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PMC – NMDCAT Lecture by Sheikh Shahbaz Ali A.PHYSICAL CHEMISTRY 2. ATOMIC STRUCTURE

Syllabus LECTURE
● Discovery and properties of CHRONICLE
William Crooks = 1879, J.J Thomson established charge = 1897, Discovery of e
proton(Positive Rays) and p discovered = 1886, Discovery of n = 1932(James Chadwick), Noble prize
Practice # 1 to Chadwick = 1935, Properties of n and p discovered = 1895, Prediction of n =
1). Charge on one million electrons is: 1920(Rutherford), Name of electron = Stoney, Name of proton = Rutherford,
A). 1.6022x10 – 19 C C). 1.6022x10 – 25 C Eugene Goldstein experiment = 1886
B). 1.6022x10 – 13 C D). 1.6022x10 – 6 C
2). The charge to mass ratio of positive rays of PART CHARGE RELATIVE MASS ( kg ) MASS
which of the following gas is maximum? ICLE (Coul ) CHARGE ( amu )
A). Oxygen C). N2 1
+1 H 1.6726 x 10 – 27 1.0073
B). CO D). CH4 (+) 1.6022x10 – 19
+1 Or
3). Charge on one mole of electron is: 1836 x
A). 96500C C). 1.6022x10 – 19C 9.1095 x 10 – 31
B). 1.00C D). 2 C n1
0 0 1.6750 x 10 – 27 1.0087
0
4). When α – particle is emitted from the 0
nucleus of a radioactive element, the mass – 1 e or –1 9.1095 x 10 – 31 5.4858x10 – 4
number of atom: –1 β
0
(–) 1.6022x10 – 19 = 0.00055
4
A). increases C). decreases +2 He +2 7344 x 4
B). does not change D). remains constant (+) 3.2044x10 – 19 9.1095 x 10 – 31
5). Gamma rays carry: (4x1.661x10–27)
A). positive charge C). negative charge 0
0
n 0 0 0 0
B). no charge D). partial positive charge
6). When electrons are passed through a
e/m of electron = 1.7588x 1011C/kg(always greater than that of proton)
magnetic field, they bend towards:
e/m of proton = 1.6022x10 – 19 /1.6726 x 10 – 27 C/kg (1836 smaller than e/m
A). north pole C). south pole
B). Both poles D). none of the poles value of electron)
7). Electrons and protons have: Order of mass = alpha particle > neutron>proton > electron = beta particle >
A). Equal charge and equal mass. Quarks
B). Different charge and different mass. Charge on one mole of electron = 96,500C
C). Equal charge and different mass. Order of e/m value = electron> proton > alpha particle
D). Different charge and equal mass. Order of e/m value of positive rays = H2 > He > CH4 > N2 > O2
8). The e/m values for the positive rays is Material nature of electron = Paddle Wheel experiment
maximum for: NUCLEAR REACTIONS
1). 42 He + 94 Be 126 C + 10 n If magnetic field is into the
A). hydrogen C). helium
(Source of alpha particle=Radium and polonium) plane of paper and
B). oxygen D). nitrogen
cathode rays are moving
9). Mass proton is (in kg):
A). +1.6 x 10 – 19 C). –1.6 x 10 – 19 towards right then they
2).10 n 1+1 P + 0– 1e + 00 n are deflected downward
B).1.672 x 10 – 27 D). 9.1 x 10 – 31
9). What will be the correct nuclear reaction and positive rays upwards
3).147 N + 10 n 115 B + 42 He
for the production of neutron?
(Fast neutron, E = 1.2MeV)
A). 42 He + 94 B 126 C + 10 n

B). 42 He + 94 Be 126 C + 10 n
4). 6529 Cu + 10 n 6629 Cu + hv (γ-radiations or photons)
C). 42 He + 94 Be 126 C + 01 n
(Slow neutron, E < 1eV)
D). 2 He + 4 Be
4 9 6
12 C + 10 n 6629 Cu 6630 Zn + 0-1 e

Practice # 2
1). Which is the heaviest particle?
A). Electron C). Neutron
B). Proton D). Beta particle
2). Which prediction of Dalton(1808) about atom is still considered correct?
A). Atoms are very small in size
B). No atom can be split into simpler parts
C). All the atoms of a particular element have same mass
D). All the atoms of one element are different in mass from the atoms of the other elements
3). Which of the following ray are non material:
A). alpha rays C). beta rays
B). gamma rays D). cathode rays
4). When beryllium is bombarded with fast moving alpha particle the sub –atomic particle produced is:
A). Electron C). Proton
B). Neutron D) . Beta particle
5). The negative charge on the cathode rays was established by:
A). William crooks C). J Perrin
B). J.J Thomson D). Hittorf
6). An isotope of 29
66Cu decays to give particle
–1
0 X. What is this particle?

A). Electron C). Neutron

B). Ion D). Proton
7). What data of properties corresponds to alpha particle

PARTICLE CHARGE (Coulomb) RELATIVE CHARGE MASS (kg) MASS(amu)

A +1.6022x10 – 19 +1 1.6726x10 – 27 1.0073
B 0 0 1.6750x10 – 27 1.0087
C – 1.6022x10 – 19 –1 9.11x10 – 31 0.00055
D +3.2044x10 – 19 +2 6.6904x10 – 27 4.0292
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PMC – NMDCAT Lecture by Sheikh Shahbaz Ali A.PHYSICAL CHEMISTRY 2. ATOMIC STRUCTURE

REVISION LECTURE SYMBOL OF NUCLIDE

● Symbol of Element
●Spectral Regions of radiation C
Where E = Element
Practice # 1 A = Mass number or number of nucleons(proton plus neutrons)
1). In which species are the number Z = Atomic number (number of protons)(proton number)
of electron and neutrons are equal? C = Number of protons minus electrons
A). 4 9Be C). 11 23Na+ Atomic number is determined graphically by Moseley’s Law
B). 9 F
19 D). 8 18O2 –
2). What is the composition of v = a (Z – b)
phosphide ion 15 32P3 – Number protons = Number of electrons ( in the neutral atom)
Protons Neutrons Electrons ANION
A). 15 17 18
Number of electrons = Charge on anion + Number of protons
C). 15 17 32
The anions with same number of electrons are isoelectronic anions
B). 17 15 17
D). 32 17 15 EXAMPLE: N – 3 O – 2 F–
3). In which species are the number CATION
of protons, neutrons and electrons Number of electrons = Number of protons – Charge on cation
all are different? The cations with same number of electrons are isoelectronic cations
A). 5 11B C). 9 19F EXAMPLE: Li + Be+2 B+3 C+4
B). 11 Na
23 + D). 12 24Mg2+ PERIODIC TABLE
4). What ion contains maximum 1H He
number of electrons? Li Be B C N O F Ne
A). K+ C). Sr+2 Na Mg Al Si P S Cl Ar
B). Na + D). Mg2+ K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br 36Kr
5). The nucleon number of the most
stable isotope(most abundant) of ELECTROMAGNETIC RADIATIONS
neon is: R M I V U X G [C]
A). 20 C). 22
B). 21 D). 20.18 ROYG B I V
Energy, Frequency, Wavenumber Increases
Wavelength Decreases
6). Which property is the same for the two nuclides 18 40Ar and 19 40K ?
A). the number of electrons C). the number of nucleons
B). the number of neutrons D). the number of protons
7). “ X ” is a particle with 18 electrons and 20 neutrons. What could be the symbol for “ X “?
A). 18 38Ar C). 19 39K+
B). 20 Ca
40 2+ D). All of these.
8). Which set do all the species contain the same number of electrons?
A). Co2+ , Co3+, Co4+ C). Na+ , Mg2+, Al3+
B). F , Br , I
– – – D). N3 – , S 2 – , Cl –
9). Maximum number of neutron are present in :
A). Tritium B). Helium – 3 C). Deutrium D). Protium
10). The number of neutron present in 3 9 1 9 K is:
A). 39 B). 18 C). 20 D). 19
11). How many neutrons are there in the most abundant isotope of neon?
A). 20 B). 12 C). 10 D). 11

Practice # 2
1). What does 3 5 Cl● contains :
Protons Neutrons Electrons
A). 17 18 16
B). 17 18 17
C). 18 17 17
D). 18 17 18
2). Which ion has more electron than protons and more protons than neutrons ( D = 12H )
A). D – B). He+ C). H3O+ D). OH –
3). Consider the ionic reaction and tell about the composition of H3 :+ H 2 + H 2+ H + H 3+
Protons Neutrons Electrons
A). 2 1 1
B). 2 1 2
C). 3 0 1
D). 3 0 2
4). Which one of the following has the same number of electrons as an alpha particle?
A). H B). H2 C). H+ D). He
5). Which of the following formulae represents a particle with the composition 1 proton, 1 neutron and 2 electron (D is
deuterium)?
A). D B). H – C). D – D). He
6). Which of an element of 3 period of Periodic Table contains the same number of neutrons as 16 32S?
rd

A). 11 23Na B). 14 28Si C). 12 24Mg D). 15 31P

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PMC – NMDCAT Lecture by Sheikh Shahbaz Ali A.PHYSICAL CHEMISTRY 2. ATOMIC STRUCTURE

Practice # 2
7). Which statement about 66 27Co is correct?
A). It contains 27 neutrons C). Its nucleus has relative charge of 60+
B). Its nucleus has relative charge of 54+ D). Its nucleus has relative charge of 27+
8). What statement about the composition of an atom of 68Ga is correct ?
A). It has four electrons in its outer p - orbital C). It has 13 electrons in the outer shell
B). It has 37 neutrons D). It proton number is 32

Syllabus LECTURE QUANTUM NUMBERS

Quantum Numbers
Concept Of Orbital Principal Q # Azimuthal Q # Magnetic Q # Spin Q #
Electronic configuration Size of shell, Subsidiary Q # The space orientations The direction of spin
Distance of shell, Energy of subshell (orbitals) in a subshell in of electron
Practice # 1 Energy of electron, and electron, shape magnetic field, Zeeman Doublet line
1). From which quantum number is Line spectrum of orbital, Fine or effect and Stark effect spectrum of alkali
the shape of an orbital determined: seen by ordinary multiple line metals in magnetic
A). principal C). azimuthal spectrometer spectrum seen by field
B). magnetic D). spin high resolving power
2). No two electrons in an orbital spectrometer
can have the same set of four n= 1,2,3,4---∞ ℓ=0,1,2,3 ---(n–1) m = 0,±1,±2,±3----±ℓ S = +½ , –½
quantum numbers is the statement K,L,M,N…… (Anti) (Clock)
of:
s,p,d,f……. “m” cannot be > ℓ
A). Newton’s Law n ≠ 0 or – ve ℓ cannot be > n
B). Coulomb’s law No. of electron No. of electrons No. of space
C). Hund’s rule in s shell=2n2 in a subhell orientations(orbitals)
D). Pauli’s exclusion principle =2(2ℓ+1) in a subhell = 2ℓ+1
3). Azimuthal quantum number
describes an orbital in terms of its: Orbital m
A). size C). orientation Orbital m
px +1
B). shape D). spin dx2 – y2 +2
pZ 0
4). If the configuration of 14 7N is dxz +1
written as 1 s 2 , 2 s 2 , 2 p x2, 2 p y 1 , Subshells in increasing py –1
dz2 0
2 p z 0,then which of following rule energy order according to
is violated: dyz –1
(n+ℓ)Rule or Aufbau
A). Paul’s exclusion principle dxy –2
Principle
B). (n + ℓ ) rule
1s < 2s < 2p < 3s < 3p < 4s
C). Aufbau principle SHAPES OF ATOMIC ORBITALS
D). Hund’s rule < 3d < 4p < 5s < 4d < 5p <
s – orbital spherical shaped No. of nodes in “s”
5). When 6d orbital is complete 6s < 4f < 5d < 6p < 7s < 5f
orbital = n – 1
the entering electron goes into: < 6d < 7p < 8s
“1s” has zero node and “2s” has one node
A). 7f C). 7s [The penetration order of
the subshell is the p – orbital dumb – bell shaped, Triply degenerate in the
B). 7p D). 7d absence of magnetic field. The are oriented along the
6). An Atomic orbital has ℓ = 1, m reversal of the above one]
corresponding axes.
= +1, 0, – 1 n = 3 then which one of d – orbital sausage shaped or complicated shape, Five – fold degenerate in the absence of
the following atomic orbital has magnetic field. The two d – orbitals , dx2 –y2 , dz2 are oriented along the axes and three d –
such values. orbitals, dxz , dyz , dxy are oriented between the corresponding axes.
A). 2s C). 2p f – orbital complicated shaped
B). 3p D). 3d
7). Maximum number of electrons
( n + ℓ)RULE
in a sub – shell is given by: 9F = 1s2 2s2 2p5 (Obeyed) 9F = 1s2 2p6 2s1 (Violated)
A). 2ℓ + 1 C). 2ℓ –1 AufBau PRINCIPLE
B). 2(2ℓ + 1) D). 2(2ℓ – 1)
6 C= 1s2 2s2 2p2 (Obeyed) 6C = 1s2 2s1 2p3 (Violated)
8). Which of these orbitals will be
filled first. PAULI’S EXCLUSION PRINCIPLE
A). 5d C). 4s 6C = 1s2, 2s2, 2p2 (Obeyed) 6C = 1s2, 2s3, 2p1 (Violated)
B). 4f D). 3d HUND’S RULE
9). If the value of ℓ = 1, the atomic
6C = 1s2,2s2, 2px 1 ,2py1, 2pz0 (Obeyed) 6C = 1s2, 2s2, 2px 2,2py0, 2pz0 (Violated)
orbital is:
A). s C). d
B). p D). f 7N = 1s , 2s , 2px ,2py , 2pz (Obeyed) 7N = 1s , 2s, 2px ,2py, 2pz (Violated)
10). Quantum number values for
“2p” orbital are: 3d series Unpaired Shapes of “s” orbitals
A). n = 2, ℓ = 1 C). n = 1, ℓ = 2 Electrons
B). n = 1, ℓ = 0 D). n = 2, ℓ = 0 21Sc [Ar] 3d14s2 1
11). The electronic configuration of 22Ti [Ar] 3d24s2 2
an atom is 1s2 2s2 2p4.The number
23V [Ar] 3d34s2 3 1s 2s 3s
of unpaired electron in this atom is:
A). 0 C). 2 24Cr [Ar] 3d54s1 6 Nodes 0 1 2
B). 4 D). 6 25Mn [Ar] 3d54s2 5 Shapes of “p” orbitals
12). The two lobes of any one “p” 26Fe [Ar] 3d64s2 4
orbital lies along the:
27Co [Ar] 3d74s2 3
A). one axis C). two axes
B). three axes D). all the axes 28Ni [Ar] 3d84s2 2
13). Which orbital has maximum 29Cu [Ar] 3d104s1 1
number of lobes : 30Zn [Ar] 3d104s2 0
A). a “s” orbital C). a “p” orbital
B). a “d” orbital D). a “f” orbital
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PMC – NMDCAT Lecture by Sheikh Shahbaz Ali A.PHYSICAL CHEMISTRY 2. ATOMIC STRUCTURE

Practice # 1
(14). The number of electron in the outermost shell of Magnesium is:
A). 4 C). 12
B). 2 D). 1
15). Partially filled orbitals in nitrogen atom are:
A). one C). two
B). three D). four
16). An orbital which is spherical and
symmetrical is:
A). s – orbital C). p – orbital Shapes of d orbitals
B). d – orbital D). f – orbital
17). What is the value of (n+ℓ) for the 3d subshell is:
A). 2 C). 1
B). 5 D). 3
Practice # 2
1). Which value of magnetic quantum number for “3p” subshell is not allowed ?
A). – 1 C). +1
B). 0 D). +2
2). What value of azimuthal quantum number is not permissible for “M” shell of the atom ?
A). 0 C). 2
B). 1 D). 3
3). Quantum numbers is a set of numerical values that gives solution to Schrodinger wave equation ?
A). Real C). Rejectable
B). Acceptable D). applicable
4). What set of numerical values of four quantum number is not possible ?
A). n =3 , ℓ=2, m = 1 , s = + ½ C). n =3 , ℓ=1, m = 1 , s = – ½
B). n =3 , ℓ=3, m = 1 , s = – ½ D). n =3 , ℓ=2, m = – 2 , s = + ½
5). Quantum numbers for the 24th electron in 24Cr = [Ar] 4s1 , 3d5 ?
A). n =3 , ℓ=2, m = 4 , s = + ½ C). n =3 , ℓ=1, m = 1 , s = – ½
B). n =3 , ℓ=3, m = 1 , s = – ½ D). n =3 , ℓ=2, m = 2 , s = + ½
6). How many “d” orbital are oriented along the axes ?
A). 1 C). 3
B). 2 D). 4
7). What is the order of increasing energy of the listed orbitals in the atom of titanium ?
A). 3s 3p 3d 4s C). 3s 4s 3p 3d
B). 3s 3p 4s 3d D). 4s 3s 3p 3d
8). A simple cation ‘ X+’ has eight protons. What is electronic configuration of ‘X+’ ?
A). 1s2 2s1 2p6 C). 1s2 2s2 2p3
B). 1s2 2s2 2p5 D). 1s2 2s2 2p7
9). Which of the following electronic configuration represents an element that forms a simple ion with charge of – 3 ?
A). 1s2 2s2 2p6 3s2 3p1 C). 1s2 2s2 2p6 3s2 3p6 3d1 4s2
B). 1s 2s 2p 3s 3p
2 2 6 2 3 D). 1s2 2s2 2p6 3s2 3p6 3d3 4s2
10). Which of the following contains an unpaired electron ?
A). Ca2+ C). Ti4+
B). Cu2+ D). K+
11). For which element does its ground state atom have no paired ‘p’ electrons ?
A). C C). Ne
B). O D). Mg
12). The electronic configuration of gallium is: 3 1 Ga = [Ar] 3d10 ,4s2 ,4p1. In which order are the electrons lost in forming Ga4+ ion ?

1st 2nd 3rd 4th

A 3d 4p 4s 4s
B 3d 4s 4s 4p
C 4s 4s 4p 3d
D 4p 4s 4s 3d

13). The diagram show the energy levels of various electronic subshells of an atom of a transition element in the fourth period of the Periodic
Table : y
x

3p
3s

2p
2s

1s E
What are subshells ‘x’ and ‘y’?
A). x = 3d , y = 4s C). x = 4s , y = 4p
B). x = 4s , y = 3d D). x = 4p, y = 3d
14). What kind of orbital must an electron with Principal Quantum Number ‘2’ occupy ?
A). a spherically – shaped orbital C). the orbital closest to the nucleus
B). either a spherically shaped or a dumb – bell shaped orbital D). a dumb – bell shaped orbital
15). Which of the following ions contains five unpaired d – electrons ?
A). Cr3+ C). Mn3+
B). Fe 3+ D). Ni2+
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UHS – MDCAT Lecture by Sheikh Shahbaz Ali A.PHYSICAL CHEMISTRY 2. ATOMIC STRUCTURE

Practice # 2
16). Which diagram best shows the shapes and relative energies of ‘2s’ and ‘2p’ orbitals in carbon atom ?

A 2p
Energy

2s
B 2p
Energy

2s
C 2p
Energy

2s
D 2p
Energy

2s

17). Which of the following particles would, on losing an electron , have a half filled set of ‘p’ orbitals?
A). C – C). N –
B). N D). O +
18). Which ion does have noble gas electronic structure?
A). I+ C). Sn2+
B). Rb+ D). Sr+
19). Which pair do both have one electron only in an ‘s’ orbital in their ground state?
A). Ca, Sc C). H, He
B). Cu, Be D). Li, Cr
20). What would be the proton number of an element that has three unpaired electron in each of its atom?
A). 5 C). 15
B). 13 D). 21
21). Which element has an equal number of electron pairs and unpaired electrons within orbitals of principal quantum number 2?
A). Be C). N
B). C D). O
22). In which pair do the both species have the same number of unpaired ‘p’ electrons?
A). O and Cl+ C). P and Ne+
B). F+ and Al+ D). C and O –
23). Which of the following particle would, on gaining an electron, have half – filled ‘p – orbital’?
A). C C). N –
B). N D). O +

LECTURE
Planck’s Quantum Theory of Radiation(1900)
INTRODUCTION
It explains the emission and absorption of radiation.
1). Energy is not emitted or absorbed continuously. It is emitted or absorbed in the form of wave packets. Wave packet or quantum
has a definite amount of energy. Quantum of energy of light is often called photon.
2). Energy associated with a quantum of radiation is proportional to the frequency (v) of the radiation.
E α v E = hv h = Plank’s constant = 6.625 x 10 – 34 Js= 6.625 x 10 – 27 erg.s = Ratio of energy and frequency
3). A body can emit or absorb energy only in terms of integral multiple of a quantum.
E = nhv Where, n = 1,2,3 ………………..
The emitted or absorbed energy is hv, 2hv, 3hv and so on. It is never as 1.5hv, 3.27hv, 5.9hv etc.

RELATED DEFINITIONS
(1).frequency
Frequency is the number of waves passing through a point per second.
Units cycle.s – 1or s – 1 or Hertz.
v = c/λ Where c = 3 x 108m.s – 1 E = hv
Greater the wavelength, smaller will be the frequency of photon. Greater the frequency of the photon greater will be its energy.

(2). WAVELENGTH Wavelength is the distance between the two adjacent crests or troughs.
UNITS A0, nm or pm. (1A0 = 10 –10m, 1nm = 10 – 9 m, 1pm = 10 – 12m).
λ = c/v
E = hc/ λ
(3).WAVE NUMBER
Wave number ( v ) is the number of waves per unit length, and is reciprocal to wavelength.
UNITS m – 1.

v = 1/ λ v= v/c or v = c v
E = hc / λ
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PMC – NMDCAT Lecture by Sheikh Shahbaz Ali A.PHYSICAL CHEMISTRY 2. ATOMIC STRUCTURE

LECTURE
BOHR’S MODEL OF ATOM HYDROGEN ATOM
Expression for angular momentum [It is quantized] [ fixed number that depend on “n”]
nh
mvr = Angular momentum of electron in hydrogen atom Where n = 1, 2, 3, …………… so on.
2π
h 2h h 3h
The permitted values of angular momenta are, Therefore, , = , ………….so on.
2π 2π π 2π
Expression for radius
є0n2h2 є0h2
r = = ( ) n2 Where Z = 1 for hydrogen atom
πmZe 2 πme 2

є0 = Permittivity of vacuum = 8.84 x 10 – 12 C2 J – 1 m – 1 or C2 N – 1 m – 2

rn = 0.529(n2)A0 rn = 5.29x10 – 11 (n2)m

Distance between orbits of H – atom goes on increasing as we move from 1st orbit to higher orbits. The orbits are not
equally spaced. r2 – r1 < r3 – r2 < r4 – r3 < …………………so on.
r2 = 4r1 , r3 = 9r1, r4 = 16r1
Expression for Potential Energy of Electron
Z e2 Energy of electron at infinity distance
Work done = Epotential =– in the Hydrogen atom:
4 πє0 r
Expression for Total Energy of electron 1313.315
E = Ekinetic + Epotential En =– kJ / mol
∞2
Z2e4m En = 0
En = –
8 πє02n2h2
UNITS OF ENERGY
Z2 m e4 1 1 J = 0.2390 cal = 107 erg
En =– [ ]
8 є02h2 n2 1cal = 4.184 J

1 1 ev/atom = 1.6021892 x 10 – 19 J/atom = 96.484 kJ/mol

En = – 2.18 x 10 – 18 [ ] J. atom – 1
n2

1313.315
En =– kJ / mol [This energy is associated with 1.008g of H – atoms or 1 mole of H – atoms]
2
n
The differences in the values of energy go on decreasing from lower to higher orbits
E2 – E1 > E3 – E2 > E4 – E3 > ---------------------- so on. (Decreases sharply)
IONIZATION OF ENERGY The energy difference between first and infinite levels of energy is calculated as:
E∞ – E1 = 0 – ( –1 313.315) = 1313.315 kJ / mol

•Energy difference of two orbits. •Frequency of radiation emitted during transition.

– Z2 e4m 1 1 Z2 e4m 1 1
∆E = E2 – E1 = – J.atom–1 v = – s– 1
8ε02 h2 n1 n2 8ε02 h3 n12 n22

∆E = 2.18x10 – 18 [1/n1 – 1/n2] J.atom–1 v = Z2e4m/ 8ε02 h3[1/n1 – 1/n2]cycles.s – or s – or

Hertz.
•Wavenumber of radiation emitted during transition. Z2 e4m
•Rydberg’s constant =
Z2 e4m 1 1 8ε02 ch3
v = – m–1 •Rydberg’s constant = 1.09678x107 m– 1
8ε02 ch3 n12 n22 (if Z =1)
• Ions having one electron = He+1, Li+2 and Be+3.
v = 1.09678x107 [1/n1 – 1/n2] m– 1 (if Z =1) • Hα – line in the Balmer series splits into five
components lines in the magnetic field.
• Motion of electron is not in the single plane.
Hydogen Spectrum n2 – n1(n2 – n1+1)
No. of spectral lines =
2
Where n1 = Orbit of low energy and n2 = Orbit of high energy
If electron jumps from higher orbits to n1=6 then spectral lines of Humphrey series are obtained.6lines are obtained
when electrons jumps from 4 to 1 and 15 lines are obtained when electron jumps from 7 to 2.
Shahbaz Ali Associate Professor of Chemistry Govt. Graduate College for boys Satellite Town, Gujranwala