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CA – 1
ATOMIC STRUCTURE
C1
Atomic Hypothesis (Dalton)
It assumed that –
C2
(i)
an atom cannot be subdivided.
(ii)
atoms are neither created not destroyed during chemical reactions.
(iii)
atoms of the same element are alike; in particular all atoms of an element have the same mass.
(iv)
atoms of different elements are not alike; in particular, their masses are different.
Properties of Electron, Proton and Neutron
Symbol
C3
C4
C5
Proton (p)
Neutron (n)
–27
Mass
1.67252 × 10
kg
Charge
1.60210 × 10–19 C
1.67482 × 10
Electron (e)
–27
kg
9.1091 × 10–31 kg
0
1.60210 × 10–19 C
Mass relative to 1836
the electron
1839
1
Charge relative
to the proton
+1
0
–1
Discovery
Goldstein
Chadwick
Thomson
Some Sub-nuclear Particles
Particle
mass
Charge
Antiproton
Same as that of proton
Negative
µ-meson (muon)
210 times that of an electron
Positive and negative
-meson
276 times that of an electron
Positive and negative
(pion)
265 times that of an electron
Zero
Neutrino
Very much less than that of an
electron
Zero
Positron
Same as that of an electron
Positive
Rutherford’s Nuclear Model of Atom
(i)
Rutherford’s scattering experiments disproved the Thomson model and led to the nuclear model
of the atom in which positive charge is spread over a sphere of radius 10–15m, the so-called
nucleus, and that the outer most electron clouds are about 10–10 m from the centre of the nucleus.
(ii)
The positive charge of a nucleus is due to the positively charged particles called protons. But
mass of the nucleus is due to the protons and neutrons.
(iii)
The total number of protons and neutrons in a nucleus determine the nuclear mass.
Nature of Light and Electromagnetic Waves
Wave theory consides light to be a form of wave motion of wavelength , related to frequency v and
velocity of light c be equation v 
c
.

Light waves are also considered electromagnetic in nature (i.e, they are oscillations of electric and
magnetic fields in space). Various types of electromagnetic radiations having various wavelengths
(or frequencies) are known and they constitute the so-called electromagnetic spectrum.
By quantum theory put forward by planck E = nh =
nhc
.

where n is number of photons.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 2
Practice Problems :
1.
Number of photons of light of wavelength 4000 Å required to provide 1.00 J of energy is
(a)
2.
(b)
12.01 × 1031
(c)
1.35 × 1017
(d)
none is correct
One quantum is absorbed per molecule of gaseous iodine for converting into iodine atoms. If light
absorbed has wavelength of 5000 Å, the energy required in kJmol–1 is
(a)
3.
2.01 × 1018
2.38 × 10–5
(b)
2.38 × 105
(c)
1.38 × 102
(d)
none
A near ultraviolet photon of 300 nm is absorbed by a gas and then re-emitted as two photons. One
photon is red with wavelength 760 nm. The wavelength of second photon is
(a)
300 nm
(b)
460 nm
(c)
760 nm
(d)
495.65 nm
[Answers : (1) a (2) b (3) d]
C6A Bohr Theory of H-atom
This theory is used to calculate the radius (r) and energy (E) of a permissible orbit for one-electron species
like H, He+, Li2+ etc.
C6B
Postulates of Borh Theory
(i)
The electrons continue revolving in their respective orbits without losing energy. Thus each orbit
is associated with a definite energy hence it is also called fixed energy levels.
(ii)
Angular momentum (mvr) of an electron in a given orbit is quantised.
mvrn 
nh
.....
2
[m is the mass of electron, v is the velocity of electron, r is radius of orbit in which it is revolving,
n is the number of orbit]
(iii)
Energy is emitted or absorbed by an atom only when an electron moves from one level to an
other. Thus
E  ( E n 2  E n 1 ) 

Thus wave number v is
hc

where
En 2 = energy of the n2 level, En1 = energy of the n1 level
 1 E
v 
 hc
C6C Results of Bohr Theory
1.
Radius of nth orbit
rn  0.53
2.
Velocity of the electron in the nth orbit
vn 
3.
n2
Å where Z = atomic number
Z
Z c 

 where c = 3 × 108 m/s
n  137 
Energy of the electron in the nth orbit
E n  13.6
Z2
n2
ev, E n  ( 2.18  10 18 )
ET = K + U, K = -ET = 
Einstein Classes,
Z2
n2
(J )
U
, U = 2E = –2K
2
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 3
4.
Wavelength of photon emitted for a transition from n2 to n1
1
1
1 
 RHZ2  2  2 

 n1 n 2 
where R = 1.096 × 107 m–1 (Rydberg’s constant)
5.
Ionisation Energy
Energy required to remove the electron from the outermost orbit of the atom in gaseous phase is
called ionisation energy (I.E.).
E = 0, thus the difference between the ground state of an atom and the excited state that
correspond to n2 =  is called ionisation energy.
(IE) z 
6.
13.6 Z 2
n2
eV / atom
Hydrogen Spectrum
Series
n1
n2
Region of Spectrum
Lyman
1
2, 3, ... 
Ultraviolet
Balmer
2
3, 4, ... 
Visible
Paschen
3
4, 5, ... 
Infrared
Brackett
4
5, 6, ... 
Infrared
Pfund
5
6, 7, ... 
Infrared
Practice Problems :
1.
The ratio of the energy of the electron in ground state of hydrogen to that of the electron in first
excited state of Be3+ is :
(a)
2.
(b)
1:8
(c)
1 : 16
(d)
16 : 1
If the shortest wavelength of H atom in Lyman series is x. then longest wavelength in Balmer series
of He+ is :
(a)
3.
1:4
9x
5
(b)
36 x
5
(c)
x
4
(d)
5x
4
The wavelength of the first line in Lyman series of hydrogen and that of the first line in Balmer series
of lithium are in the ratio
(a)
9:4
(b)
3:5
(c)
5:3
(d)
3:2
[Answers : (1) a (2) a (3) a]
C7A Particle and Wave Nature (Dual nature of Electron)
de Broglie based on Millikan’s oil drop experiment (which showed particle nature) and diffraction study
(which showed wave nature) suggested the dual nature of electron, both as a material particle and as a
wave. According to de Broglie’s equation.

h
h


p mv
h
2m(KE)
where p (= mv =
, also  
h
2m e (eV )
2m( KE ) ) is called momentum of the particle of mass m moving with velocity v..
The circumference of nth orbit is equal to n times of wavelength of electrons.
2rn = n
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 4
Practice Problems :
1.
How fast is an electron moving if it have a wavelength equal to the distance it travels in one
second ?
(a)
2.
m
h
(b)
h
p
(c)
(d)
h
2(KE )
A hydrogen molecule at 2000C is moving with a speed of 2.4 × 105 cm per sec. The de-Broglie’s wave
length is of the order of
(a)
3.
h
m
10,000 Å
(b)
1. Å
(c)
5.000 Å
(d)
5Å
The accelerating potential is needed to produce an electron beam with an effective wavelength of
0.090 Å is
(a)
(7.33 × 10–23)/meqe
(b)
(7.33 × 10–23)2/meqe
(c)
(7.33 × 10)2/meqe
(d)
(9.33 × 10–23)2/meqe
[Answers : (1) a (2) b (3) b]
C7B
Photoelectric Effect : When a beam of light of suitable wavelength or frequency is allowed to fall on the
surface of metal, the electrons are emitted from the surface of metal. This phenomena of emittion of
electrons is known as photoelectric effect. It was suggested that some portion of the photon energy is used
up in removing the electrons from the surface by overcoming the attractive force of nucleus (known as
threshold energy or work function of the metal) and rest portion is utilised in imparting velocity of these
electrons (into K.E. of the photoelectrons). h  W0 
1
mv 2 ...[m, v is the mass and velocity of electron]
2
W0 = h0. [where 0 is the threshold frequency]
Practice Problems :
1.
When a certain metal was irradiated with a light of frequency 3.2 × 1016 Hz, the photoelectrons
emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was
irradiated with light frequency 2.0 × 1016 Hz. Hence threshold frequency is
(a)
2.
1.6 × 1016 Hz
(b)
0.8 × 1015 Hz
(c)
8 × 1015 Hz
(d)
8 × 1016 Hz
The threshold frequency for photo electric emission of electrons from platinum is 1.3 × 1015 sec–1.
What is the minimum energy that photons of a particulars radiation must possess to produce the
photo electric effect with platinum metal ?
(a)
energy must be equal to 6.63 × 10–34 × 1.3 × 1015 J
(b)
energy must be greater than 6.63 × 1034 × 1.3 × 1015 J
(c)
energy corresponding to 400 nm
(d)
none of these
[Answers : (1) c (2) b]
C8
Heisenberg’s Uncertainity Principle
It is not possible to determine precisely both the position and the momentum (or velocity) of a small moving
particle (e.g. electron, proton etc.)
h
4
h
x . v 
4 m
x . p 
where x, p and v are the uncertainties with regard to position, momentum and velocity respectively.
In terms of uncertainty in energy, E and uncertanty in time t, this principle is written as,
E . t 
Einstein Classes,
h .
4
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 5
Practice Problems :
1.
The uncertainly in the momentum of an electron is 1 × 10–6 kg m sec–1. The uncertainly in position
would be
(a)
2.
1.05 × 10–23 m
(b)
2.1 × 10–25 m
(c)
1.05 × 10–27 m
(d)
1.05 × 10–25 m
If a 1.0 gm body is travelling along x-axis at 100 m sec–1 within 1 m sec–1. The theoretical uncertainity
in its position is
(a)
103/4h
(b)
103h/4
(c)
h/4
(d)
none
[Answers : (1) d (2) b]
C9A Schrodinger wave mechanical model :
As it was given by the Heisenberg that it is impossible to determine the position and velocity of
e– simultaneously. To overcome this a new model of atom was introduced which was based on dual behaviour
of matter. It was introduced by Erwin Schrodinger.
He introduced a new concept for determining the position of e– i.e., orbital.
C9B
Orbital : It is the region in space where there is high probability of finding the electron.
Difference between orbit and orbital
Orbit
Orbital
1.
It is circular or eliptical path traced by an electron
while revolving round the nucleus of atom giving
fixed value of the distance of e– from the nucleus.
1.
It is the region in space where there is
high probability of finding the electron.
2.
It violates the Heisenberg principle
2.
It does not violate the Heisenberg
principle
3.
It is not in accordance to the dual character
of matter
3.
It is in accordance of dual character of
matter.
C9C Quantum numbers : Each orbitals is designated by three quantum number n, l and m.
The principle quantum number (n) :
1.
It determines the size and to a large extent the energy of the orbital.
2.
The larger the value of n, the larger the energy of the orbital.
3.
Principle quantum number also identifies the shell
i.e.,
n  1 , 2 , 3 , 4 shell.




K
L
M
N
4.
There are n2 orbitals in a shell.
5.
All the orbitals of a given volume of n constitute a single shell of atom.
6.
Each shell consists of one or more subshells or sublevels.
7.
The number of subshells in a principal shell is equal to the value of n.
Azinuthal or Subsidiary quantum number : (l)
Each subshell in a shell is designated by l. l can have n values ranging from o to (n – 1) e.g. when n = 1,
l = 0, n = 2, l = 0, 1 etc. i.e. for n = 1, l = 0 it means there is one subshell for n = 2, l = 0, 1. It means there
are two subshell.
Subshells corresponding to different values of l are represented by following symbols
l=
0, 1, 2, 3, 4, 5
notations
s, p, d, f, g, h
Each subshell consists of one or more orbitals.
The number of orbitals in a subshell is given by (2l + 1).
e.g. In any l = 0 subshell, there are 2(0) + 1 = 1 orbitals.
For l = 1 subshell there are 2(1) + 1 = 3 orbitals.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 6
In any l = 2 subshell, there are 2(2) + 1 = 5 orbitals.
In other words,
subshell notation
=
s, p, d, f, g
value of l
=
0, 1, 2, 3, 4
Number of orbitals
=
1, 3, 5, 7, 9
The orbital angular momentum =
l (l  l )
h
2
The quantum number l also gives the shape of the orbital in the subshell.
Magnetic quantum number (m) :
It gives information about the orientation of the orbital.
For any subshell (defined by l values), (2l + 1) values of m are possible and these values are given as
m = –l, –(l – 1), ...... O, ....... + (l – 1), +l
e.g. for l = 2(d), m can have total = 2(2) + 1 = 5 values.
These values are –2, –1, 0, +1, +2 [Five orbitals]
For l = 1(p) m = 2(1) + 1 = 3 values.
These values are –1, 0, +1 [Three orbitals]
i.e., there are three preffered orientation of p in space.
Thus each orbital is defined by set of values of n, l and m
e.g.
If 4s is given then n = 4, l = 0
If 5p is given then n = 5, l = 1
Spin quantum number : Electrons spin about the axes.
Some spinning in one direction and some in other direction.
Two orientation of electron which is possible is one is in clockwise direction and other one is in anticlockwise
direction.
They are represented by two arrows  (spin up) and  (spin down)
The two spins have either +½ value or –½ value.
Practice Problems :
1.
2.
3.
Correct set of quantum numbers for the unpaired electron in chlorine atom is
(a)
n = 2, l = 1, m = 0
(b)
n = 2, l = 1, m = 1
(c)
n = 3, l = 1, m = 1
(d)
m = 3, l = 0, m = 0
Arrange the electrons represented by the following sets of quantum numbers in the decreasing order
of energy for a multi-electron atom
(i)
n = 4, l = 0, m = 0, s = +½
(ii)
n = 3, l = 1, m = 1, s = –½
(iii)
n = 3, l = 2, m = 0, s = +½
(iv)
n = 3, l = 0, m = 0, s = – ½
(a)
(iii) > (ii) > (i) > (iv)
(b)
(iii) > (iv) > (i) > (ii)
(c)
(i) > (ii) > (iii) > (iv)
(d)
(iii) > (i) > (ii) > (iv)
(c)

For a d-electron the orbital angular momentum is
(a)
4.
(b)
2
(d)
2
(d)
2s, 2p
Among the following the possible orbitals are 1p, 2s, 2p and 3f.
(a)
5.
6
1p, 2s
(b)
1p, 3f
(c)
3f, 2p
The number of electrons in an atom having the following quantum numbers are respectively
(1)
n = 4,
(a)
16, 9
Einstein Classes,
ms = –½
(2)
(b)
32, 18
n = 3, l = 0
(c)
16, 2
(d)
9, 16
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 7
6.
The lowest value of n that allows g orbitals to exist is
(a)
7.
2
(b)
3
(c)
4
(d)
5
An electron is in one of the 3d orbital. The possible values of n, l and ml for this electron are
respectively
(a)
3, 1, 1
(b)
3, 2, –2
(c)
3, 0, 2
(d)
3, 1, –1
[Answers : (1) c (2) d (3) a (4) d (5) d (6) d (7) b]
C9D Electronic configuration of atoms : The distribution of electrons into orbitals of atom is called its
electronic configuration.
The filling of electron into different orbitals takes place according to following three rules :
1.
Aufbau Principle : In the ground state of the atoms, the orbitals are filled in order of their increasing
energies.
In other words,electrons first occupy the lowest energy orbital available to then and then enther to higher
energy orbital.
The order of increase of energy of orbitals can be calculated by (n + l) rule.
Lower the value of (n + l) for an orbital, the lower is its energy.
If two orbitals have the same (n + l) value, the orbital with lower value of n has the lower energy.
2.
Pauli Exclusion Principle : The number of electrons to be filled in various orbitals is restricted by this
principle.
No two electrons in an atom can have the same set of four quantum numbers.
e.g. If an electron in an atom has particular set of quantum number say n = 1, l = 0, m = 0, s = +½ then no
other electron in an atom can have this set of quantum number.
3.
Hund’s Rule of maximum multiplicity : This rule deals with the filling of electrons into orbitals
belonging to same subshell.
Orbital of same subshells like px, py, pz are of equal energy and these are called as degenrate orbitals.
Similarly d has five orbitals of same energy i.e., dxy, dyz, dzx, d x 2  y 2 , d z 2 , f has 7 degenrate orbitals etc.
Hund’s Rule States that : pairing of electrons in the orbitals belonging to the same subshell (p, d or f) does
not takes place until each orbital belonging to that subshell has got one electron each i.e. is singly occupied.



e.g. p x p y p z as there are 3-orbitals the pairing of electron starts in p with the entry of 4th electron.
4.
Magnetic Moment µ =
n(n  2) , n  number of unpaired electron.
Practice Problems :
1.
A compound of vanadium has magnetic moment of 1.73 B.M. The electronic configuration of
vanadium in the compound is
(a)
2.
3.
[Ar]3d1
(b)
[Ar]3d3
(c)
[Ar]3d2
(d)
[Ar]3d0
The ground state electronic configuration of Nitrogen (z = 7) atom. Can be represented as
(a)
(b)
(c)
(d)
If Hund’s rule is not followed, magnetic moment of Fe2+, Mn+ and Cr all having 24 electrons will be
in order :
(a)
Fe2+ < Mn+ < Cr
(b)
Fe2+ = Cr < Mn+
(c)
Fe2+ = Mn+ < Cr
(d)
Mn2+ = Cr < Fe2+
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 8
4.
Magnetic moment of Xn+ (Z = 26) is
respectively are :
(a)
4, 2
(b)
24 B.M. Hence number of unpaired electrons and value of n
2, 4
(c)
3, 1
(d)
0, 2
[Answers : (1) a (2) a (3) b (4) a]
C10
Isotopes
Atoms of the same element having same atomic numbers but different mass numbers are called isotopes
e.g., 92U238 92U235, 84PO213, 84PO216.
Isobars : Atoms of different elements having same mass numbers but different atomic numbers are called
isobars, e.g., Po216 and 85At216, 88Ra228, 89Ac228 and 90Th228.
Isosters : Molecules having same number of atoms and same number of electrons are called Isosters.
For e.g., CO and N2, each has two atoms and the total number of electrons are 14.
Isodiaphers : Atoms having the same difference of neutrons and protons or same isotopic numbers. It is
noticed that nucleide and its decay product after  emission are isodiaphers. e.g., 92U235 and 90Th231, 29Cu65
and 24Cr55.
Isotones : Nucleides having same no. of neutrons are known as isotones 2N14 & 8O15, 54Xe136 & 56Ba139,
Ce140 59Pr141.
50
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 9
SINGLE CORRECT CHOICE TYPE
1.
2.
3.
4.
Which electronic level would allow the hydrogen
atom to absorb a photon but not to emit a photon
(a)
3s
(b)
2p
(c)
1s
(d)
3d
Number of photons of light of wavelength 4000 Å
required to provide 1.00 J of energy is
8.
(b)
10–7
(c)
0.5 × 10–8
(d)
5 × 10–10
(b)
12.01 × 1031
(c)
1.35 × 1017
(d)
none is correct
(a)
3, 1, 1
(b)
3, 2, –2
The energy of an electron in the first Bohr orbit of
H-atom is –13.6 eV. The possible energy values of
the excited state(s) for electrons in Bohr orbit of
hydrogen is (are) :
(c)
3, 0, 2
(d)
3, 1, –1
–3.4 eV
(b)
–4.2 eV
(c)
– 6.8 eV
(d)
6.8 eV
For a 2p-electron the orbital angular momentum is
6
13.
Maximum number of electrons present in an
orbital having n + l = 4
(a)
8
(b)
4
(c)
6
(d)
2
Which of the following statement is correct ?
(a)
The electronic configuration of Cr
(z = 24) is [Ar] 3d5 4s
(b)
The magnetic quantum number may
have a fractional.
(c)
In silver atom (z = 47), 23 electrons have
a spin of one type and 24 of opposite
type
(d)
The oxidation state of Nitrogen in NH3
is 3
2
(b)

12.
(d)
2
h
is the angular momentum of the electron in the

_______________orbit of He+.
7.
2.5 × 10–8
2.01 × 1018
(c)
6.
(a)
(a)
(a)
11.
Two particles A and B are in motion ; if wavelength
of A is 5 × 10 –8 m, the wavelength of B as its
momentum is double of A is
An electron is in one of the 3d orbital. The possible
values of n, l and m l for this electron are
respectively
(a)
5.
10.
(a)
1
(b)
2
(c)
3
(d)
4
14.
Calculate the work function for Na if threshold
frequency is 4.39 × 1014 sec–1
(a)
hv0
(b)
h × c/0
(c)
6.63 × 10–34 × 4.39 × 1014
(d)
all of these
The ionization energies of H, He+ and Li2+ are in
the ratio of
(a)
1:4:9
(b)
1:2:3
(c)
1 : 1/4 : 1/9
(d)
1:1:1
15.
The nucleus of an atom is located at x = y = z = 0. If
the probability of finding an s-orbital electron in a
tiny volume around x = a, y = z = 0 is 1 × 10–5, what
is the probability of finding the electron in the same
sized volume around x = z = 0, y = a ?
(a)
1 × 10–5
(b)
1 × 10–5 × a
(c)
1 × 10–5 × a2
(d)
1 × 10–5 × a–1
Correct set of four quantum numbers for the
valence (outermost) electron of rubidium (Z = 37)
is
(a)
5, 0, 0, 
1
2
(b)
5, 1, 0, 
1
2
(c)
5, 1, 1, 
1
2
(d)
6, 0, 0, 
1
2
The lyman series of the hydrogen spectrum can be
represented as  = 3.2881 × 10
15
1 
 1
 2  2  wheree
n 
1
n = 2, 3...........
16.
An isotone of
The maximum wavelength of line in this series is
9.
76
32 Ge
is
(a)
1.21 × 10–7
(b)
2.20 × 10–7
(a)
77
32 Ge
(b)
77
33 As
(c)
3.28 × 1015
(d)
2.46 × 1015
(c)
77
34 Se
(d)
none
Then the wavelength of an electron moving with a
velocity of 2.05 × 107 m sec–1 is
(a)
(c)
3.53 × 10–10 m
3.53 × 10
–11
m
Einstein Classes,
(b)
3.53 Å
(d)
None
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
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17.
18.
19.
20.
21.
22.
23.
24.
25.
When a certain metal was irradiated with a light of
frequency 3.2 × 1016 Hz, the photoelectrons emitted
had twice the kinetic energy as did photoelectrons
emitted when the same metal was irradiated with
light frequency 2.0 × 10 16 Hz. Hence threshold
frequency is
(a)
1.6 × 1016 Hz
(b)
0.8 × 1015 Hz
(c)
8 × 1015 Hz
(d)
8 × 1016 Hz
26.
27.
A photon was absorbed by a hydrogen atom in its
ground state and the electron was promoted to the
fifth orbit. When the excited atom returned to its
ground state, visible and other quanta was
emitted. Other quanta are :
(a)
21
(b)
52
(c)
31
(d)
41
The wave number of the first line in the Balmer
series of hydrogen is 15200 cm–1, wave number of
the first line in Balmer series of Be3+ is :
(a)
2.43 × 105 cm–1 (b)
1.87 × 105 cm–1
(c)
2.43 × 105 m–1
1.87 × 105 m–1
(d)
28.
pz
(b)
dyz
(c)
dzx
(d)
px
1
(b)
3
(c)
2
(d)
0
10+12
(b)
10+15
(c)
10+8
(d)
10+20
The threshold frequency for photo electric
emission of electrons from platinum is
1.3 × 1015 sec–1. What is the minimum energy that
photons of a particulars radiation must possess to
produce the photo electric effect with platinum
metal ?
(a)
energy must be equal to
6.63 × 10–34 × 1.3 × 1015 J
(b)
energy must be greater than
6.63 × 1034 × 1.3 × 1015 J
(c)
energy corresponding to 400 nm
(d)
none of these
The Rydberg relation between all hydrogen atoms
spectral lines is
The shortest wave length of a photon that can be
emitted when an electron jumps from the n = 4 state
is
The radial distribution curve of 2s sublevel
consists of x nodes, x is :
(a)
(a)
1
 R[1 / n12  1 / n 22 ]

Which of the following orbitals has/have zero
probability of finding the electron in xy plane ?
(a)
If nuclear radius is 10–13 cm and atomic radius is
10 –8 cm, the ratio of atomic volume to nuclear
volume would be
29.
If each orbital can hold a maximum of 3 electrons,
the number of elements in 4th period of periodic
table (long form) is :
(a)
1/R
(b)
16/15 R
(c)
36/5R
(d)
144/7R
Consider the four orbitals in a calcium atom :
2p, 3p, 3d and 4s. These orbitals are arranged in
order of increasing energy is
(a)
2p < 3p < 3d < 4s
(a)
48
(b)
54
(b)
2p < 3p < 4s < 3d
(c)
27
(d)
36
(c)
2p < 4s < 3p < 3d
(d)
4s < 2p < 3p < 3d
Which orbital gives an electron the greatest
probability of being found close to the
nucleus ?
(a)
3p
(b)
3d
(c)
3s
(d)
equal
30.
Which describes orbital :
(a)

(b)
2
(c)
|2|
(d)
none
Suppose 10–17 J of energy is needed by the human
eye to see an object. How many photons of green
light ( = 550 nm) are needed to genrate this
minimum amount of energy ?
(a)
14
(b)
28
(c)
39
(d)
42
Einstein Classes,
31.
The quantum numbers +½ and –½ for the electron
spin represent.
(a)
Rotation of the electron in clockwise and
anti clockwise direction respectively.
(b)
Rotation of electron in anticlockwise and
clockwise direction respectively.
(c)
Magnetic moment of the electron
pointing up and down respectively
(d)
Two quantum mechanical spin states
which have no classical analogue.
One quantum is absorbed per molecule of gaseous
iodine for converting into iodine atoms. If light
absorbed has wavelength of 5000 Å, the energy
required in kJmol–1 is
(a)
2.38 × 10–5
(b)
2.38 × 105
(c)
1.38 × 102
(d)
none
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
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32.
Assume that the ions acts as point charges, each
with a magnitude of 1.6 × 10–19 C. Then how much
energy is released when a sodium ion and chloride
ion originally at infinite distance are brought
together at a distance of 2.76 Å is
(a)
(c)
33.
34.
35.
8.34 × 10
1.3 × 10
–10
–19
ANSWERS
(SINGLE CORRECT CHOICE TYPE)
1.
c
14.
a
27.
b
2.
a
15.
a
28.
b
3.
a
16.
b
29.
b
4.
b
17.
c
30.
d
5.
b
18.
a
31.
b
–19
(b)
8.34 × 10
(d)
1.9 × 10–10
When light with a wavelength of 400 nm falls on
the surface of sodium electrode with a kinetic
energy of 1.05 × 10 5 J mol –1 are emitted. The
minimum energy needed to remove an electron
from sodium is
(a)
3.22 × 10–19 J/atom
6.
d
19.
a
32.
b
(b)
1.22 × 10–15 J/atom
7.
a
20.
a
33.
a
(c)
4.22 × 10–10 J/atom
(d)
5.22 × 10–15 J/atom
8.
a
21.
a
34.
a
9.
c
22.
c
35.
a
10.
a
23.
c
11.
b
24.
a
12.
c
25.
b
13.
c
26.
b
The mass of an electron is me. If its K.E. is E, then
its wavelength is
(a)
h/2meE
(b)
h/2E
(c)
h/2me
(d)
h/meE
If uncertainities in the measurement of position and
momentum of an e– are equal. Then the uncertainity
in the measurement of velocity is
(a)
(c)
1
me
h
4
(b)
v 
h
4
(d)
v 
Einstein Classes,
v 
h
4me
v 
h

Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 12
EXCERCISE BASED ON NEW PATTERN
1.
2.
COMPREHENSION TYPE
MATRIX-MATCH TYPE
Comprehension-1
Matching-1
A spherical particle of unit density and radius of
1.00 × 10–4 cm is trapped between two parallel
charged plates 2.40 cm apart. To keep the particles
at rest, a potential difference of 2052 volts has to be
applied across the plates. From these data find
Column - A
The applied electric field in volts cm–1
(a)
555 volts cm–1
(b)
655 volts cm–1
(c)
755 volts cm–1
(d)
855 volts cm–1
The number of unit charges of electricity on the
particle. One electrostatic unit of potential is equal
to 300 volts.
(a)
1
(b)
2
(c)
3
(d)
4
(A)
(B)
5.
6.
7.
(Q)
Zero
(R)
1.95
(S)
5.82
2+
Mn
(D)
+
Cu
Matching-2
(A)
Column - A
Column - B
A certain metal when
(P)
5.09 × 1014
(Q)
20 : 80
(R)
4v0
(S)
8 × 1015
irradiated to light
(v = 3.2 × 1016 Hz)
emits photoelectrons
with twice kinetic energy
as did photo electrons
Light of wavelength 2000 Å falls on aluminium
surface (work function of aluminium 4.2 eV).
Calculate :
4.
3.8
(C)
Comprehension-2
3.
(P)
3+
Cu
Cr
Column - B
2+
when the same metal is
irradiated by light
The kinetic energy of the fastest and slowest
emitted photoelectrons.
(a)
2.0 eV, 0
(b)
3.0 eV, 0
(c)
4.0 eV, 0
(d)
5.0 eV, 0
(v = 2.0 × 1016 Hz). The
v0 of metal is
(B)
line of Balmer series
Stopping potential.
(a)
5.0 V
(b)
4.0 V
(c)
3.0 V
(d)
2.0 V
The frequency of first
in hydrogen atom is v0.
The frequency of
corresponding line
Cut off wavelength for aluminium.
emitted by singly ionised
(a)
2670 Å
(b)
2770 Å
(c)
2870 Å
(d)
2970 Å
helium atom is
(C)
Naturally occurring
Comprehension-3
Boron consists of two
The series limit for Balmer series of H spectrum
occurs at 3664 Å, calculate
isotopes of atomic weight
10.01 and 11.01. The
Ionisation energy of H atom
(a)
21.79 × 10–12 erg
(b)
31.79 × 10–12 erg
(c)
41.79 × 10–12 erg
(d)
51.79 × 10–12 erg
atomic weight of Boron
is 10.81. The ratio of
percentages of each
isotopes (i.e. B10.01 : B11:01)
Wavelength of the photon that would remove the
electron in the ground state of the H atom.
in Boron is
(D)
A portion of the light
(a)
616 Å
(b)
716 Å
emitted by a sodium
(c)
816 Å
(d)
916 Å
vapour lamp has a wave
length of 589 nm. What
is the its frequency ?
Einstein Classes,
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Matching-3
(B)
Column - A
(A)
Column - B
The wave number of
(P)
4.23 × 10
–8
first line of Balmer series
(C)
–1
electron wave function
quantum
obeying Pauli’s principal
number
Shape, size and
(R)
Magnetic
quantum
The wave number of the
like atomic orbitals
number
2+
(D)
In a hydrogen atom, the
(Q)
1.33 × 10–11
transition takes place from
Probability density
(S)
Electron spin
of electron at the nucleus
quantum
in hydrogen-like atom
number
MULTIPLE CORRECT CHOICE TYPE
n = 3 to n = 2. It Rydberg’s
constant is 1.097 × 10 m ,
For an electron, magnetic quantum number
m = ±2 the electron may be present in
the wavelength of the
(a)
4s-subshell
(b)
4p-sub shell
emitted radiation is
(c)
4d-subshell
(d)
4f-sub shell
7
1.
–1
When is the change in
(R)
136, 800
2.
the orbit radius when
the electron in the
hydrogen atom
(Bohr model) undergoes
the first Paschen transition
(D)
Azimuthal
orientation of hydrogen
ion is
(C)
(Q)
of hydrogen is 15200 cm .
first Balmer line of Li
(B)
A hydrogen-like one-
The speed of a proton
(S)
6564 × 10–8
3.
is one hundredth of the
speed of light in vacuum.
Which of the following set of quantum numbers
are correct
(a)
n = 4, I = 2, m = + 2, s = 0
(b)
n = 5, I = 4, m = 0, s = ± ½
(c)
n = 3, I = 3, m = + 3, s = + ½
(d)
n = 3, I = 2, m = 1, s = 1 ½
Which of the following statements are correct ?
(a)
The electronic configuration of
Cr (z = 24) is [Ar] 3d5 4s1
(b)
The magnetic quantum number may
have a negative value.
(c)
In silver atom (z = 27), 23 electrons have
a spin of one type and 24 of opposite type
(d)
The oxidation state of Nitrogen in NH3
is –3
What is its de-Broglie
wavelength ? Assume that
one mole of protons has
a mass equal to one gram
Matching-4
4.
According to Bohr’s theory
E n = Total energy, K n = Kinetic energy,
Vn = Potential energy, rn = Radius of nth orbit
The ground state electronic configuration of
Nitrogen (z = 7) atom. Can be represented as
(a)
Column - A
Column - B
(A)
Vn/Kn = ?
(P)
0
(B)
If radius of nth orbit
(Q)
–1
(c)
(R)
–2
(d)
(b)
x
n
E ,x=?
(C)
Angular momentum in
lowest orbital
(D)
1
r
n
5.
 Zy , y = ?
(S)
1
Matching-5
(A)
Column - A
Column - B
Orbital angular
(P)
Principal
momentum of the electron
quantum
in a hydrogen-like atom
number
Einstein Classes,
6.
Which of the following species have identical bond
order
(a)
CN–
(b)
O 2–
(c)
NO+
(d)
CN+
Which of the following statements are incorrect
(a)
there are five unpaired electrons in Fe3+
(z = 26)
(b)
Fe2+, Mn+ and Cr all having 24 electrons
will have same value of magnetic moment
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7.
8.
9.
10.
11.
12.
(c)
Copper (I) chloride is coloured salt
(d)
every coloured ion is paramagnetic
13.
Which of the following properties are proportional
to the energy of electromagnetic wave
If Aufbau’s principle and Hund’s Rule are not
followed, which of the following statements are
correct
(a)
Cu+2 is colourless ion
(b)
Fe2+ has electronic configuration
(a)
wave length
(b)
wave number
(c)
number of photons
(c)
K+ 1s of d block
(d)
frequency
(d)
Magnetic moment of Mn is 3 BM.
[Ar]
Which of the following statements are incorrect
Assertion-Reason Type
(a)
The shape of an orbital is determined by
angular wave function
(b)
 2 gives the probability of finding the
electron in a given function
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
(c)
for n = 2, the total value of m are equal to
three
(d)
for l = 0, m does not have any value
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
Which of the following are correct having ns1
electron
I.
Cr(z = 24)
II.
Cu(z = 29)
(C)
Statement-1 is True, Statement-2 is False
III.
Nb(z = 41)
IV.
Mn(z = 25)
(D)
Statement-1 is False, Statement-2 is True
(a)
I, II
(b)
III, IV
(c)
I, II, III
(d)
I, II, IV
About angular momentum of an electron present
in 2p- and 3p-orbital, pick up correct alternates
(a)
for 2p and 3p-orbital, l = 1
(b)
angular momentions of electron depend
upon the value of n
(c)
angular momentum of the electron in
both the orbitals is same
(d)
angular momentum of electron depend
upon the value of 1
Out of the following values for quantum numbers,
which are not possible for an electron in an atom
n
l
m
I.
4
0
–1
II.
4
2
–3
III.
4
2
0
IV.
4
3
–1
(a)
III, IV
(b)
I, II
(c)
I
(d)
II, III
1.
STATEMENT-1 : Electronic configuration of Cu
(z = 29) is [Ar] 3d04s2
STATEMENT-2 : It obey Aufbaue principle
2.
STATEMENT-1 : KE of photo electrons is directly
proportional to the intensity of the incident
radiation.
STATEMENT-2 : Each photon of light causes the
emission of one photo electron
3.
STATEMENT-1 : Pairing of electron in
degenerate orbitals will not start unless each of the
orbitals occupy electrons singly with parallel spins
STATEMENT-2 : The statement is related to
(n + 1) rule
4.
STATEMENT-1 : Dipositive zinc exhibit
paramagnetism due to the loss of two electrons from
3d orbitals of neutral atom
STATEMENT-2 : Paramagnetism-is due to the unpaired electron.
Which of the following statements are incorrect
(a)
Zn (II) salt are coloured
(b)
Fe3+ is light green in colour
(c)
Fe3+ has five unpaired electrons in
(n – 1) d subshell
(d)
Mn+, Fe+ both having 24 electrons will
have same paramagnetism
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 15
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
d
7.
d
2.
c
3.
a
4.
d
5.
d
6.
a
3.
[A-R, B-S, C-P, D-Q]
MATRIX-MATCH TYPE
1.
[A-R, B-P, C-S, D-Q]
2.
[A-S, B-R, C-Q, D-P]
4.
[A-R, B-Q, C-P, D-S]
5.
[A-Q, B-P, Q, R, S, C-P, Q, R, D-P, Q, R]
MULTIPLE CORRECT CHOICE TYPE
1.
c, d
2.
b, d
3.
a, b, c, d
4.
a, d
5.
a, c
6.
b, c
7.
b, c, d
8.
b, c, d
9.
a, c
10.
a, b, c
11.
b, c
12.
a, b, c
13.
b, c
3.
B
ASSERTION-REASON TYPE
1.
D
2.
D
4.
D
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
3.
4.
5.
6.
One quantum is absorbed per molecule of gaseous
iodine for converting into iodine atoms. If light
absorbed has wavelength of 4500 Å, calculate
K.E. energy of iodine atoms (kJ/mol) required in
kJ mol–1.
[B.E. of I2 = 240 kJ/mol]
(a)
Estimate the difference in energy between
the Ist and IInd Bohr orbit for
hydrogen atom.
(b)
At what minimum atomic number, would
a transition from n = 2 to n = 1 energy
level result in the emission of X-rays with
 = 3 × 10–8 m.
(c)
Which Hydrogen atom like species does
this atomic number correspond to ?
How much energy is released when a sodium ion
and chloride ion originally at infinite distance are
brought together at a distance of 2.76 Å. Assume
that the ions acts as point charges, each with a
magnitude of 1.6 × 10–19 C.
What electron transition in a hydrogen atom
starting from the n = 7, will produce infrared light
of wavelength 2170 nm ? (RH = 1.09677 × 107 m–1).
Calculate the velocity of electron placed in the third
orbit of hydrogen atom. Also calculate the number
of revolutions per second that this electron makes
around the nucleus ?
The work function for cesium is 3.43 × 10–19 J. What
is the kinetic energy of an electron liberated by
radiation of 550 nm ? What is the stopping
voltage ? How many electrons are generated if the
total energy absorbed at 550 nm is 1.00 × 10–3 J ?
Einstein Classes,
7.
8.
In the hydrogen spectrum emission lines from
n > 3 to n = 3 are known as the Ritz series.
Calculate the wavelengths and frequencies (s–1) of
the Ritz series. What is the frequency limit (i.e., the
highest possible frequency) for the Ritz series ? (RH
(Rydberg constant = 1.09677 × 107 m–1))
The normalised wave function of 1s orbital is :
  Ne  Zr / a 0
and the radial distribution function is = 4r 2  2
Where
9.
10.
11.
12.
13.
N
Z3
a 03
Calculate the most probable distance at which the
1s electron of hydrogen-like atom which atomic
number Z is to be found.
Calculate the wavelength associated with an
electron accelerated from rest by a potential of
10000 volt.
The mass of an electron is 9.1 × 10–31 kg. If its K.E.
is 3 × 10–25 J, Calculate its wavelength.
Calculate the energy emitted when electrons of
1.0 g atom of H undergo transition giving the
spectral line of lowest energy in the visible region
of its atomic spectrum.
In an X-ray photoelectron experiment, a photon of
wavelength 150 pm ejects an electron from the
inner shell of an atom and it emerges with a speed
of 2.14 × 107 m s–1. Calculate binding energy of the
electron.
The wavelength of -line of the Balmer series is
4815 Å. What is the wavelength of -line of the
Balmer series of the same atom ?
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 16
14.
Calculate the longest wavelength of light that will
be needed to remove an electron from the third Bohr
orbit of He+ ion.
15.
Calculate the frequency, energy and wavelength of
the radiation corresponding to spectral line of
lowest frequency in Layman series in the spectra
of H-atom. Also calculate the energy of the
corresponding line in the spectra of Li2+.
16.
Assuming the spherical shape of nucleus of F
(Fluorine). Calculate the radius of F nucleus, also
find out the nuclear density of F nucleus of mass
number 19.
17.
If a 1 × 10–3 kg body is travelling along the x-axis at
1m/sec within 0.01 m/sec. Calculate the theoretical
uncertainty in its position.
18.
A compound of vanadium has a magnetic moment
of 1.73 BM. Work out the electronic configuration
of the vanadium ion in the compound.
19(a). Write the electronic configuration and names with
symbols of elements with At. numbers 25, 28, 29
and 36.
(b)
Why are the electronic configurations of atoms with
atomic numbers 24 and 29 analogous ?
(c)
How are the following orbitals designated :
(i)
n = 2, l = 1
(ii)
n = 5, l = 3
(iii)
n = 4, l = 0
(iv)
n = 3, l = 2
–28
20(a). An electron has mass = 9.1 × 10 g and is moving
with a velocity of 105 cm/sec. Calculate its kinetic
energy and wavelength when, h, the Planck’s
constant = 6.625 × 10–27 ergs-sec.
(b)
Calculate the de Broglie wavelength associated with
an -particle having an energy of 7.7 × 10–3 J and a
mass of 6.6 × 10–24 g.
(c)
Allowing 0.1% error in the measurement of a
moving dust particle having a mass equal to 10–11 g
and diameter equal to 10–4 cm while moving with a
velocity of 10–4 cm/sec, calculate the uncertainty in
its position.
1.
Find the quantum numbers of the excited state of
electrons in He+ ion which on transition to the
ground state and also on transition to the first
excited state emits two photons of wavelengths
30.4 nm and 108.5 nm respectively.
5.
2.
The lyman series of the hydrogen spectrum can be
represented as
O 2 undergoes photochemical dissociation into
1 normal oxygen atom (O) and more energetic
oxygen atom O*. If (O*) has 1.967 eV more energy
than (O) and normal dissociation energy of O2 is
498 kJ mol–1, what is the maximum wavelength
effective for the photochemical dissociation of O2 ?
6.
With what velocity should an -particle travel
towards the nucleus of a copper atom so as to
arrive at a distance of 10–13m from the nucleus of a
copper atom.
7.
Consider the hydrogen atom to be a proton
embedded in a cavity of radius a0 (Bohr radius)
whose charge is neutrilised by the addition of an
electron to the cavity in vaccum, infinetely slowly.
FINAL STEP EXERCISE
(SUBJECTIVE)
1 
 1
 = 3.2881 × 1015  2  2  where n = 2, 3...........
l
n


3.
4.
(a)
Calculate the maximum and minimum
wavelength of lines in this series.
(b)
In what portion of the electric magnetic
spectrum will this series be found ?
(c)
What value of n corresponds to a
spectral line at 95.0 nm ?
(d)
Is there a line in Lyman Series at
108.5 nm ? Explain.
How many mol of photon would contain sufficient
energy to raise the temperature of 225 gm of water
from 24.50C to 99.50C ? Specific heat of water is
4.18 J gm–1 k–1 and frequency of light radiation used
is 2.45 × 109 sec–1.
An electron in a H-atom in its ground state absorbs
1.5 times as much as energy as the minimum
required for its escape (13.6 eV) from the atom.
Calculate the wavelength of emitted electron ?
Einstein Classes,
Estimate the average total energy of an electron in
its ground state in the hydrogen atom as the work
done in the above neutralisation process. Also if the
magnitude of average of K.E. is half the magnitude
of the average P.E. Find the average PE.
8.
What transition in the hydrogen spectrum would
have the same wavelength as the Balmer transition
n = 4 to n = 2 of He+ spectrum.
9.
Two 1 gm carbon disks 1.00 cm apart have equal
and opposite charges. If force of attraction between
them is 1 × 10–5 N. Calculate the ratio of excess
electrons to total atoms on the negatively charged
disk.
10.
1 mol of He+ ion is excited. Spectral analysis showed
the existance of 50% ions in 3rd level, 25% in 2nd
level and remaining 25% in ground state. Ionisation
energy of He+ is 54.4 eV; calculate total energy
evolved when all the ions return to ground state.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CA – 17
11.
A doubly ionised lithium atom is hydrogen like with
atomic number 3.
(a)
Find the wavelength of the radiation
required to excite the electron in Li2+ from
the first to the third Bohr orbit. Ionisation
energy of hydrogen is 13.6 eV atom–1.
(b)
How many spectral lines are observed in
the emission spectrum of the above
excited system ?
12.
The dissociation energy of H2 is 430.53 kJ mol–1. If
H 2 is exposed to light energy of wavelength
253.7 nm what % of light energy will be converted
into kinetic energy ?
13.
The wavelength of -line of the Balmer series is
4815 Å. What is the wavelength of -line of the
Balmer series of the same atom ?
14.
The dye, when dissolved in water, has its light
absorption at 4530 Å and its maximum fluorescence
emission at 5080 Å. The number of fluorescence
quanta is, on the average, 53% of the number of
quanta absorbed. Using the wavelength of
maximum adsorption and emission, what
percentage of absorbed energy is emitted as
fluorescence ?
15.
The kinetic energy of a sub-atomic particle is
5.625 × 10–25 J. Calculate frequency of the particle
wave.
ANSWERS SUBJECTIVE (Initial Step Exercise)
1.
A = 0.216 × 10–19 J
2.
(a)
3.
8.3 × 10–19 J
6.
3.62 × 10–19 J, 1.9 × 10–20 J, 0.12 V, 2.76 × 1015 photons
7.
3.66 × 1014 s–1
8.
r
9.
0.123 Å
10.
8.93 × 10–7 m
11.
182.5 kJ (for visible spectrum, n1 = 2)
12.
1.12 × 10–15 J
13.
6500 Å
14.
4.69 × 10–7 m
15.
 = 2.46 × 1015 Hz, EH = 16.36 × 10–19,  = 1215.6 Å, Li2+ = 14.7 × 10–18
16.
7.616 × 1013 g/cm3
19.
(a)
Mn : [Ar] 4s2 3d5, Ni : [Ar] 4s2 3d8, Cu : [Ar] 4s1 3d10
(b)
The two configurations are analogous with respect to 4s1 orbital.
(c)
(i) 2p (ii) 5f (iii) 4s (iv) 3d
(a)
4.55 × 10–18 ergs, 7.28 × 10–5 cm
(b)
6.56 × 10–16 cm
20.
10.2 eV
1.
(c)
5 × 10
(n2 = 2, n2 = 5)
3.
–10
(b)
Z=2
(c)
He+
4.
n=4
5.
7.275 × 105 m/sec
a0
Z
2.636 × 10–30 m
17.
18.
([Ar] 4s2)
ANSWERS SUBJECTIVE (Final Step Exercise)
cm
max = 121.67 nm, min = 91.24 nm
2.
(a)
7.22 × 104 mol
4.
4.70 Å
5.
 = 174 nm
7.
  e2 


 2 a 
0 0 

8.
n1 = 1, n2 = 2
9.
4.17 × 10–14 electron/atom
10.
331.13 × 104 J
11.
(a)
(b)
Three
13.
6500 Å
14.
47.26 % [Hint : fraction of energy emitted =
15.
1.709 × 109 s–1
Einstein Classes,
114 Å
6.
12.
8.97 × 106 m/sec
8.68 %
emitted energy
 0.53 J ]
absorbed energy
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111