 protons and charge of a proton is same as that of an electron.
Therefore, the charge on the nucleus of a helium atom is (+2)×4.8×10⁻¹⁰=+9.6×10⁻¹⁰ esu

In case of hydrogen atom, there is no neutron. It consists of only one proton

Neutron is a neutral practice. Hence it will not be deflected on passing through an electric field.

According to de Broglie equation,\(\lambda =\frac { h }{ mv } \)
Mass of electron = 9.1 x 10^-31 kg; Planck's constant = 6.626 x 10^-34 kgm²s^-1
Velocity of electron = 1% of speed of light = 3.0 x 10⁸ x 0.01 = 3 x 10⁶ me-1
Wavelength of electron (\(\lambda \)) =\(\frac { h }{ mv } =\frac { (6.626x10^{ -34 }kgm^{ 2 }s^{ -1 }) }{ (9.1x10^{ -31 }kg)x(3x10^{ 6 }ms^{ -1 }) } \)
= 2.43 x 10^-10 m.

For n = 3, the possible values of l are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0 and ml = 0); there are three 3p orbitals (n = 3, l = 1 and ml = –1, 0, +1); there are five 3d orbitals (n = 3, l = 2 and ml = –2, –1, 0, +1+, +2).
Therefore, the total number of orbitals is 1+3+5 = 9
The same value can also be obtained by using the relation; number of orbitals = n², i.e. 3² = 9

(i) Total electrons if n = 4 = \(2n^{2}\),2 x \(4^{2}=32\)
half of the total electrons,i.e 16 electrons have ms = – ½
(ii) n = 3, l = 0 it is 3s-orbital and it can have two electron.

Given, m = 100g = 0.1kg
v = 100km/h = \(\frac { 100\times 1000 }{ 60\times 60 } =\frac { 1000 }{ 36 } { ms }^{ -1 }\)   
From de-Broglie equation, wavelength,
\(\lambda =\frac { h }{ mv } =\frac { 6.626\times { 10 }^{ -34 }{ kgm }^{ 2 }{ s }^{ -1 } }{ 0.1kg\times \frac { 100 }{ 36 } { ms }^{ -1 } } =238.5\times { 10 }^{ -36 }m\)
As the wavelength is very small so wave nature cannot be detected.

(i) Aufbau Principle: In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. In other words, electrons first occupy the lowest-energy orbital available to them and enter into higher energy orbitals only after the lower energy orbitals are filled.
The order in which the energies of the orbitals increase and hence the order in which the orbitals are filled is as follows:
15, 25, 2p, 3s, 3p, 4s, 3d, 4p, 55, 4d, 5p, 6s, 4f, 5d, 6p, 75, Sf, 6d, 7p .......
(ii) Pauli Exclusion Principle: An orbital can have maximum of two electrons and
these must have opposite signs.
For example: Two electrons in an orbital can be represented by

The two electrons have opposite spin, if one is revolving clockwise, the other is revolving anticlockwise or vice versa.
(iii) Hund's Rule of Maximum Multiplicity: Electron pairing in p, d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is single occupied.
For example: For the element nitrogen which contains 7 electrons, the following configuration can be written.

Total spin of unpaired electrons \(=\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =1\frac { 1 }{ 2 } \)

1 mole of NH₃ =10 moles of protons
= \(6.022\times 10^{ 23 }\times 10 \ protons\)
1 mole of NH₃ (or 17g) contains 6.022 x 10²⁴ protons
34 mg or 34 x 10^-3 g NH₃ will contain
\(=\frac { 34\times { 10 }^{ -3 }\times 6.022\times { 10 }^{ 24 } }{ 17 }\)
\(=12.044\times 10^{ 21 } \ protons\)
\(=1.2044\times 10^{ 22 } \ protons\)
Mass of 1 proton = 1.6726x10^-27kg
Mass of 1.2044 x 10²² x 1.6726 x 10^-27kg
= 2.01447 x 10^-5kg

We know that, mass of electron = 9.11 x 10^-31 kg
h = 6.626 x 10^-34 Js
Wavelength, 
\(\lambda =\frac { h }{ mv } =\frac { 6.626\times{ 10 }^{ -34 }kg \ { m }^{ 2 }{ s }^{ -1 } }{ 9.11\times{ 10 }^{ -31 }kg\times2.19\times{ 10 }^{ 6 }\times m{ s }^{ -1 } }\)
\( \lambda =3.32\times{ 10 }^{ -10 }m=332pm\)

1p is not possible because if n = 1 then l = 0 only and for p, l = 1
2s is possible because if n = 2 then l = 0, 1 and for s, l = 0
2p is possible because if n = 2 then l = 0, 1 and for p, l = 1
3f is not possible because if n = 3 then l = 0, 1 and 2 and for f, l = 3.

(a) Kinetic energy of an ejected electron,
KE = hv - hv₀
hv = 3.10 eV (Energy of striking photon)
hv₀ = W₀ = 2.13 eV (Work function of the metal)
KE \(=\frac { 1 }{ 2 } m{ v }^{ 2 }\) = 3,10 - 2.13 = 0.97 eV
(b) \(KE=\frac { 1 }{ 2 } m{ v }^{ 2 }=0.97 \ eV\)
\( \frac { 1 }{ 2 } m{ v }^{ 2 }=0.97\times1.602\times{ 10 }^{ -19 }\)
\( [\because \ 1 \ eV=1.602\times{ 10 }^{ -19 }J]c\)
\( \frac { 1 }{ 2 } \times9.11\times{ 10 }^{ -31 }kg\times{ v }^{ 2 }=0.97\times1.602\times{ 10 }^{ -19 }J\)
\( [\because Mass \ of \ 1{ e }^{ - }=9.11\times{ 10 }^{ -31 }kg]\)
\( { v }^{ 2 }=\frac { 0.97\times1.602\times{ 10 }^{ -19 }\times2J }{ 9.11\times{ 10 }^{ -31 }kg } =0.341\times{ 10 }^{ 12 }\)
\( v=0.584\times{ 10 }^{ 6 }=5.84\times{ 10 }^{ 5 }m{ s }^{ -1 }\)