q. 1 which is not basic postulate of dalton’s atomic...
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Q. 1 Which is not basic postulate of Dalton’s atomic theory?
Option 1 Atoms are neither created nor destroyed in a chemical reaction
Option 2 In a given compound, the relative number and kinds of atoms are constant.
Option 3 Atoms of all elements are alike, including their masses.
Option 4 Each element is composed of extremely small particles called atoms.
Correct Answer 3
Explanation Dalton souid atoms of same element will be alike but atoms of different elements will
be different in all aspect including their masses.
Q.2 The number of electrons in a neutral atom of an element is equal to its:
Option 1 Atomic weight
Option 2 Atomic number
Option 3 Equivalent weight
Option 4 Electron affinity
Correct Answer 2
Explanation For a neutral atom, number of electrons = number of protons = atomic number
Q.3 The e/m for positive rays in comparison to cathode rays is:
Option 1 Very low
Option 2 High
Option 3 Same
Option 4 none
Correct Answer 1
Explanation Cathode rays contains electrons and positive rays contain positive charged gaseous
atom mass of electron is negligible in comparison to positive charge atom
e
m∴∴∴∴ ratio for cathode rays is very low.
Q.4 Which has highest e/m ratio?
Option 1 He2+
Option 2 H+
Option 3 He+
Option 4 H
Correct Answer 2
Explanation e/m ratio for
+2 1He = =
4 2
2+ +1
H = = 11
+
+1 1He = =
4 4
+ 0
H = = 01
Q.5 Cathode rays have:
Option 1 Mass only
Option 2 Charge only
Option 3 Neither mass nor charge
Option 4 Mass and charge both
Correct Answer 4
Explanation Both mass and charge according to the Properties of cathode rays.
Q.6 Mass of neutron is _____ times the mass of electron.
Option 1 1840
Option 2 1480
Option 3 2000
Option 4 None of these
Correct Answer 1
Explanation Massneutron = 1840 Melectron
Q.7 Positive rays or canal rays are:
Option 1 Electromagnetic waves
Option 2 a steam of positively charged gaseous ions
Option 3 A stream of electrons
Option 4 neutrons
Correct Answer 2
Explanation Positive rays are a steam of positively charged gaseous ions.
Q.8 The triad of nuclei that is isotonic is:
Option 1 C, N, F14 15 176 7 9
Option 2 C, N, F12 14 196 7 9
Option 3 C, N, F14 14 176 7 9
Option 4 C, N, F14 14 196 7 9
Correct Answer 1
Explanation For Isotonic species, they should have equal number of neutrons.
C = 14 - 6 = 8
N = 15 - 7 = 8
F = 17 – 9 = 8
Number of neutrons = Mass number – atomic number
Q.9 The ion that s isoelectronic with CO is :
Option 1 CN-
Option 2 O +2
Option 3 O2
Option 4 N+2
Correct Answer 1
Explanation For Isoelectronic species, they should have equal number of electrons.
CO = 6 + 8 = 14
CN = 6 + 7 + 1 = 14
due to negative charge
Q.10 An isotone of 32Ge76
is
(i) 32Ge77
(ii) 33As77
(iii) 34Se77
(iv) 34Se78
Option 1 Only (i) and (ii)
Option 2 Only (ii) and (iii)
Option 3 Only (ii) and (iv)
Option 4 (ii), (iii), and (iv)
Correct Answer 3
Explanation For Isotonic species , Number of neutrons should be same
Ge = 76-32 = 44 As = 77-33 = 4476 7732 33
Ge = 77-32 = 45 Se = 78-34 = 4476 7832 34
34Se77
= 77 – 34 = 43
Q.11 Which of the following atoms and ions are isoelectronic i.e. have the same number of
electron with the neon atom
Option 1 F-
Option 2 Oxygen atom
Option 3 Mg
Option 4 N-
Correct Answer 1
Explanation For isoelectronic species number of electrons should be equal
Ne = 10
F- = a + 1 = 10
0 = 8
Mg = 12
N- = 7 + 1 = 8
Q.12 When a gold sheet is bombarded by a beam of –particlesαααα , only a few of them get
deflected whereas most go straight, undeflected. This is because
Option 1 The force of attraction exerted on the –particlesαααα by the oppositely charged
electrons is not sufficient.
Option 2 A nucleus has a much smaller volume than that of an atom.
Option 3 The force of repulsion acting on the fast moving –particlesαααα is very small.
Option 4 The neutrons in the nucleus do not have any effect on the –particlesαααα
Correct Answer 2
Explanation
Only a few gut defeated whereas most of them go straight, undefeated because most
of the space in the atom is empty and nucleus has very small volume because of which
diffraction occurs.
Q.13 Discovery of the nucleus of an atom was due to the experiment carried out by
Option 1 Bohr
Option 2 Mosley
Option 3 Rutherford
Option 4 Thomson
Correct Answer 3
Explanation Nucleus was discovered during the a-rays scattering experiment performed by
Rutherford
Q.14 Rutherford’s scattering experiment is related to the size of the
Option 1 Nucleus
Option 2 Atom
Option 3 Electron
Option 4 Neutron
Correct Answer 1
Explanation Nucleus
Q.15 The velocity of light is 3.0 10 ms×××× 8 -1. Which value is closest to the wavelength in
nanometers of a quantum of light with frequency of 8 10 s×××× 15 -1
Option 1 3 10×××× 7
Option 2 2 10×××× -25
Option 3 5 10×××× -18
Option 4 3.7 10×××× 1
Correct Answer 4
Explanation c = 3 10 m / sec×××× 8
. = cλ υλ υλ υλ υ
3 10 mlsec=
8 10 sec
××××λλλλ
××××
8
15 -1
3= 10 m
8λ ×λ ×λ ×λ × -7
= 0.37 10 10λ × ×λ × ×λ × ×λ × ×-7 9
= 3.7 10nmλ ×λ ×λ ×λ ×
Q.16 The frequency of a wave of light is 12 10 s×××× 14 -1 . The wave number associated with this
light is
Option 1 5 10 m×××× -7
Option 2 4 10 cm×××× -8 -1
Option 3 2 10 m×××× -7 -1
Option 4 4 10 cm×××× 4 -1
Correct Answer 4
Explanation . = cλ υλ υλ υλ υ
3 10 1= = 10
412 10
××××λ ×λ ×λ ×λ ×
××××
8-6
14
1= wave number = = 4 10 mλ ×λ ×λ ×λ ×
λλλλ6 -1
4 10
cm100
×××× 6-1
≃
4 10 cm×××× 4 -1≃
Q.17 Rank the following types of radiations from the highest energy to the lowest.
Ultraviolet/visible/X-ray/microwave/infrared
Option 1 X-ray, ultraviolet, microwave, infrared, visible
Option 2 Ultraviolet, X-ray, visible, infrared, microwave
Option 3 infrared , microwave, ultraviolet, visible , X-ray
Option 4 X-ray, ultraviolet, visible , infrared, microwave
Correct Answer 4
Explanation Highest Energy →→→→ lowest Energy
X-ray, ultraviolet, visible , infrared, microwave
Q.18 The frequency of a green light is 6 10 Hz×××× 14 . Its wavelength is :
Option 1 500 nm
Option 2 5 mm
Option 3 50,000 nm
Option 4 None of these
Correct Answer 1
Explanation . = cυ λυ λυ λυ λ
3 10=
6 10
××××λλλλ
××××
8
14
1= 10 m
2λ ×λ ×λ ×λ × -6
= 0.5 10 10 nmλ × ×λ × ×λ × ×λ × ×-6 9
= 500nmλλλλ
Q.19 Which wave property is directly proportional to energy of electromagnetic radiation:
Option 1 Velocity
Option 2 Frequency
Option 3 Wave number
Option 4 All of these
Correct Answer 4
Explanation hc 1E = h = = hc. = hcλ υλ υλ υλ υ
λ λλ λλ λλ λ
Q.20 The number of photons of light of = 2.5 10 mν ×ν ×ν ×ν × 6 -1 necessary to provide 1 J of energy
are
Option 1 2 10×××× 18
Option 2 2 10×××× 17
Option 3 2 10×××× 20
Option 4 2 10×××× 19
Correct Answer 1
Explanation E = nhυυυυ
1J = nhc.υυυυ hc = 2 10 J.m×××× -25∵
1 Jn = 2.5 10 m
J.m2 10× ×× ×× ×× ×
××××6 -1
-25
1n =
5 10×××× -19
n = 0.2 10×××× 19
n = 2 10×××× 8
Q.21 The number of photons emitted in 10 hours by a 60 W sodium lamp
( λλλλ of photon = 6000 A����
)
Option 1 6.50 10×××× 24
Option 2 6.40 10×××× 23
Option 3 8.40 10×××× 23
Option 4 3.40 10×××× 23
Correct Answer 1
Explanation EnergyPower =
time
hc = 2 10 J.m×××× -25∵
E60 =
10 60 60sec× ×× ×× ×× ×
E = 2160000 Joules
nhcE =
λλλλ
2160000 6000 10= n = 6.40 10
2 10
× ×× ×× ×× ×××××
××××
-1024
-25
Q.22 Minimum number of photons of light of wavelength 4000 A
����
, which provide 1J energy:
Option 1 2 10×××× 18
Option 2 2 10×××× 9
Option 3 2 10×××× 20
Option 4 2 10×××× 10
Correct Answer 1
Explanation nhcE =
λλλλ
n 2 101J =
4000 10
× ×× ×× ×× ×
××××
-25
-10
4 10n =
2 10
××××
××××
-7
-25
n = 2 10×××× 18
Q.23 The energy E∆∆∆∆ corresponding to intense yellow line of sodium of λλλλ 589 nm is:
Option 1 2.10 eV
Option 2 43.37 eV
Option 3 47.12 eV
Option 4 2.11 kcal
Correct Answer 1
Explanation hcE =∆∆∆∆
λλλλ
1240 cv.nmE =
589nm∆∆∆∆
E = 2.10ev∆∆∆∆
Q.24 The relation between energy of a radiation and its frequency was given by:
Option 1 De Broglie
Option 2 Einstein
Option 3 Planck
Option 4 Bohr
Correct Answer 3
Explanation Planck
Q.25 Which is not characteristics of Planck’s quantum theory of radiation?
Option 1 Radiation is associated with energy.
Option 2 Energy is not absorbed or emitted in whole number or multiples of quantum.
Option 3 The magnitude of energy associated with a quantum is proportional to the frequency.
Option 4 Radiation energy is neither emitted nor absorbed continuously but in small packets
called quanta.
Correct Answer 2
Explanation Plank did not say that energy is absorbed or emitted in whole number or multiple of
quantum
Q.26 Which of the following is not a characteristics of Planck’s quantum theory of radiation?
Option 1 Energy is not absorbed or emitted in whole number or multiples of quantum.
Option 2 Radiation is associated with energy.
Option 3 Radiation is associated with energy emitted or absorbed continuously but in the form
of small packets called quanta.
Option 4 The magnitude of energy associated with quantum is proportional to frequency.
Correct Answer 1
Explanation Energy is not absorbed or emitted in whole number or multiples of quantum.
Q.27 Einstein’s theory of photoelectric effect is based on:
Option 1 Newtons corpuscular theory of light
Option 2 Huygen’s wave theory of light
Option 3 Maxwell’s electromagnetic theory of light
Option 4 Planck’s quantum theory of light
Correct Answer 4
Explanation Planck’s quantum theory of light
Q.28 In photoelectric effect the number of photo-electrons emitted is proportional to :
Option 1 Intensity of incident beam
Option 2 Frequency of incident beam
Option 3 Velocity of incident beam
Option 4 Work function of photo cathode
Correct Answer 1
Explanation According to photoelectric effect frequency decides whether there will be current or
not but the amount of current (no. of photon ejected out) will be decided by the
intensity of light
Q.29 Increase in the frequency of the incident radiations increase the :
Option 1 Rate of emission of photo-electrons
Option 2 Work function
Option 3 Kinetic energy of photo-electrons
Option 4 Threshold frequency
Correct Answer 3
Explanation h -h = K.Eυ υυ υυ υυ υ0
So if we increase υυυυ then K.E. will increases as υυυυ0 or h - wυυυυ0 0 is constant for a given
metal
Q.30 Threshold wavelength depends upon:
Option 1 Frequency of incident radiation
Option 2 Velocity of electrons
Option 3 Work function
Option 4 None of the above
Correct Answer 3
Explanation Work function =
hcw =
λλλλ0
0
(Threshold wavelength)
∴∴∴∴ it only depends on work function.
Q.31 Photoelectric effect shows
Option 1 Particle-like behavior of light
Option 2 Wave-like behavior of light
Option 3 Both wave-like and particle behaviour of light
Option 4 Neither wave-like nor particle-like behaviour of light
Correct Answer 1
Explanation Particle-like nature of light. (Theroy Part)
Q.32 Ultraviolet light of 6.2 eV falls on aluminium surface (work function = 4.2 eV). The
kinetic energy (in joule) of the fastest electron emitted is approximately :
Option 1 3 10×××× -21
Option 2 3 10×××× -19
Option 3 3 10×××× -17
Option 4 3 10×××× -15
Correct Answer 2
Explanation h -h = K.Eυ υυ υυ υυ υ0 max
6.2 - 4.2 = kE
k.E = 2ev
= 2 1.6 10 Joules× ×× ×× ×× × -19
= 3 10×××× -19
Q.33 The threshold wavelength for photoelectric effect on sodium is 5000 A
����
. Its work
function is :
Option 1 4 10 J×××× -19
Option 2 1 J
Option 3 2 10 J×××× -19
Option 4 3 10 J×××× -10
Correct Answer 1
Explanation hcw =
λλλλ0
0
2 10 J.mw =
5000 10 m
××××
××××
-25
0 -10
2= 10 10
5× ×× ×× ×× ×-25 7
= 4 10×××× -18
= 4 10 Joules×××× -19
Q.34 The kinetic energy of the photoelectrons does not depend upon
Option 1 Intensity of incident radiation
Option 2 Frequency of incident radiation
Option 3 Wavelength of incident radiation
Option 4 Wave number of incident radiation
Correct Answer 1
Explanation Intensity of incident radiation
Q.35 If the threshold frequency of a metal for photoelectric effect is νννν0 , then which of the
following will not happen?
Option 1 If the frequency of the incident radiation is νννν0 , the kinetic energy of the electrons
ejected is zero.
Option 2 If the frequency of the incident radiation is νννν , the kinetic energy of the electrons
ejected will be h – hν νν νν νν ν0
Option 3 If the frequency is kept same at νννν but intensity is increased, the number of electrons
ejected will increase.
Option 4 If the frequency of incident radiation is further increased, the number of
photoelectrons ejected will increase.
Correct Answer 4
Explanation h -h = K.Eυ υυ υυ υυ υ0
a) if - ,KE = 0υ υυ υυ υυ υ0
b) If - then KE = h -hυ υ υ υυ υ υ υυ υ υ υυ υ υ υ0
c) no. of electrons ejected intensity of light but Independent of frequency
Q.36 The line spectrum of two elements is not identical because
Option 1 They do not have same number of neutrons
Option 2 They have dissimilar mass number
Option 3 They have different energy level schemes
Option 4 They have different number of valence electrons
Correct Answer 3
Explanation Elements can have same no. of neutrons →→→→ Isotones
Elements can have same mass no. →→→→ Isobar
Elements can have same no of valence e⊙
→→→→ Elements in a group
But Elements cannot have same Energy level and because of that line spectrum of two
elements is different
Q.37 The energy of electron in 3rd
orbit of hydrogen atom is
Option 1 -1311.8 kJ mol-1
Option 2 -82.0 kJ mol-1
Option 3 -145.7 kJ mol-1
Option 4 -327.9 kJ mol-1
Correct Answer 3
Explanation (((( ))))(((( ))))
-1312 zE = KJ / mol
n
2
2
(((( ))))(((( ))))
-1312 1E =
3
××××2
3 2
E3 = -145.7 kJ/mol
Q.38 The ionization energy of H atom is 13.6 eV. The ionization energy of Li2+
ion will be
Option 1 54.4 eV
Option 2 40.8 eV
Option 3 27.2 eV
Option 4 122.4 eV
Correct Answer 4
Explanation (((( )))) E +2 = E Z××××2
Ionisation Li ionisation H
(((( ))))= 13.6 3××××2
= 122.4ev
Q.39 The ratio of the difference in energy between the first and the second Bohr orbit to
that between second and third Bohr orbit is
Option 1 1
2
Option 2 1
3
Option 3 27
5
Option 4 4
9
Correct Answer 3
Explanation (((( ))))
(((( ))))
1 1-13.6 z -
E -E 1 2=
1 1E -E-13.6 z -
2 3
2
2 21 2
22 32 2
3
4=5
36
3 36=
4 5××××
27=
5
Q.39 Energy of electron of hydrogen atom in second Bohr orbit is
Option 1 -5.44 10 J×××× -19
Option 2 -5.44 10 kJ×××× -19
Option 3 -5.44 10 cal×××× -19
Option 4 -5.44 10 eV×××× -19
Correct Answer 1
Explanation zE = -13.6
n
2
2
(((( ))))(((( ))))1
E = -13.62
××××2
2
-13.6E = = -3.4ev
4
E = -3.4 1.6 10× ×× ×× ×× × -19
= -5.44 10 Joules×××× -19
Q.41 The energy of second Bohr orbit in the hydrogen atom is -3.4 eV. The energy of fourth
orbit of He+ ion would be
Option 1 -3.4 eV
Option 2 -0.85 eV
Option 3 -13.64 eV
Option 4 +3.4 eV
Correct Answer 1
Explanation -R zE =
n
2H
H 2
(((( ))))(((( ))))
-R 1-3.4 =
2
2
H
2
RH = 13.6
zE = -RH
n
2
+ 2He
(((( ))))(((( ))))2
= -13.64
××××2
2
= -3.4ev
Q.42 The energy of an electron in the first Bohr orbit of H atom is -13.6 eV. The possible
energy value(s) of the excited state(s) for electrons in Bohr orbits to hydrogen is (are)
Option 1 -3.4 eV
Option 2 -4.2 eV
Option 3 -6.8 eV
Option 4 +6.8 eV
Correct Answer 1
Explanation -13.6zE =
n
2
2
For
Q.43 The ionization energy of hydrogen atom is 13.6 eV. The energy required to excite the
electron in a hydrogen atom from the ground state to the first excited state is
Option 1 1.69 10 J×××× -18
Option 2 1.69 10 J×××× -23
Option 3 1.69 10 J×××× 23
Option 4 1.69 10 J×××× 25
Correct Answer 1
Explanation First excited state means n = 2.
Ionisation Energy = -E st1 state
Ei = -13.6ev
(((( ))))(((( ))))
(((( ))))-13.6 1
E =E -E = - -13.6 =10.2ev2
∆∆∆∆2
2 1 2
10.2 1.6 10⇒⇒⇒⇒ × ×× ×× ×× × -19
1.6 10 J⇒⇒⇒⇒ ×××× -18
Q.44 In a Bohr’s model of atom when an electron jumps from n = 1 to n = 3, how much
energy will be emitted or absorbed (1erg = 10+J)
Option 1 2.15 10 erg×××× -11
Option 2 0.1911 10 erg×××× -10
Option 3 2.389 10 erg×××× -12
Option 4 0.239 10 erg×××× -10
Correct Answer 2
Explanation E =E -E∆∆∆∆ 3 1
1 1= -13.6z -
3 1
2
2 2
(((( )))) 1 - 9= -13.6 1
9
××××
2
8= 13.6ev
9××××
=12.08 1.6 10 Joules× ×× ×× ×× × -19
=19.34 10 10 erg× ×× ×× ×× ×-19 7
=19.34 10×××× -12
= 0.1934 10 erg×××× -10
Q.45 An electron in H-atom is moving with a kinetic energy of 5.45 10 J×××× -19 . What will be
energy level for this electron?
Option 1 1
Option 2 2
Option 3 3
Option 4 None of these
Correct Answer 2
Explanation T.E. = -K.E
T.E. = -5.45 10 J×××× -19
(((( ))))-13.6 1-5.45 10= ev =
1.6 10 n
××××
××××
2-19
-19 2
13.6 1.6= n =
5.45
××××2
n = 3.99 42≃
n = 2
Q.46 The energy required to dislodge electron from excited isolated H-atom, IE1 = 13.6 eV is
Option 1 = 13.6 eV
Option 2 13.6 eV>>>>
Option 3 13.6 and 3.4 eV< >< >< >< >
Option 4 3.4 eV≤≤≤≤
Correct Answer 4
Explanation IE1 = +13.6eV
IE2 = +3.4ev →→→→ 1st
excited state
IE3 = +1.51ev →→→→ 2nd
excited state
3.4 eV∴≤∴≤∴≤∴≤
Q.47 The radius of first Bohr’s orbit for hydrogen is 0.53 A
����
. The radius of third Bohr’s orbit
would be
Option 1 0.79 A
����
Option 2 1.59 A
����
Option 3 3.18 A
����
Option 4 4.77 A
����
Correct Answer 4
Explanation nr = 0.529 A
z
20
(((( ))))(((( ))))
0.529 3r =
1
2
3
= 0.53 9××××
= 4.77A����
Q.48 The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 A
����
. The
radius for the first excited state (n = 2) orbit is
Option 1 0.13 A
����
Option 2 1.06 A
����
Option 3 4.77 A
����
Option 4 2.12 A
����
Correct Answer 4
Explanation First excited state n = 2
(((( ))))0.53 2r = = 2.12A
Z
××××2
����
Q.49 According to Bohr model, angular momentum of an electron in the 3rd orbit is :
Option 1 3h
ππππ
Option 2 1.5h
ππππ
Option 3 3
h
ππππ
Option 4 9h
ππππ
Correct Answer 2
Explanation nhmvr =
2ππππ (According to Bohr’s Model)
For 3rd
orbit
3h 1.5hmvr = =
2π ππ ππ ππ π
Q.50 Electronic energy is a negative energy because
Option 1 Electron carries negative charge
Option 2 Energy is zero near the nucleus and decrease as the distance form the nucleus
increases.
Option 3 Energy is zero at an infinite distance from the nucleus and decreases as the electron
comes closer to the nucleus.
Option 4 There are interelectronic repulsions.
Correct Answer 3
Explanation Energy is zero at n= ∞∞∞∞ , it decreases as we go close to the nucleus
∴∴∴∴ It is negative
Q.51 Ratio of frequency of revolution of electron in the second excited state of He+ and
second state of hydrogen is
Option 1 32
27
Option 2 27
32
Option 3 1
54
Option 4 27
2
Correct Answer 1
Explanation velocity 2.18Z zf = =
Distance n 0.529 n××××
×××× 2
(((( ))))z
fn
∝∝∝∝2
3
(((( ))))(((( ))))
(((( ))))(((( ))))
2 2f3 ofHe 32= =
27f2 ofH 3 1××××
2 3rd +
nd 3 2
Q.52 The line spectrum observed when electron when electron jumps from higher level to
M level is known as
Option 1 Balmer series
Option 2 Lyman series
Option 3 Paschen series
Option 4 Brackett series
Correct Answer 3
Explanation M level means n = 3
It comes under Paschen series
Q.53 How many spectral lines are produced in the spectrum of hydrogen atom from 5th
energy level?
Option 1 5
Option 2 10
Option 3 15
Option 4 4
Correct Answer 2
Explanation Number of spectral lines
(((( ))))n n-1=
2
where, n = higher Energy level
n = 5
(((( ))))5 5-1=
2
20= =10
2
Q.54 What transition in He+ ion shall have the same wave number as the first line in Balmar
series of H atom?
Option 1 7 5→→→→
Option 2 5 3→→→→
Option 3 6 4→→→→
Option 4 4 2→→→→
Correct Answer 3
Explanation First line of Balmer series means 3 2→→→→
wave number 1 1 1
= = Rz -2 3
λλλλ
2
2 2
(((( ))))5 5R= Rz = z = 1
36 36
υυυυ
2∵
a) 7 5→→→→ for He+
1 1
= Rz -25 49
υυυυ
2 _________ X
b) 5 3→→→→ For He+
(((( )))) 1 1= R 2 -
9 25
υυυυ
2 _________ X
c) 6 4→→→→ For He+
(((( )))) 1 1= R 2 -
16 36
υυυυ
2
5
= 4R36 16
××××××××
5R
=36
Q.55 An electron jumps from 6th
energy level to 3rd
energy level in H-atom, how many lines
belong to visible region?
Option 1 1
Option 2 2
Option 3 3
Option 4 zero
Correct Answer 4
Explanation 6 3→→→→
In visible region (lines which comes to n = 2)
∴∴∴∴ No lines will come in visible region
Q.56 The wavenumber for the shortest wavelength transition in the Balmer series of atomic
hydrogen is
Option 1 27420 cm-1
Option 2 28420 cm-1
Option 3 29420 cm-1
Option 4 12186 cm-1
Correct Answer 1
Explanation Shortest wavelength for balmer series 2→ ∞ →→ ∞ →→ ∞ →→ ∞ →
1 1 1= = Rz -
2
∴ υ∴ υ∴ υ∴ υ λλλλ ∞∞∞∞
2
2 2
109670 1=
4
×××× 2
= 27417.5cm-1
Q.57 The difference in wavelength of second and third lines of Balmer series in the atomic
spectrum is
Option 1 131 A
����
Option 2 524 A
����
Option 3 324 A
����
Option 4 262 A
����
Correct Answer 2
Explanation second line of balmer series 4 2→→→→
1 1 1 3R= Rz - =
4 16 16
λλλλ
2
1 16=
3Rλλλλ
3rd
line of Balmer series 5 2→→→→
1 1 1 21R= Rz - =
4 25 100
λλλλ
2
1 100=
3 21R
16 100 1- = -
3 21 R
λ λ ×λ λ ×λ λ ×λ λ ×
2 3
1= 911.5A
R∵
����
112 -100 12= 911.5 = 911.5
21 21× ×× ×× ×× ×
=520.8A����
Q.58 The third line in Balmer series corresponds to an electronic transition between which
Bohr’s orbits in hydrogen atom
Option 1 5 3→→→→
Option 2 5 2→→→→
Option 3 4 3→→→→
Option 4 4 2→→→→
Correct Answer 2
Explanation 3rd
line in Balmer series is 5 2→→→→
1 1 1= Rz -
4 25
λλλλ
2
1 21R=
100λλλλ
Q.59 When electrons in N shell of excited hydrogen atom return to ground state, the
number of possible lines spectrum is :
Option 1 6
Option 2 4
Option 3 2
Option 4 3
Correct Answer 1
Explanation N shell means n = 4
no. of spectral line (((( ))))n n-1
=2
(((( ))))4 4 -1=
2
= 6
Q.60 The wave number of the first line of Balmer series of H atom is 15200 cm-1
. What is the
wave number of the first line of Balmer series of Li2+
ion?
Option 1 15200 cm-1
Option 2 6080 cm-1
Option 3 76000 cm-1
Option 4 136800 cm-1
Correct Answer 4
Explanation First line of balmer 3 2→ →→ →→ →→ →
1 1 1= Rz -
4 9
λλλλ
2
1 5R= = 15200
36λλλλ
15200R = 36
5×××× (z = 1 for hydrogen) … (i)
∴∴∴∴ First line of Balmer for Li+2
1 1= RZ -
4 9
1111 λλλλ
2
(((( ))))1 5=R 3
36××××
λλλλ2
15200 36 5= = 9
5 36
×××××××× (after putting volume of R from (i))
= 136800cm-1
Q.61 In hydrogen spectrum, the series of lines appearing in ultra violet region of
electromagnetic spectrum are called
Option 1 Balmer lines
Option 2 Lyman lines
Option 3 Pfund lines
Option 4 Brackett lines
Correct Answer 2
Explanation Lyman lines (Theory)
Q.62 The wave number of the first line of Balmer series of hydrogen is 15200 cm-1
. The wave
number of the first Balmer line of Li2+
ion is
Option 1 15200 cm-1
Option 2 60800 cm-1
Option 3 76000 cm-1
Option 4 136800 cm-1
Correct Answer 4
Explanation First line of Balmer for H 3 2→ →→ →→ →→ →
(((( ))))1 1 1 5R= R 1 - = = 15200
4 9 36
λλλλ
2
First line of Balmer for Li+2
3 2→ →→ →→ →→ →
(((( ))))1 1 1 5R= R 3 - = 9 = 15200 9
4 9 36
× ×× ×× ×× × λλλλ
2
= 136800cm-1
Q.63 A certain transition in H spectrum from an excited state to the ground state in one or
more steps gives rise to a total of 10 lines. How many of these belong to the UV
spectrum?
Option 1 3
Option 2 4
Option 3 6
Option 4 5
Correct Answer 2
Explanation Number of spectral lines
(((( ))))n n-1=
2
n -n10 =
2
2
n2- n – 20 = 0
n2 – 5n + 4n - 20 = 0
(n - 5) (n + 4) = 0
n = 5 as n = -4 is not possible
For uv spectrum line has to come form any level to n = 1.
Q.64 The de-Broglie wavelength associated with a material particle is
Option 1 Directly proportional to its energy
Option 2 Directly proportional to momentum
Option 3 Inversely proportional to its energy
Option 4 Inversely proportional to momentum
Correct Answer 4
Explanation h h= =
mv pλλλλdb
p →→→→ momentum of particle
λ →λ →λ →λ →db de-broglie wavelength
Q.65 The de Broglie wavelength of a tennis ball of mass 66 g moving with the velocity of 10
metres per second is approximately
Option 1 10-35
metres
Option 2 10-33
metres
Option 3 10-31
metres
Option 4 10-36
metres
Correct Answer 2
Explanation h=
mvλλλλ
6.6 10= 10
66 10
××××××××
××××
-34
-3
= 10 10××××-34
= 10 m-33
Q.66 The wavelength of a cricket ball weighing 100 g and travelling with a velocity of 50 m/s
is
Option 1 1.3 10 m×××× -28
Option 2 1.3 10 m×××× -37
Option 3 1.3 10 m×××× -34
Option 4 1.3 10 m×××× -30
Correct Answer 3
Explanation h=
mvλλλλ
6.6 10= 50
100 10
××××××××
××××
-34
-3
6.6= 10
5×××× -34
= 1.32 10 m×××× -34
Q.67 An electron has kinetic energy 2.8 10 J×××× -23 de-Broglie wavelength will be nearly
( = 9.1 10 kg)×××× -31me
Option 1 9.24 10 m×××× -4
Option 2 9.24 10 m×××× -7
Option 3 9.24 10 m×××× -8
Option 4 9.24 10 m×××× -10
Correct Answer 3
Explanation h h= =
mv 2mkEλλλλdb
6.6 10=
2 9.1 10 2.8 10
××××
× × × ×× × × ×× × × ×× × × ×
-34
-31 -23
6.6 10=
50.96 10
××××
××××
-34
-54
6.6 10=
50.9 10
××××
××××
-34
-27
= 0.942 10×××× -7
= 9.42 10 m×××× -8
Q.68 A cricket ball of 0.5 kg is moving with a velocity of 100 ms-1
. The wavelength associated
with its motion is
Option 1 1cm
100
Option 2 66 10 m×××× -34
Option 3 1.32 10 m×××× -35
Option 4 6.6 10 m×××× -28
Correct Answer 3
Explanation h=
mvλλλλdb
6.6 10=
0.5 100
××××
××××
-34
66= 10
5×××× -36
= 13.2 10×××× -36
= 1.32 10 m×××× -35
Q.69 An electron with velocity v is found to have a certain value of de Broglie wavelength.
The velocity that the neutron should possess to have the same de Broglie wavelength
is
Option 1 V
Option 2
1840
v
Option 3 1840 v
Option 4 1840
v
Correct Answer 2
Explanation =λ λλ λλ λλ λelectron neutron
h h=
m v m ve e n n
mev = v
m××××n
n
1m = m
1840∴∴∴∴ e n
vv =
1840n
Q.70 If uncertainty in the position of an electron is zero, the uncertainty in its momentum
would be
Option 1 Zero
Option 2 h
4≥≥≥≥
ππππ
Option 3 h
4ππππ<
Option 4 Infinite
Correct Answer 4
Explanation hx. p
4∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥
ππππ
x 0∴∆ =∴∆ =∴∆ =∴∆ =
hp
4∆ ≥ ∞∆ ≥ ∞∆ ≥ ∞∆ ≥ ∞
ππππ≃
Q.71 For an electron, if the uncertainty in velocity is ∆ν∆ν∆ν∆ν , the uncertainty in its position ( )∆∆∆∆x
is given by :
Option 1
2π ∆νπ ∆νπ ∆νπ ∆ν
hm
Option 2 2ππππ
∆ν∆ν∆ν∆νhm
Option 3
4π ∆νπ ∆νπ ∆νπ ∆ν
h
m
Option 4 2ππππ
∆ν∆ν∆ν∆ν
m
h
Correct Answer 3
Explanation hx. p
4∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥
ππππ
hx. mv
4∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥
ππππ
P = mv∵
hx. v
4 m∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥
ππππ
hx
4 m v∆ ≥∆ ≥∆ ≥∆ ≥
π ∆π ∆π ∆π ∆
Q.72 A ball of mass 200 g is moving with a velocity of 10m sec-1
. If the error in measurement
of velocity is 0.1% , the uncertainty in its position is:
Option 1 3.3 10 m×××× -31
Option 2 3.3 10 m×××× -27
Option 3 5.3 10 m×××× -25
Option 4 2.64 10 m×××× -32
Correct Answer 4
Explanation v = 0.1∆∆∆∆ %
0.1= 10 = 0.01
100××××
hx
4 m v∆ ≥∆ ≥∆ ≥∆ ≥
π ∆π ∆π ∆π ∆
6.6 10
4 3.14 200 10 0.01
××××≥≥≥≥
× × × ×× × × ×× × × ×× × × ×
-34
-3
= 0.2627 10 10× ×× ×× ×× ×-34 3
= 2.62 10 m×××× -32
Q.73 The Heisenbergs uncertainty principle states that______.
Option 1 No two electrons in the same atom can have the same set of four quantum numbers
Option 2 Two atoms of the same element must have the same number of protons
Option 3 It is impossible to determine accurately both the position and momentum of an
electron simultaneously
Option 4 Electrons of atoms in their ground states enter energetically equivalent sets of orbitals
singly before they pair up in any orbital of the set
Correct Answer 3
Explanation It is impossible to determine accurately both the position and momentum of an
electron simultaneously
By theory
Q.74 If the uncertainty in the position of an electron is zero, the uncertainty in its
momentum be
Option 1 Zero
Option 2
2ππππ
h
Option 3
4ππππ
h
Option 4 Infinity
Correct Answer 4
Explanation hx. p
4∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥
ππππ
x = 0∆∆∆∆
p∆ ∞∆ ∞∆ ∞∆ ∞≃
Q.75 If uncertainty in the measurement of position and momentum of an electron are equal
then uncertainty in the measurement of its velocity is approximately:
Option 1 8 10 m s×××× 12 -1
Option 2 6 10 m s×××× 12 -1
Option 3 4 10 m s×××× 12 -1
Option 4 2 10 m s×××× 12 -1
Correct Answer 1
Explanation hx p
4∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥∆ ∆ ≥
ππππ≃≃≃≃
x = p∆ ∆∆ ∆∆ ∆∆ ∆
(((( )))) hp
4∆ ≥∆ ≥∆ ≥∆ ≥
ππππ2
(((( )))) hm v
4∆ ≥∆ ≥∆ ≥∆ ≥
ππππ2
hv
4 m∆ ≥∆ ≥∆ ≥∆ ≥
ππππ2
2
(((( ))))6.6 10
v =
4 3.14 9.1 10
××××∆∆∆∆
× × ×× × ×× × ×× × ×
-34
2-31
6.6= 10 10
1040× ×× ×× ×× ×-34 62
6.6= 10
1040×××× 14
= 0.079 10×××× 14
= 8 10×××× 12
Q.76 In the Schrodinger’s wave equation ΨΨΨΨ represents
Option 1 Orbit
Option 2 Wave function
Option 3 Wave
Option 4 Radial probability
Correct Answer 2
Explanation ΨΨΨΨ represents wave function
Q.77 For each value of ℓℓℓℓ , the number of ms values are
Option 1 2ℓℓℓℓ
Option 2 nℓℓℓℓ
Option 3 2 + 1ℓℓℓℓ
Option 4 -n ℓℓℓℓ
Correct Answer 3
Explanation 2 + 1ℓℓℓℓ (Theory)
Q.78 A subshell with n = 6, = 2ℓℓℓℓ can accommodate a maximum of
Option 1 10 electrons
Option 2 12 electrons
Option 3 36 electrons
Option 4 72 electrons
Correct Answer 1
Explanation n = 6, = 2ℓℓℓℓ
means 6d →→→→ will have 5 orbitals.
∴∴∴∴ max 10 electrons can be aceomalid as each orbital can have maximum of 2
electrons.
Q.79 Which of the following sets of quantum number is correct for an electron in 4f orbital?
Option 1 +1= 3, = 2, = -2, =
2n m sℓℓℓℓ
Option 2 -1= 4, = 4, = -4, =
2n m sℓℓℓℓ
Option 3 +1= 4, = 3, = +1, =
2n m sℓℓℓℓ
Option 4 +1= 4, = 3, = +4, =
2n m sℓℓℓℓ
Correct Answer 3
Explanation +1= 4, = 3, = +1, =
2n m sℓℓℓℓ
4f orbital means
= 4, = 3, = -3,-2,-1,0,1,2,31n mℓℓℓℓ
ms = 1
2±±±±
+1= 4, = 3, = +1, =
2∴∴∴∴n m sℓℓℓℓ
Q.80 For a d-electron, the orbital angular momentum is
Option 1 h6
2
ππππ
Option 2 h2
2
ππππ
Option 3 h
2ππππ
Option 4 Zero
Correct Answer 1
Explanation orbital angular momentum = (((( ))))
h+1
2ππππℓ ℓℓ ℓℓ ℓℓ ℓ
for d electron = 2ℓℓℓℓ
(((( )))) h= 2 2+1
2ππππ
h= 6
2ππππ
Q.81 The correct designation of an electron with n = 4, l = 3, m = 2, and s = 1/2 is :
Option 1 3d
Option 2 4f
Option 3 5p
Option 4 6s
Correct Answer 2
Explanation n = 4, l = 3, m = 2, and s = 1/2
means electron is present in 4f
Q.82 A 3d-electron having s = +1/2 can have a magnetic quantum no :
Option 1 +2
Option 2 +3
Option 3 -3
Option 4 +4
Correct Answer 1
Explanation For 3d
n = 3, l = 2 m can be -2, -1, 0, 1, 2
Q.83 The electrons identify by n and I
(i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n = 3, l = 2 (iv) n = 3, l = 1
Can be placed in order of increasing energy, from lowest to highest
Option 1 (iv) (ii) (iii) (i)< < << < << < << < <
Option 2 (ii) (iv) (i) (iii)< < << < << < << < <
Option 3 (i) (iii) (ii) (iv)< < << < << < << < <
Option 4 (iii) (i) (iv) (ii)< < << < << < << < <
Correct Answer 1
Explanation n = 4 l = 1 →→→→ 4p n + l = 5
n = 4 l = 0 →→→→ 4s n + l = 45
n = 3 l = 2 →→→→ 3d n + l = 5
n = 3 l = 1 →→→→ 3p n + l = 4
n + l rule state the shell will less value of n + l will have lesser energy and in case if n + l
is same for two orbitals we have to give preference to n,
3p 4s 3d 4p∴ < < <∴ < < <∴ < < <∴ < < <
Q.84 According to (n + l) rule after completing ‘np’ level the electron enters to :
Option 1 (n - l) d
Option 2 (n + l) s
Option 3 Nd
Option 4 (n + l) p
Correct Answer 2
Explanation Configuration order is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p
∴∴∴∴ we can see that after 2p
There is 3s
after 3p there is 4s and so on
∴∴∴∴ after np level electron will enter (n + 1)S.
Q.85 The correct ground state electronic configuration of chromium atom (Z = 24) is
Option 1 [Ar] 3d5 4s
1
Option 2 [Ar] 3d4 4s
2
Option 3 [Ar] 3d6 4s
0
Option 4 [Ar] 4s1 4p
5
Correct Answer 1
Explanation z = 24
chromium is exceptional electronic configuration
n[Ar] 3d5 4s
1
Q.86 In manganese atom, Mn (Z = 25), the total number of orbitals populated by one or
more electrons (in ground state) is
Option 1 15
Option 2 14
Option 3 12
Option 4 10
Correct Answer 1
Explanation
Q.87 The correct set of quantum numbers for the unpaired electron of chlorine atom is
n ℓℓℓℓ m
Option 1 2 1 0
Option 2 2 1 1
Option 3 3 1 1
Option 4 3 0 0
Correct Answer 3
Explanation Cl 1s 2s 2p 3s 3p→→→→ 2 2 6 2 5
Q.88 The maximum number of 4d-electrons having spin quantum number
1= +
2s are
Option 1 10
Option 2 7
Option 3 1
Option 4 5
Correct Answer 4
Explanation 4d will have 5 orbitals
5 electrons will have 1
+2
spin
rust 5 will have 1
-2
spin
Q.89 Which of the following has maximum number of unpaired electrons?
Option 1 Mg2+
Option 2 Ti3+
Option 3 V3+
Option 4 Fe3+
Correct Answer 4
Explanation Mg+2
1s2 2s
2 2p
6 ⇒⇒⇒⇒ 0
Ti+3
1s2 2s
2 2p
6 3s
2 3p
6 3d
1 ⇒⇒⇒⇒ 1
V+3
1s2 2s
2 2p
6 3s
2 3p
6 3d
2 ⇒⇒⇒⇒ 2
Fe+3
1s2 2s
2 2p
6 3s
2 3p
6 3d
5 ⇒⇒⇒⇒ 5
Q.90 Azimuthal quantum number for the last electron in Na atom is
Option 1 1
Option 2 0
Option 3 2
Option 4 3
Correct Answer 2
Explanation Na →→→→ 1s22s
22p
63s
1
↓↓↓↓
l = 0 for 3s
Q.91 Presence of three unpaired electrons in phosphorus atom can be explained by
Option 1 Pauli’s rule
Option 2 Uncertainty principle
Option 3 Aufbau’s rule
Option 4 Hund’s rule
Correct Answer 4
Explanation P= 1s2 2s
2 2p
6 3s
2 3p
3
Hund’s rule (Theory)
Q.92 Consider the following ion
1. Ni2+
2. Co2+
3. Cr2+
4. Fe3+
(Atomic number : Cr = 24, Fe = 26, Co = 27 and Ni = 28)
The correct sequence of increasing number of unpaired electrons in these ions is
Option 1 1, 2, 3, 4
Option 2 4, 2, 3, 1
Option 3 1, 3, 2, 4
Option 4 3, 4, 2, 1
Correct Answer 1
Explanation
1, 2, 3 , 4
Q.93 The quantum numbers for the outermost electron of an element are given below
1= 2, = 0, = 0, = +
2n m mℓℓℓℓ s
The atom is
Option 1 hydrogen
Option 2 lithium
Option 3 beryllium
Option 4 boron
Correct Answer 2
Explanation n = 2 l = 0 m = 0 ms =
1+
2
2s
H 1s1
Li 1s22s
1
I3 1s
22s
22p
1
Q.95 Which electronic configuration does not follow the Pauli’s exclusion principle?
Option 1 1s2, 2s
22p
4
Option 2 1s2, 2s
22p
4, 3s
2
Option 3 1s2, 2p
4
Option 4 1s2, 2s
2 2p
6, 3s
2
Correct Answer 4
Explanation Pauli exclusion principle says an orbital can have max. of 2 electrons
∴∴∴∴ 3s3 is not possible
∴∴∴∴ 1s2, 2s
22p
4, 3s
2
Q.96 Magnetic quantum number for the last electron in sodium is :
Option 1 3
Option 2 1
Option 3 2
Option 4 zero
Correct Answer 4
Explanation Na →→→→ 1s22s
22p
63s
1
↓↓↓↓
n = 3
l = 0
m = 0
Q.97 Nitrogen has the electronic configuration 1 ,2 2 2 22 2 1 1 1
s s p p px y z and not
1 ,2 2 2 22 2 2 1 0
s s p p px y z . It was proposed by :
Option 1 Aufbau principle
Option 2 Pauli’s exclusion principle
Option 3 Hund’s rule
Option 4 Uncertainty principle
Correct Answer 3
Explanation Hund’s rule
Q.98 Which of the following has maximum number of unpaired electron (atomic number of
Fe 26)
Option 1 Fe
Option 2 Fe(II)
Option 3 Fe(III)
Option 4 Fe(IV)
Correct Answer 3
Explanation
Fe(III)
Q.99 A compound of vanadium has magnetic moment of 1.73 B.M. The electron
configuration of the vanadium ion in the compound is
Option 1 [Ar] 4s03d
1
Option 2 [Ar] 4s13d
0
Option 3 [Ar] 4s23d
0
Option 4 [Ar] 4s03d
3
Correct Answer 1
Explanation Magnetic moment (((( ))))= n n+2⇒⇒⇒⇒µµµµb
(((( ))))1.73 = n n+2
3 = n2 + 2n
n2 + 2n - 3 = 0
n2 + 3n - n - 3 = 0
(n + 3) (n - 1) = 0
n = 1
∴∴∴∴ one unpaired electron should be there
V = [Ar]4s23d
3
For one unpaired e⊙⊙⊙⊙ will have to remove 4 electrons
∴∴∴∴ V+4
= [Ar]3d1
Q.101 How many spherical nodes are present in 4s orbital in a hydrogen atom?
Option 1 0
Option 2 2
Option 3 3
Option 4 4
Correct Answer 3
Explanation spherical nodes = n - l – 1
For 4s = 4 - 0 - 1
= 3
Q.102 The number of nodes possible in radial wave function of 3d orbital is
Option 1 1
Option 2 2
Option 3 0
Option 4 3
Correct Answer 3
Explanation Radial nodes = n - l - 1
For 3d = 3 - 2 - 1
= 0
Q.103 The d-orbital with the orientation X and Y axes is called :
Option 1 2dz
Option 2 dzy
Option 3 dyz
Option 4 2 2
-dx y
Correct Answer 4
Explanation
2 2-
dx y
Q.104 Which d-orbital does not have four lobes?
Option 1 2 2
-dx y
Option 2 dxy
Option 3 2dz
Option 4 dxz
Correct Answer 3
Explanation 2dz
Q.105 3py orbital has ____ nodal plane
Option 1 XY
Option 2 YZ
Option 3 ZX
Option 4 Any
Correct Answer 3
Explanation
Q.106 The number of angular nodes in a 3s atomic orbital is
Option 1 0
Option 2 1
Option 3 2
Option 4 3
Correct Answer 1
Explanation Angular nodes = l
For 3s = 0
Q.107 The number of radial nodes in a 3s atomic orbital is
Option 1 0
Option 2 1
Option 3 2
Option 4 3
Correct Answer 3
Explanation radial nodes = n-l-l
For 3s = 3-0-1
= 2
Q.108 The quantum number not obtained from Schrodinger equation is
Option 1 n
Option 2 l
Option 3 m
Option 4 s
Correct Answer 4
Explanation s
Theory (n,l,m can be derived from schrodinger equation)