Atomic Structure

Atomic Structure

INTEXT EXERCISE: 1 hc
16. I.E. of one sodium atom = & I.E. of one mole Na atom
m
1. Hydrogen atom contains 1 proton, 1 electron and no neutrons. hc 6.62 # 10 # 3 # 108 # 6.02 # 10 23
34
NA =
2. Cathode Rays have constant b l ratio. m 242 # 10 -9
e
m = 494.65 kJ. mol.
(e/m) e e/m e 3672 17. Energy absorbed = 40 J/s
3.
(e/m) a = 2e/(4 # 1836me) = 1 Energy released (as EM waves) =0.8×40= 32 J/s
4. Speed of cathode rays is less than speed of light. nhc
m 32 × 20 =
5. e = 1.5 × 10 8
8
e = 1.6 × 10 19 m
m = 1.5 × 10 × 1.6 × 10 19 = 2.4 × 10 27 kg n = 2 × 1021
m = 2.4 × 10 24 g 18. 4 waves are passing through a given point in 2s.
6. Charge on oil drop = 6.39 × 10 19 C a Frequence = no. of waves / sec.
a × C is charge on one electron Frequency = 4/2 = 2 Hz
× C is charge on 19. According to Planck’s Quantum Theory the minimum energy
- 6.39 # 10 -19 hc
= = 4 electrons. which can be absorbed from an EM wave =
- 1.602 # 10 -19 m
hc
20. Eabsorbed =
7. Factual. 500 = Ereleased
- -
8. rN + 10 13 cm and ratom + 10 8 cm hc hc
Volume of nucleus
Now, Ereleased =
800 + m
Volume fraction = Total vol. of atom 1 1 1
= -
(4/3) r (10 -13) 3 m 500 800
= = 10 -15 1 3
(4/3) r (10 -8) 3 =
m 4000
9. R = RoA1/3 = 1.3 × 641/3 = 5.2 fm
= 1333.3 nm
10. From particle scattering experiment, distance of closest
approach of particle with nucleus came out to be of the INTEXT EXERCISE: 3
order of 10 14 m. This scattering followed Coulomb’s law and 21. Energy is absorbed when electron moves from lower to
gave the size of the nucleus to be < 10 14m. higher orbit.
INTEXT EXERCISE: 2 22. mvr = 2r
nh
c 3 # 10 8
13.6Z 2
11. m = y = = 0.75 m 23. Total energy is negative value -
400 # 106 n2
12. Violet colour has minimum wavelength so maximum energy. 24. It can explain the spectra of unielectronic species only.
b n l As Z increases, radius of 1st orbit decreases.
1 2
13. y = 25. r
m Z
1 n2 c
y= = 2 × 106 m 1 26. Radius = 0.529 z A = 10 # 10 -9 m
500×10 -9 m
14. E \
1 So, n2 = 189 or n . 14
m 1 2
= 2000 and = 4000 27. E1 (H) =- 13.6 # 2 =- 13.6 eV ;
1 2 1
E1 m2 22
E 2 = m1 = 2 E 2 (He +) =- 13.6 # 2 =- 13.6 eV
2
15. = 6 × 1014 Hz 2+ 32
Now, c = E3 (Li ) =- 13.6 2 =- 13.6 eV ;
3
1 y 42
y= = 3+
E 4 (Be ) =- 13.6 # 2 =- 13.6 eV
m c 4
6×1014 E1 (H) = E 2 (He +) = E3 (Li 2+) = E 4 (Be3+)
y= = 2×106 m -1
3×108 Z
28. v = 2.188 # 10 6 n m/s
Z Z1 /n1 3/3
Now, v \ n So, =
Z /n = 1/1 = 1
2 2
v3(Li2+) = v1 (H)

17
Chemistry
29. rn1 - rn2 = 24 # (r1) H 47. If > 0, emission takes place with a maximum K.E. of h
h o. The K.E. is independent of the intensity but the number
0.529 # n 2
1 0.529 # n 2
2
of photoelectrons increases with intensity.
1 - 1 = 24 # 0.529
48. Number of lines in Balmer series = 2
(n12 - n 22) = 24
n = 4 (lines will be 4 2, 3 2).
So, n1 = 5 and n2 = 1 KE of ejected photoelectrons
30. a) Energy of ground state of He+ 13.6
= Ephoton En = 13 - = 13 0.85 = 12.15 eV.
= –13.6×2 = –54.4eV
2
(iv) 42
b) Potential energy of 1st orbit of H atom 49. P.E.E. occurs only if in > o
= –27.2×1 2 = –27.2eV (ii) K.E. in
not velocity of photoelectron.
c) Kinetic energy of 2nd excited state of He+ 50. K.E.max = h in

= 13.6×
22
= 6.04 eV (i)
INTEXT EXERCISE: 6
32
51. According to de Broglie equation m = h or h
d) Ionisation potential of He+ = 13.6×2 2 = 54.4 V (iii) mv p
h
52. m = or h h De Broglie equation.
INTEXT EXERCISE: 4 p mv or mc
31. An electron jumps from L to K shell energy is released. 53. According to De Broglie equation
32. 9E = E3 - E 2 = 13.6 < 1 2 - 1 2 F = 1.9 eV
h
m = mv , p = mv , m = hp , m = mc h
(2) (3)
33. In H spectrum, all lines in Lyman series lie in the UV 54. m = h = 6.625 ×10 -34 = 6.625 × 10 29 m
mv -6
spectrum. 10 ×10
55. According to de–Broglie
34. In H spectrum, all lines in Balmer series lie in the visible
range of the spectrum. h = 6.62 # 10 -27 erg.sec
m = mv 2
Z2 (1) 2 # 5 # 10 4 cm/sec
35. E n = E1 E5 =- 13.6 # =- 0.54 eV 6.023 # 10 23
n2 (5) 2 6.62 ×10 -27 × 6.023 ×10 23 -8
36. When electron falls from n to 1, total possible number of = cm = 4 ×10 cm = 4Å .
2 × 5 ×10 4
lines = n – 1. h 6.625 # 10 -34 #
37. Visible lines Balmer series (5 2, 4 2, 3 2) 56. m = mv =
0.2 # 5 3600 . 10 -30 m.
So, 3 lines. m1 V 200 2
57. = V2 = 50 = 1 .
38. (He+)2 4 = (Li 2+) n 4 " n3 m2 1
n n n 2 4
Z1 n2 n1 2 2 4 58. m \ Z ` Z1 = Z2 or 3 = 6 (n = 4 of C5+ ion)
Z 2 = n 4 = n3 or
3 = n 4 = n3 1 2

n4 = 3 and n3 = 6. h
59. For a charged particle m =
Transition in Li2+ ion = 3 6 1 2mqV
m\ .
39. y1 = Rc Z 2 c m = Rc Z 2 ,
1 1 V
-
n12 3 2 h
y 2 = Rc Z 2 c 2 - 2 m = 4 Rc Z 2 .
1 1 3 60. m = mv
1 2 m p ma va
y3 = Rc Z 2 c 2 - 2 m = 4 Rc Z 2
1 1 1 = m v m = 4mp
ma p p
2 3
= 3. mp 4m v 1 va vp 8
1 2 = m av a 2 = 4 # vp va = 1
40. Visible lines Balmer series only in H atoms. ma p p

In H like species, e.g. He+, we cannot make generalizations INTEXT EXERCISE: 7

of certain lines lying in the visible region.
h
61. Orbital angular momentum = , (, + 1)
INTEXT EXERCISE: 5 2r = 0.
= 0 (s orbital).
41. Photo electric emission occurs when > o
62. Cu: 1s 2 2s 2 2p 6 3s 2 3p 6 3d10 4s1 .
42. Refer Q. 41 above
43. V I B G Y OR Cu 2+: 1s 2 2s 2 2p6 3s 2 3p6 3d9 or 6Ar@ 3d9 .
decreasing frequency
63. Magnetic moment = n (n + 2) = 24 B.M.
So only V, I, B, G can cause emission
No. of unpaired electron = 4.
44. It establishes the existence of quantum of energy, which is a
X26 : 1s22s22p63s23p63d84s2.
packet of energy.
45. h ( 1 ) = K.E.
0
3d6.
1
No. of electrons lost = 2 (from 4s2).
46. For photoelectric effect to take place,
n = 2.
hc hc
Elight > $ or m # m 0 .
m m0

18
 Atomic Structure
64. Zn2+ : [Ar]3d10 (0 unpaired electrons). 5. Urea CO (NH2)2
Fe2+ : [Ar]3d6 (4 unpaired electrons) maximum. Atomic no. of C=6
Ni3+ : [Ar]3d7 (3 unpaired electrons). O=8
Cu+ : [Ar]3d10 (0 unpaired electrons). 2N = 7 x 2 = 14
h 2H2 = 4
65. Orbital angular momentum = , (, + 1)
2r 6 + 8 + 14 + 4 = 32
Now for 2s orbitals, l = 0
6. A nucleus has a much smaller volume than that of an atom,
Orbital angular momentum = 0
66. 17Cl : [Ne]3s23p6
7. The fraction of particles undergoing a deviation of 180o
Last electron enters 3p orbital. is proportional to the ratio of volume of nucleus to that of the
, = 1 and m = 1, 0, - 1. atom.
67. d7 : 3 unpaired electrons 8. Photons denote the particle nature of light.
n 3 9. A packet of energy of EM waves is called quantum.
Total spin = !
2 =! 2 .
10. = 600 nm
68. X23 : 1s22s22p63s23p63d34s2. c
No. of electron with , = 2 are 3 (3d 3) . m
69. Cr (Zn = 24) 3 ×108
=
1s 2 2s 2 2p6 3s 2 3p6 4s1 3d5 600 ×10 -9
So, no. of electron in , = 1 i.e., p subshell is 12 and no. of = 5 × 1014 Hz
electron in , = 2 i.e., d subshell is 5. 11. In the equation E = h , E is the energy of a packet which
70. n = 3 can have = 0, 1, 2 discribes the particle nature and is the frequency of the EM
wave and describes the wave nature.
if = 0 orbital angular momentum = 0
12. Red has the largest wavelength and hence smallest frequency
2h among the visible light.
if =1 orbital angular momentum =
2r
Z2
6h 13. E = 13.6
if =2 orbital angular momentum = n2
2r 1
= 13.6 ×
INTEXT EXERCISE: 8 9 = 1.51 eV
14. L shell — n = 2
71. Spin quantum number is not the outcome of Schrodinger
E = 3.4 eV
wave equation.
P.E. = 2 × 3.4 eV = 6.8 eV
72. D = 2.dV = 2.4 r2dr
15. The charge on the nucleus of atomic no. Z
73. dxy orbital is oriented between x and y axis.
16. Bohr said that electrons neither gain nor lose energy when
74. For 6g orbitals, n = 6 and = 4.
they are at particular orbits and hence these orbits are called
Radial nodes = n 1 = 6 4 1 = 1 and Angular stationary states.
nodes = = 4
n2
Total 5 nodes. 17. rn = 0.529 × rn \ n2
Z
75. The radius of maximum probability of 1s orbital of H atom r1 : r2 : r3 = 1 : 4 : 9
has a value of 0.529Å. 1
18. K.E. = T.E. = -
76. Dumbell lies at 45o to x & y axis represents dxy orbital. 2 P.E.
77. XZ is the nodal plane of px orbital For n = 3 and Z = 1
78. Spherical node = n 1 T.E. = 1.51 eV
non spherical = . K.E. = + 1.51 eV and P.E. = 3.02 eV
79. i) Electron density in the XY plane in 3d x2 - y2 orbital is 19. rn = ron2 r3 = 9ro
not zero 20. E at n = 1 = 13.6 eV
ii) Electron density in the XY plane in 3d z2 orbital is not E at n = 2 = 3.4 eV
zero E at n 3 = 1.51 eV
iii) 2s orbital has one nodal surface it is a spherical node. 1st excitation potential = 10.2
iv) For 2pz orbital, XY is the nodal plane 2nd excitation potential = 12.09 eV
80. Factual. 21. I.E. from n = 3 is 1.51 eV
EXERCISE 1 E3 = 1.51 eV
E1 = 1.51 × 9 = 13.6 eV
1. Cathode rays are made up of electrons and have mass and 13.6
E2 =
charge both 4 = 3.4 eV
2. It is average isotropic weight 22. 3.4 eV (When n = 2). The possible value of energy are
13.6
3. Isotones same value of A Z i.e. same number of neutrons. - eV where ‘n’ is an integer.
n2
4. Nucleon are the particles in the nucleus. 23. Electrons reaching n = 4 from any state emits EM waves in
the Brackett series.

19
Chemistry
24. Max number of spectral lines 38. In option (c) if = 2 then ‘m’ cannot be 3.
(n 2 - n1) (n 2 - n1 + 1) ]4 - 1g]4g 39. a) s orbital is spherically symmetric
= 2 = 2 =6
25. As atom is excited so energy of electron > b) shape of orbital is given by ‘ ’ not ‘m’.
13.6 eV c) Since = 0, therefore orbital angular momentum = 0 for
1s, 2s, 3s etc.
hc
26. E = h = = hc y d) Electrons in different orbitals have different velocities.
m
hc
Also, Etot = n , but energy of 1 quantum is independent of 40. For 3d orbitals: n = 3, = 2, m = 2, 1, 0, 1, 2 and s = + 0.5
n.
m
41. If ‘ ’ of two orbitals are same, then their orbital angular
27. Bohr’s model doesn’t talk about probabilities. Hence momentum is also same.
statement (a) is incorrect.
42. b) if n = 3 then can be 0 or 1 or 2.
28. Bohr derived that
d) if = 2 then m can be 0 or + 1 or + 2.
n2 Z Z2
rn \ Z , v n \ n and freq \ 3 43. All 3 statements are correct.
n
44. All 3 statements are correct.
Kq q KZe 2 Z3
]n /Zg
Now, Force = &F\ 2 2F\ 4
1 2

2
r n 45. electron cloud density of px orbital

29. 2nd I.E. of He = 13.6 × 22 eV 46.

3rd I.E. of Li = 13.6 × 32 eV 47. (b) For n = 2 we can have 4 orbitals but they all do not have
1st I.E. of H = 13.6 eV different energy levels. Out of the 4, 3 are degenerate
o and have different energy from the 4th.
9 o
r3 of H = 0.529 × 9 A and r3 of Li2+ = 0.529 ×
3A (c) M n = 3 and can accommodate a maximum of 18
n, electrons.

Z2 Z2 48. Quantum numbers are derived by solving the Schroedinger’s

2 , P.E. \ - 2 , rn = ro $ n
2
30. K.E. K.E. \ wave equation. For a multi electron species, the energy of an
n n
electron can be estimated by the value of (n + ). Azimuthal
Energy levels or orbits are not equally spaced. The distance
Q.N. does not tell us anything about the motion of the
between the orbits keeps increasing with n.
electron.
31. Bohr’s model can be applied to all single electron species.
49. Shape of orbital is given by azimuthal Q.N. Also, electrons in
n3 an atom move with different velocities.
32. Frequency \ 2
Z 50. Due to the presence of electrons in them the orbital ex orient
33. For electronic transition from n 1, no. of spectral lines is
n ]n - 1g 51. Spin of electron does not affect the angular momentum and
given by 1 + 2 + 3 + ..... (n 1) =
2 can be represented as clockwise and anticlockwise.
34. In H atom, or any other H like species, the energy of the
52. Factual.
electron depends only on the value of ‘n’. Thus, we can say
that for such species (E4s = E4p = E4d = E4f) > E3s = E3p = E3d 53. E4s < E3d. If n = 2 the there are 4 orbitals which can
and so on. accommodate a maximum of 8 electrons. If n = 5 then
maximum no. of electrons in it is 2n2 = 50.
35. Factual.
54. If n = 4 and m = 2, it may be a 4d or 4f orbital. The spin of
36. Heisenberg’s uncertainty principle is valid for both 1 1
microscopic and macroscopic moving objects. However, it is the electron may be +
2 or - 2 .
55.
h electron p q x y z
37. 3 x $3 p $
4r
n 3 3 3 3 3
dp
Now,
dt = dF dp = dF.dt m 0 0 1 1 1
dE 0 0 1 +1 0
And dF $ dx = dE & dx =
dF s 1 +1 2 +1 2 +1 2 +1 2
2
dE
dp $ dx = dF $ dt $ dF = dE $ dt 56. K.E. of photo electron varies with and but is constant
with intensity.
Heisenberg’s principle may be written as
h 57. A1 = r12 An = rn2 and rn = r1n2
3 E $3 t $ 4r
b A n l = rr1 n2 & ln b A n l = 4ln (n)
2 4

A1 rr1 A1

20
 Atomic Structure

Graph of ln b
An l 54. p orbital is dumbell shaped.
A1 vs ln (n) is a straight line with slope 55. For m = 2, can be from 2 to 2.
= 4 and this line passes through the origin.
56. If n = 2 then can be 0 or 1.
58. Energy of electron depends on (n + 1) in case of multi 57. An orbital can have maximum 2 electrons.
] g
electron species while in case of single electron species, it h
depends on only the value of ‘n’. 58. Orbital anular momentum =
2r , , + 1
P.E.1 = 2 K.E.1 and P.E.2 = 2 K.E.2 | P.E.|= 2| K.E.| 59. No. of orbitals in a subshell = 2 + 1. Since each orbital can
It is practically not possible for an electron in an excited state have a maximum of 2 electrons therefore, maximum no. of
to absorb another photon. electrons = 2 (2 + 1).
60. n = 2 may have one s and 3 p orbitals with a maximum of 8
59. \ particles being much heavier than the electrons have a electrons.
high penetrating power, which is also because of the high 61. spin Q.N. is not obtained from .
speed of \ particles.
62. Theoretical
60. All 4 statements were conclusions drawn from the \
63. For principal Q.N. n, number of possible values of is also
scattering experiment. 36. E in L shell (n =
n. Hence, number of subshells = n.
2) = 3.4 ev. If it loses 10.2 eV then new energy = ( 3.4
10.2) eV = 13.6 eV. 64. No. of subshells = n = 3.
37. Photo electric emission occurs when > o No. of orbitals in 3s, 3p, 3d = 1 + 3 + 5 = 9.
38. Refer above question. 65. 19th electron goes to 4s orbital in Chromium.
39. It establishes the existence of quantum of energy, which is a 66. Fe3+ : [Ar] 3d5
packet of energy. Co2+ : [Ar] 3d7
40. n = n (n + 2)
V I B G Y OR
41. n Fe 3+
5×7
decreasing frequency nCo = 3×5
2+

So only V, I, B, G can cause emission

42. h ( 1 0
) = K.E. = 35 : 15
1
then K.E. - 67. For g subshell =4
1242 No. of e = 2 ( 2 × 4 + 1) = 18.
43. E =
m (nm) 68. Two spectral lines of an element cannot have the same wave
1242 number.
4=
m (nm) 69. s orbital is spherically symmetric. The probability
(nm) = 310.5 nm = 3105 Å distribution is uniform in all directions independent of the
angle.
44. KE1 = E1
70. No. of radial nodes = 4 2 1 = 1.
KE2 = 2E1
Thus (d) is correct.
K.E.2 = 2K.E.1
45. According to uncertainty principle, the product of uncertainties EXERCISE 2
of the position and momentum, is Dx × Dp $ h/4r .
h 1. rnu = 1.25 × 10 13
× 641/3 cm = 5 × 10 13
cm
46. Dx.Dp $ 4r , if D x = 0 then Dp = 3 ratom = 10 8 cm
47. According to Dx × Dp = h VNu 3
r nu 125×10 -39
4r = = = 1.25 × 10 13
Dx =
h
=
6.62 ×10 -34 Vatom r atom
3
10 -24
-5
Dp × 4r 1×10 × 4 × 3.14 2. In an element ZA X ,
= 5.27 × 10 -30 m . no. of protons = electrons = Z
48. r1 = x r4 = 42 x = 16 x. no. of neutrons = A Z
circumference = 2 (16x) = 4 3 28×60 + 29×30 + 30×20
Mavg = = 28.64.
=8 x 110
3
49. According to Heisenberg it is impossible to determine 3. Atomic electrically neutral as well as nucleus contains
both the position as well as the momentum of a particle neutrons but neutrality is due to equal number of protons and
simultaneously and accurately. electrons.
50. d orbital has = 2 4. particles are positively charged, since silicon has lowest
number of protons, so it will exert least repulsion to
51. For 2 electrons in same orbital, n, , m are same but s is
particles.
different.
/ % abundance × Atomic weight
52. n=2 = 1 for 2 p orbitals. 5. Avg. at wt. = 100
53. f subshell has 7 orbitals and can accommodate a maximum
of 14 electrons.
/ % abundance × Atomic weight
6. Avg. at wt. = 100

21
Chemistry
hc hc hc 4E E hc
7. Eabsorbed = Ereleased =
m
+
m1 m 2 3 - E = 3 = m1 ..... (2)
hc hc hc mm Divide (1) and (2)
= + &m= 1 2
m m1 m 2 m1 + m 2 m1
8. Rest mass is not property of photons as they never come to 3=
m
rest. ` m1 = 3m (d)
hc 0.529×16
9. E of quanta able to ionize Na = 21. r4 (Z) =
2414 ×10 -10 Z
hc 6.02×10 23 r1 (H) = 0.529
I. E. of Na = -10 × kJ/mol
2414×10 103 Z = 25 from the given options
. 497 kJ/mol. Note: for Z = 16 it overlaps and doesn’t fit inside the 1st Bohr orbit
10. Let no. of quanta absorbed and emitted be na and ne of H.
respectively. 22. Since EC + EB = EC
B A A
n a hc n e hc
Ea = 4500 and E e =
5000 hc hc hc
+ =
ne 1 na m1 m 2 m3
Now, Ee = 0.5 Ea &
5000 = 2 × 4500 m1 + m 2 1
=
ne 5000 5 m1 m 2 m3
na = 9000 = 9 = 0.55 mm
11. P.E. = 2 T.E. T.E. = 3.4 eV. m3 = 1 2
m1 + m 2
- 13.6
3.4 = 23. He(g) — H(g)+ + e 24.5 eV (given)
n2
n=2 He (g)
+
— H(g)2+ + e 54.4 eV (Bohr’s Model)
1st excited state. He(g) — H(g)+ + 2e 78.9 eV (Adding the 2)
12. N shell n = 4 24. Second series is Balmer series (n1 = 2) fourth line would be 6
- 13.6 × 4 2 2.
E4 (Bi ) =
3+
= 13.6 eV
42 25. 4th excited state is n = 5. So, 5 3 and 4 3 i.e., 2 lines in
Now, P.E. = 2 T.E. = 27.2 eV the Paschen series.
- 13.6
13. P.E. of n = 2 of H atom = 2 × = 6.8 eV nh 2h
4
6.8
26.
2r = r n=4
Now, for He+ if K.E. = n=4
2 , T.E. = 3.4
- 13.6×4 n2 = 4 n1 = 3 is the transition that takes place.
=- 3.4 & n = 4
= R ×1 : 9 - 16 D
n2 1 1 1
0.529×9 o 0.529×16 o m
14. r3 (He+) = A and r4 (He+) = A 1 7R
2 2 =
0.529 o o m 144
r4 r3 =
2 (16 9) A = 1.851 A 144
m = 7R
ro $ 4 RZ
15. r2 = R =
Z & ro - 4 27. I.E. = x E1 = x
x x
E2 = -
4 and E3 = - 9
RZ 9 9
r3 =
4 × Z = 4 R = 2.25R
x x 5x
16.
n2
0.529× 4 = 0.529
E2 3 =
4 - 9 = 36
`n=2 28. n2 n1 = 2 and n2 + n1 = 4
17. rn \ n 2 n2 = 3 and n1 = 1
Spacing is maximum when ^n 22 - n12 h is maximum. 1
= R4 ; 2 - 2 E & = 9
1 1 1 32R
18. More the energy of an electron, for ther a way from the m 1 3 m
nucleus it can stay.
= RZ 2 ; 2 - ]n + 1g2 E
1 1 1
19. E3 E1 = 12.09 eV for H atoms. 29.
m n
= RcZ 2 c 2 ]n + 1g2 m
Other permissible values of energy absorbed may be: c 2n + 1
or y =
Z = 2 : 12.09 × 4 = 48.36 eV m n
Z = 3 : 12.09 × 9 = 108.81 eV if n > > 1 then (n + 1) . n and (2n + 1) . 2n
and so on. 2RcZ 2
y=
20. From III to I, n3
= R ;1 - 2 E & 2 = 1 -
hc 1 1 1 1 1
2E E=E= .....(1) 30.
m m n n Rm
or, 2 = b l& n =
From II to I 1 Rm - 1 Rm
n Rm Rm - 1

22
 Atomic Structure
31. K.E.1 = h i = and K.E.2 = 2h i h2
K.E.2 = 2 K.E.1 + 4m 2A = m 2B = 2m × K ` E A = 4eV
B
Now, as intensity is doubled, the number of ejected electrons EB = 4.5eV
increases and hence photo current increases but it might not 1 K
` 4 = K B ` K A = 4K B VA = 2V, VB 0.5 V
necessarily double. A

32. As is doubled, Eincident is halved and hence stopping potential 0.101 c

42. For an particle ( ), m = A.
decreases provided Eincident > . V
43. r1 = x r4 = 42 x = 16 x.
33. The number of photons emitted per unit time increases
incident as higher intensity EM waves of suitable are circumference = 2 (16x) = 4 3

3
=8 x
hc
34. = + 3qVo
m 44. No. of electrons for given value of = 2 (2 + 1)
hc and can very from 0 to (n 1)
= + qVo
2m , = n-1
hc
Solving we get
4m
= o
=4 No. of electrons = / 2 (2, + 1)
,=0
hc 45. Mn = 1s 2 2s 2 2p 6 3s 3p 4s 2 3d5
2 6
35. E = mc2 and also E =
m
mc2 =
hc 3.873 = n (n + 2)
m 15 = n2 + 2n
h ` n 2 + 2n - 15 = 0 n = 3 or 5
m=
cm
`n=3
36. i
= 4 × 1015 Hz Mn +x = 1s 2 2s 2 2p6 3s 2 3p6 3d3
K.E. = 10.33 eV x = + 4.
= h i K.E.
46. By convention d z2 and pz are the orbitals which are assigned
4×1015 ×6.6×10 -34
= 10.33 + m = 0 for = 2 and = 1 respectively.
1.6×10 -19
= 6.2 eV
47. No. of values of m is the no. of orbtials present in a given
h m-1
37. p= x p subshell. Thus, m = 2 +1 =
4r 2
h 48. s orbitals are spherically symmetric hence the probability
p=m v=
2 r
1 h directions.
v=
2m r 1 1
h
49. The quantum numbers +
2 and - 2 for electron spin
38. 0.1 × 10 9 = p represent two quantum mechanical spin states which have no
6.6 ×10 -34 classical analogue.
`p=
0.1×10 -9
50. Hund’s Rule says that pairing of electrons in an atomic
p = 6.6 × 10 24
kg m/s
orbital will take place only after all the degenerate orbitals
39. Eabsorbed = 1.5 × 13.6 eV have been singly occupied.
K.E. of emitted electron = 6.8 eV 51. Greater the atomic number smaller is the size of the species.
h 6.625×10 -34 Thus, the graph with the highest probability at the least value
2m K.E.g
]
=
2×9.1×10 -31 ×6.8×1.6×10 -19 of ‘r’ represents Li2+ while the one with the peak at largest
o value of ‘r’ represents H.
= 4.70 A
52. For radial node, =0
-34
h 6.625 × 10 6 ! 36 - 24
2m ]KE-7 g
40. m = = 6 6 + 2
=0 =
2×9.1×10 -31 ×4.55×10 -25 2
= 7.28 × 10 m 2rZ
= 3! 3 3! 3
41. E A - WA = K A ` E A - 2 = 4K B 3a o
EA - 2 = KA E A + 0.5 - 4 = K B 3a o ^ h 3a o ^ h
r1 =
2Z 3 + 3 and r2 = 2Z 3 - 3
EB - 4 = KB E A - 3.5 = K B
3a o 3 3 ao
h2
m a2 = 2m × K ` E A - 2 = 4E A - 14 (r1 r2) =
2Z ×2 3 = Z
A

23
Chemistry
14. 4 2 in He+ has same energy difference as 2 1 of H.
1 :ZD 2
3
v
53. } = [(v - 1) (v 2 - 8v + 12)] e - 2 15. 13.6 Z2 = 217.6 Z2 = 16
16 4 a o
Z=4
(v - 1) (v 2 - 8v + 12) = 0
v = 1 or v 2 - v6 - 2v + 12 = 0 EXERCISE 4
v (v - 6) - 2 (v - 6) = 0
1. For He+ : x = R $ 4 ; - E & Rc 2 - 2 m = x
1 1 1 1
v = 2 or v = 6 n12 n 22 n1 n 2 4
=1
= R $ 16 ; 2 - 2 E & = 16× 4 = 4x
min 1 1 1 1 x
a For Be3+ :
& rmin = 2Z0 and max
=6 m n1 n 2 m
] g
3a h 6h
& rmax = Z0 2. Orbital angular momentum =
2r , , + 1 = 2r
54. Since becomes 0 for one value of r, therefore number of 3h
angular momentum in 3rd Bohr orbit =
radial nodes = 1.
2r
6 2
55.
ratio =
3 = 3
3. 12
6 C contains 6 protons and 6 neutrons and weighs 12 amu.
EXERCISE 3 Mass of electron is negligible.
1. No. of radial nodes = (n 1) = 1 Now, if mass of neutrons is halved, total mass = 9 amu.
2. Any d subshell can have maximum of 5 unpaired electrons. 3
reduction in mass = 3 amu
1 12 × 100 = 25%
so the maximum spin possible = 5× ! = S 4. Balmer series lines of only H atoms lies in the visible region
2
5 of the spectrum
Spin multiplicity = 2 | s | + 1 = 2 ×
2 + 1 = 6. 5. Lyman series is in UV region and Balmer series is in visible
3. Longest wavelength of any series is the lowest energy line of region in H atom.
that series and has the lowest energy. For Lyman it is 2 1.
The 4 lines in uv region are : 2 1, 3 1, 4 1, 5 1
4. P.E. = 6.8 eV = 2 × T.E.
- 13.6 In the visible region we will have: 3 2, 4 2, 5 2
T.E. = 3.4 eV = 3.4 n2 = 4.
n2 In the IR region: 5 4, 5 3, 4 3 3 lines
n=2 1st excited sate.
h
h 6. Rest mass of photon =
5. 3 x = 3 p & 3 p = cm
4r
= R :1 - 4 D = 4 R
h 1 1 3
` m3v = For 1st line of Lyman:
4r m1

= R : 4 - 9 D = 36 R
1 6.625×10 -34 1 1 1 5
` 3v = × For 1st line of Balmer:
9.1×10 -31 4×3.14 m2
= 7.98 × 10+12 y = 8
m1 m2 m1 3 36 27
6. No. of volues of m = 2 + 1 m 2 = m1 & m 2 = 4 × 5 = 5
7-1
=
2 =3 7. Series limit of any series & 3 $ n1
= R :1 - 3 D &
7. 3 2, 2 1, 3 1. 1 1 1 1
8. 5.27 = + 0.27 =5
For Lyman series: =R
mL mL
= R; 2 - 3 E &
9. Cu+ has 3d10 1 1 1 1 R
bs =- 1 l = 5
For Balmer series: = 4
mB 2 mB
2 =4 B
= 1.09677×10 +7 ×Z 2 :1 - 4 D
L
1 1 1 8. Spectral lines of an element are unique and no 2 lines of an
10.
3×10 -8 element can have the same wavelength.
4
or, Z 2 = 9. 2 is the probability density i.e. probability per unit volume.
3×3×1.09677×10 -8 ×10 +7
10. In H atom
Z2 = 4 Z =2
Elyman > Ebalmer > Epaschen ..........
11. Maximum m = + 3 =3 n =4
< balmer < paschen
No. of waves by the electron in the fourth orbit = 4 lyman
< < 3< 4
12. n = 2 r n 200 = 400 n =2 1 2
o o
RH RH 11. For n = 10, r = 0.529 × 102 A = 52.9 A
13. E =-
9 =- n 2 & n = 3 . 2 × × 52.9 = 10
No. of electrons that can be accomodated in n = 3 is 2 × 32 = o
18. = 33.22 A
No. of orbitals presents in n = 3 is 9. All these orbitals are 12. I : 1 radial node 3p
degenerate as of single electron species E3s = E3p = E3d and II : Zero radial nodes 3d
so on. III : 2 radial nodes 3s

24
 Atomic Structure
13. Longest wavelength of Paschen series 4 3 26. Heisenberg’s uncertainty principle is valid for both

R $ 1 2 $ : 9 - 16 D = R $ 2 2 $ ; 2 - 2 E
1 1 1 1 microscopic and macroscopic moving objects. However, it is
n1 n 2
n1 = 6 and n2 = 8. h
27. 3 x $3 p $
1 4r
14. 4.25 A
= TA and A
\
TA dp
Now,
1 dt = dF dp = dF.dt
4.2 B
= TB and B
\
TB dE
a =2 A And dF $ dx = dE & dx =
B dF
TA = 4 TB dE
dp $ dx = dF $ dt $ dF = dE $ dt
Now, TB = TA 1.5 3 TB = 1.5
TB = 0.5 eV and TA = 2 eV Heisenberg’s principle may be written as
A
= 4.25 TA = 2.25 eV h
3 E $3 t $ 4r
and B = 4.2 TB = 3.7 eV
15. 2 is maximum at nucleus and 3 r2 2 is zero. 28. In option (c) if = 2 then ‘m’ cannot be 3.
hc 29. a) s orbital is spherically symmetric
16. E = h = = hc y
m b) shape of orbital is given by ‘ ’ not ‘m’.
hc c) Since = 0, therefore orbital angular momentum = 0 for
Also, Etot = n , but energy of 1 quantum is independent of
n.
m 1s, 2s, 3s etc.
d) Electrons in different orbitals have different velocities.
17. Bohr’s model doesn’t talk about probabilities. Hence
statement (a) is incorrect. 30. For 3d orbitals: n = 3, = 2, m = 2, 1, 0, 1, 2 and s = + 0.5
18. Bohr derived that 31. If ‘ ’ of two orbitals are same, then their orbital angular
2 2
momentum is also same.
n Z Z
rn \ Z , v n \ n and freq \ 3 32. b) if n = 3 then can be 0 or 1 or 2.
n
d) if = 2 then m can be 0 or + 1 or + 2.
Kq q KZe 2 Z3
]n /Zg
Now, Force =
1

2 &F\ 2 2F\ 4
2
33. All 3 statements are correct.
r n
34. All 3 statements are correct.
19. 2 I.E. of He = 13.6 × 2 eV
nd 2
35. electron cloud density of px orbital
3rd I.E. of Li = 13.6 × 32 eV
36.
1st I.E. of H = 13.6 eV
o
37. (b) For n = 2 we can have 4 orbitals but they all do not have
9 o different energy levels. Out of the 4, 3 are degenerate
r3 of H = 0.529 × 9 A and r3 of Li = 0.529 × A 2+
3 and have different energy from the 4th.
n,
(c) M n = 3 and can accommodate a maximum of 18
Z2 Z2 electrons.
2 , P.E. \ - 2 , rn = ro $ n
2
20. K.E. K.E. \
n n 38. Quantum numbers are derived by solving the Schroedinger’s
Energy levels or orbits are not equally spaced. The distance wave equation. For a multi electron species, the energy of an
between the orbits keeps increasing with n. electron can be estimated by the value of (n + ). Azimuthal
Q.N. does not tell us anything about the motion of the
21. Bohr’s model can be applied to all single electron species.
electron.
n3 39. Shape of orbital is given by azimuthal Q.N. Also, electrons in
22. Frequency \
Z2 an atom move with different velocities.
23. For electronic transition from n 1, no. of spectral lines is 40. Due to the presence of electrons in them the orbital ex orient
n ]n - 1g
given by 1 + 2 + 3 + ..... (n 1) = 41. Spin of electron does not affect the angular momentum and
2
24. In H atom, or any other H like species, the energy of the can be represented as clockwise and anticlockwise.
electron depends only on the value of ‘n’. Thus, we can say 42. Factual.
that for such species (E4s = E4p = E4d = E4f) > E3s = E3p = E3d
43. E4s < E3d. If n = 2 the there are 4 orbitals which can
and so on.
accommodate a maximum of 8 electrons. If n = 5 then
25. Factual. maximum no. of electrons in it is 2n2 = 50.
44. If n = 4 and m = 2, it may be a 4d or 4f orbital. The spin of
1 1
the electron may be +
2 or - 2 .

25
Chemistry
45. 56. 4d orbital has 1 radial node, as (n 1) = 1
electron p q x y z No. of peaks = no. of radial nodes + 1 = 2

b a1 l <27 - 18 r + 2r2 F 3-ar

3/2
n 3 3 3 3 3 1
57. }3s = 0 = e 0

81 3r 0 a 0 a20
m 0 0 1 1 1
0 0 1 +1 0 18r 2r2
27 - a 0 + 2 = 0
a0
- ]18a 0g ! ]- 18a 0g2 - 4 × 2 × 27a20
s 1
2 +1 2 +1 2 +1 2 +1 2
r =- 2×2
46. K.E. of photo electron varies with and but is constant
with intensity. 18a 0 ! 10.39a 0
= 4
47. A1 = r12 An = rn2 and rn = r1n2 Distance between to radial node (d)
d 2 × 10.39a 0
b A n l = rr1 n2 & ln b A n l = 4ln (n)
2 4
1.73a 0 = 4 × 1.73a 0
A1 rr1 A1 d=3

Graph of ln b
An l Radial probability density at r = a 0 R2 (a0)
A1 vs ln (n) is a straight line with slope 58.
Radial probability density at r = 0 = R2 (0)
= 4 and this line passes through the origin. r
1 -a
For 1s orbital: R(r) = 3/2 e
r b a1 l
0

48. Energy of electron depends on (n + 1) in case of multi

electron species while in case of single electron species, it 0
depends on only the value of ‘n’.
R ]a 0g ^1/ra 0h e
2 3 -2r/a 0

R2 ]0g ^1/ra30h e 0
P.E.1 = 2 K.E.1 and P.E.2 = 2 K.E.2 | P.E.|= 2| K.E.| = = e -2
It is practically not possible for an electron in an excited state 59. r3 = 0.529 × 9 Å
to absorb another photon.
2 r3 = 3 2 0.529 × 9 = 3
49. \ particles being much heavier than the electrons have a = 9.96 × 10 10 m c = 10 9 m
high penetrating power, which is also because of the high 2.178×10 -11
speed of \ particles. 60. T.E.3 = - ergs
9
50. All 4 statements were conclusions drawn from the \ a P.E. = 2 T.E.
scattering experiment. 2 -
h k h k P.E. = - ×2.18×10 11 ergs
51. m1 = = , m2 = =9 9
2mq (100) 10 2mq (81) = 4.84 × 10 12
erts
h k h
2m ]K.E.g
m3 = =7 61. m =
2mq (49)
m3 - m 2 ]1/7g - ]9/4g 10 × 2 h
]1/10g
= = 63 = 0.317 62. m =
m1 2m qV
52. 3 E =
1240 6.625×10 -34
1282.5 eV = 0.96 eV 5 × 10 =
12

2×1.67×10 -27 ×1.6×10 -19 ×V

This is the difference in levels 5 and 3 of atomic hydrogen. Solving we get V = 32.8 V
Now series limit means 3 63. m = 75 × 10 3 kg
= (912) -1 c 2 m & = (912) -1 b 9 l & m
1 1 1 1 h 6.625×10 -34
m n1 m x= = mv x=
75×10 -3 ×4
= 8208 Å
h 6.625×10 -34
v=
53. 3 E =
12400
6200 = 2 eV. 4rm 3 x = 4×3.14×75×10 -3 ×2.2×10 -33
= 0.32 m/s
2 × 1.6 × 10 19 × n = 40 × 60 × 0.5
64. Shortest wavelength is emitted for highest energy.
n = 3.75 × 1021
D has shortest wave length emitted
54. n = 2, = 0, 1, 2, 3
65. Ionization potential is the amount of energy absorbed by a
for last subshell, n = 2 and = 3.
gaseous atom to lose the aoutermost electron.
n+ =5
EB = 13.6 b
1 1 l
55. m1 = d n & V1 = 150
150 1/2 66. B = 4 3
9 - 16 eV
V1 m12 7
= 13.6 ×
150 150 9×16 eV = 1.983 eV
]1.41g2 ]1.73g2
& V1 = = 75 V and V2 = = 50 V EB = 1.983 × 1.6 × 10 19 J
Hence, potential should be dropped by 25 V. = 3.18 × 10 19
J energy is released.

26
 Atomic Structure

5. m = p ;m \ p E
22 h 1
67. a) U1, 2 = 2 × 13.6 × 2 = 13.6 × 8 eV
1
K1, 1 = + 13.6 eV p = h (constant)
Ratio = 8 : 1 So, the plot is a rectangular hyperbola.
22 1 2 ro
b) r2, 1 = ro
1 = 4 ro and r1, 2 = ro
2 = 2
Ratio = 8 : 1
2 4
c) V1, 2 = Vo
1 = 2 Vo and V2, 4 = Vo 2 = 2 Vo
Ratio = 1 : 1
= R; 2 - 3 E &
n3 T1,2 1 1 1 1 1
d) T \ 2 & 6. H atom ; Shortest Lyman : =R
z T2,2 = 8 m1 1 m1
= R ]2g2 ; 2 - 2 E
Z2 1 1 1
68. T.E. =- 13.6 2 eV as n increases, T.E. increases. He+ ion : longest Balmer :
n m2 2 3
Z2 1 1 # # 5 9m
Frequency of revolution \ 3 as n increases, frequency & = 4 36 & m2 = 5
decreases. n m2 m1
Z3 7. Based on “n + l” rule only (II) has pair of electron in
Acceleration \ as n increases, acceleration decreases degenerate orbitals.
n4
radius \
n2
as n increases, radius increases hc
Z 8. Energy of one mole of photons = × NA
m
69. a) 3 radial nodes 4s / 5p but for p orbitals, < 0 6.63 # 10 -34 ×3×108
= × 6.02 × 1023
at r = 0 300×10 -9
b) 3 radial node 4s and 5 p = 399.13 × 103 Joule/mole 399 kJ mol 1

c) For p and d orbitals angular probability distribution 9. Number of radial nodes = n – l – 1

depends on and =3–0–1=2
d) S orbital is spherically symmetric, hence no planar Therefor corresponding graph is (IV)
node. Hence answer is (c)
70. a) Orbital angular momentum 10. 2 r = n
= 2r , ], + 1g = 2 2r & , = 1 & p orbital.
h h n2
2 × a =n
b) Bohr’s model classical mechanical Z 0
c) 5 degenerate orbitals d orbitals 42
2 × a =4
d) N shell n = 4 no. of waves made by the electron = 1 0
4 = 8 a0

11. v = R H c - m cm -1 = R H - R2H = R2H + R H

Previous Year (JEE Main) 1 1
82 n2 64 n n 64
1. n = 4 can have Comparing it with general straight line equation,
4s 4p 4d 4f y = mx + c, we get
RH
Total e 2 6 10 14 Slope (m) = – RH, Intercept (c) =
64
Total e with = + 1 3 5 7 12. (i) Higher the value of ‘l’, lesser is the penetration of the
orbital. Thus, given statement is correct.
So, Ans. 16
(ii) Size of orbit depends on value of ‘n’. Thus, incorrect.
2. E = nhv
nh
(iii) Angular momentum =
= (6.022 × 1023) (6.626 × 10 34) × (2 × 1012) 2r and for ground state n=1.
= 798.03 J c 798 J Thus, given statement is correct.
5 (iv) As value of azimuthal Q.N. increases, vs r graph has
3. Tv = 90×
100 = 4.5 m/s more number of peaks and hence shift towards lower
h value of r. Thus, incorrect statement.
Tv $ Tx = 4rm
13. The graph given in option (c) is incorrect as the emission will
h 6.63×10 -34 begin only once > 0. The graph for (c) should look similar
Tx = 4rm $ Tv = 4×3.14×0.01×4.5 = 1.17 × 10 33 to that of (a) because E .
4. E = W + K.Emax 14. Radius of nth Bohr orbit in H atom = 0.53 n2 Ao
hc Radius of 2nd Bohr orbit = 0.53 × (2)2 = 2.12 Ao
K.Emax = E W= 4.41 × 10 19
m 13.6
15. E n =- eV Where, n = 1, 2, 3...
6.63 ×10 -34 × 3 ×108 n2
= 4.41 × 10 19 = 222 × 10 21 joule
300 ×10 -9 - 13.6
In excited states, E2 =
4 =- 3.4 eV

27
Chemistry
6. At d = d0, nucleus nucleus & electron electron repulsion is
Previous Year (JEE Advanced) absent.
Hence potential energy will be calculated for 2 H atoms. (P.E.
Z2
1. KE =+ 13.6 × due to attraction of proton & electron)
- Kq1 q 2 ]9 ×109g^1.6 ×10 -19h2
n2
12 P.E. = =
(a) KE1,H =+ 13.6 × 2 = 13.6 eV r 0.529 ×10 -10
1 ]Bohr radiusg
22 = 4.355 × 10 21 kJ
(b) KE1,He =+ 13.6 × 2 = 13.6 × 4 eV
+

1 For 1 mol = 4.355 × 10 21 × 6.023 × 1023

22 = 2623.249 kJ/mol
(c) KE 2,He =+ 13.6 × 2 = 13.6 eV
+

2 For 2 H atoms = 5246.49 kJ/mol

2
3 9
(d) KE 2,Li =+ 13.6 × 2 = 13.6 × 4 eV
2+

2 r
r
2. (a) Uncertainity principle talks about probability of
e e
! !

do
(b) With increase in distance of electron from the
nucleus, its energy increases.
7. n = 4
Z2 |m|=1 may be 1 | 2 | 3 and m = +1 or 1
(c) Energy of electron E n = - 13.6 × eV/atom .
n2
2 electrons each of 4p, 4d and 4f can have
Z
(d) Velocity of electron Vn = 2.19 ×106 × n m/sec . 1
n = 4, | m | = 1 and s = +
2
Total 6 electrons
- 13.6 # 22 1242
3. - 3.4 = 8. z =
n2 300 = 4.14 eV
n=4 2 Li, Na, K, Mg show P.E.E.
Subshell = 4d Angular nodes = 2 9. n = 3 can accommodate maximum of 18 electrons of which 9
Radial nodes = n 1 4 2 1=1 1 1
n2 2 and 9 will have s = - 2
will have s = +
4. For H like species: rn = 52.9 × Z pm
13.6Z2
10. & 11. KE = eV / atom
For He+: Z = 2 n2
105.8 = 52.9 × Z2
n2
n2 = 2 2 # 13.6Z2
PE = eV / atom
n2
2
n n2 %
& 26.45 = 52.9 × Z n1 = 1 Radius 0.529
ZA
= RH Z2 c 2 - 2 m
hc 1 1 nh
Now Angular momentum of electron (mvr) =
m n1 n 2 2r
–9
m = 30 nanometer. 12. 2s orbital has are radial node and (r) = 0 once
-27 13. 1s has zero radial nodes
h 6.6×10
5. Momentum of photon = = gm cm s 1 14. E2 4 = 2.55 eV E2 6 = 3.022
m 330×10 -7
Momentum of 1 mole of He atoms = m v
h
m v = NA ×
m
6×10 23 ×6.6×10 -27
4×Tv =
330×10 -7
6×6.6×10 2
Tv = 33×4 = 30 cm s 1
Change in velocity of He atoms = 30 cm s 1

28