Atomic Structure 2.

Chapter 2 Atomic Structure

INTEXT EXERCISE: 1
1. Hydrogen atom contains 1 proton, 1 electron and no m
8.
e = 1.5 × 10 e = 1.6 × 10-19
-8
neutrons.

\ m = 1.5 × 10-8 × 1.6 × 10-19 = 2.4 × 10-27 kg
2. Cathode Rays have constant b m
e l ratio.

\ m = 2.4 × 10-24 g
(e/m) e e/me
3. (e/m) = 2e/(4 # 1836m
3672
= 1 9. Charge on oil drop = - 6.39 × 10-19 C
a e)

a -1.602 × 10-19 C is charge on one electron
4. Factual.

\ - 6.39 × 10-19 C is charge on
5. rN + 10 -13 cm and ratom + 10 -8 cm
- 6.39 # 10 -19
Volume of nucleus
= = 4 electrons.

Volume fraction = Total vol. of atom - 1.602 # 10 -19
10. From a particle scattering experiment, distance of
(4/3) r (10 -13) 3

= = 10 -15 closest approach of a particle with nucleus came out
(4/3) r (10 -8) 3 to be of the order of 10-14 m. This scattering followed
6. R = RoA1/3 = 1.3 × 641/3 = 5.2 fm Coulomb’s law and gave the size of the nucleus to
7. Speed of cathode rays is less than speed of light. be < 10-14m.

INTEXT EXERCISE: 2
c 3 # 10 8 7. According to Planck’s Quantum Theory the
1. m = y = = 0.75 m minimum energy which can be absorbed from an
400 # 106
EM wave
2. Violet colour has minimum wavelength so maximum
energy. hc
=
m
hc
3. I.E. of one sodium atom = 8. E \
1
m m
& I.E. of one mole Na atom \ l1 = 2000 and l2 = 4000
hc 6.62 # 1034 # 3 # 108 # 6.02 # 10 23
N A = m
m 242 # 10 -9
E
\ E1 = 2 = 2
m1
= 494.65 kJ. mol.
2

9. u = 6 × 1014 Hz
4. Energy absorbed = 40 J/s
Now, c = ul

\ Energy released (as EM waves) =0.8×40= 32 J/s 1 y
\ y= = c

\ 32 × 20 =
nhc m
m
6×1014

⇒ n = 2 × 1021
\ y= = 2×106 m -1
3×108
1
5. y =
m hc
10. Eabsorbed = 500 = Ereleased
1

\ y= = 2 × 106 m-1 hc hc
500×10 -9 m Now, Ereleased = 800 +
m
6. 4 waves are passing through a given point in 2s. 1 1 1
= 500 - 800
a Frequence = no. of waves / sec. m
\ Frequency = 4/2 = 2 Hz ⇒
1
=
3
m 4000
\ l = 1333.3 nm
2.2 Chemistry
INTEXT EXERCISE: 3
6. S1 : Potential energy of the two opposite charge
1. r a b nZ l As Z increases, radius of 1st orbit decreases.
2

system decreases with decrease in distance,

n2 c S4 : The energy of 1st excited state of He+ ion
2. Radius = 0.529 z A = 10 # 10 -9 m
= –3.4 Z 2 = –3.4×2 2 = –13.6 eV
So, n2 = 189 or n . 14 S2 and S3 are correct statement.
12 n3 13 1
3. E1 (H) =- 13.6 # =- 13.6 eV ; T
7. T1 = 13 = 3 = 8 . (for same Z)
12 2 n2 2
a bT = v l
22 2rr so, T \ 2
n3

E 2 (He +) =- 13.6 # =- 13.6 eV
22 Z
32 r n2 R

E3 (Li 2+) =- 13.6 =- 13.6 eV ; 8. r \ n2 and r \ n3 for same Z r12 = 12 = 4R
32 n2
n 1
42 ⇒ n12 = 2

E 4 (Be3+) =- 13.6 # =- 13.6 eV
42 T n3 1
\ T1 = 13 = 8 .

\ E1 (H) = E 2 (He +) = E3 (Li 2+) = E 4 (Be3+) 2 n2
9. Angular momentum J = mvr
Z
4. v = 2.188 # 106 n m/s J2 = m2 v2 r2
or 2 = b 2 mv 2 l mr 2 or K.E. =
Z Z /n 3/3 J2 1 J2
Now, v \ n So, = Z 1 /n1 = 1/1 = 1
2 2 2mr 2
v3(Li2+) = v1 (H) 10. a) Energy of ground state of He +

= –13.6×2 2 = –54.4eV (iv)

5. rn - rn = 24 # (r1) H
1 2 b) Potential energy of 1 orbit of H-atom
st

= –27.2×1 2 = –27.2eV (ii)

0.529 # n12 0.529 # n 22
1 - 1 = 24 # 0.529 c) Kinetic energy of 2 excited state of He+
nd

22

\ (n12 - n 22) = 24 = 13.6× 2 = 6.04 eV (i)
3
So, n1 = 5 and n2 = 1 d) Ionisation potential of He+
= 13.6×2 2 = 54.4 V (iii)

INTEXT EXERCISE: 4
Z2 (1) 2 For last line of Pachen series
1. E n = E1 E5 =- 13.6 # =- 0.54 eV
y 2 = R H (3) 2 < 2 - F= R
1 1
n2 (5) 2
(3) (3) 2
hc 1
2. m = 9E ` ma 9E R
Thus, difference = 4
3. Infrared lines = total lines – visible lines – UV lines
6 (6 - 1) 8. (He+)2 → 4 = (Li 2+) n 4 " n3
Total lines = 2 = 15 Z n n 2 2 4
\ Z1 = n 24 = n13 or 3 = n 4 = n3
visible lines = 4 (6 → 2, 5 → 2, 4 → 2, 3 → 2) 2

UV lines = 5 (6 → 1, 5 → 1, 4 → 1, 3 → 1, 2 → 1) \ n4 = 3 and n3 = 6.
\ Transition in Li2+ ion = 3 → 6
4. When electron falls from n to 1, total possible
9. y1 = Rc Z 2 c 2 - 2 m = Rc Z 2 ,
1 1
number of lines = n – 1.
n1 3
5. Visible lines ⇒ Balmer series (5 → 2, 4 → 2, 3 → 2)
y 2 = Rc Z 2 c 2 - 2 m = 4 Rc Z 2 .
1 1 3
So, 3 lines.
1 2
y3 = Rc Z 2 c 2 - 2 m = 4 Rc Z 2
6. According to energy, E 4 " 1 > E3 " 1 > E 2 " 1 > E3 " 2 . 1 1 1

2 3

According to energy, Violet > Blue > Green > Red. \ u 1 - u 2 = u 3.

\ Red line ⇒ 3 → 2 transition.
10. Visible lines ⇒ Balmer series ⇒ only in H atoms.
7. For 1st line of Balmer series In H like species, e.g. He+, we cannot make
generalizations of certain lines lying in the visible
y1 = R H (3) 2 < 2 - F = 9R b 5 l = 5 R
1 1

(2) (3) 2 36 4 region.
 Atomic Structure 2.3
INTEXT EXERCISE: 5
1. For photoelectric effect to take place, 6. P.E.E. occurs only if uin > uo
Elight > f K.E. ∝ uin not velocity of photoelectron.
\
hc
$
hc
or m # m 0 . 7. K.E.max = huin - f
m m0
2. Photoelectric effect is a random phenomena. So, 8. uincident = u1
electron may come out with a kinetic energy less K. E. =
hc
-z
than or equal to (hu – w) as some energy is lost m1
hc z
while escaping out. \ stopping potential = -
m1 e e
3. If u > u0, emission takes place with a maximum K.E. when uincident = u2
of hu-huo. The K.E. is independent of the intensity hc z
but the number of photoelectrons increases with stopping potential = -
m2 e e
intensity.
\ stopping potential increase by e c
hc 1 - 1 m

hc m 2 m1
4. = 1 + z ...(1)
m
hc 9. | Max K.E. | = | stopping potential |
3# = 4 + z ...(2)
m
\ K.E.max = 1.5 eV also f = 2.5 eV
from, e.q., (1) and (2) z = 0.5 eV
\ Eincident = 1.5 + 2.5 = 4 eV
5. Number of lines in Balmer series = 2 10. K.E.mzx = Ein - f = 6 - 2.1 eV

\ n = 4 (lines will be 4 → 2, 3 → 2). = 3.9 eV
KE of ejected photoelectrons
\ stopping potential = - 3.9 eV
13.6
= Ephoton - En = 13 - 2 = 13 - 0.85 = 12.15 eV.
4
INTEXT EXERCISE: 6
-34
h 6.625 # 10 # h
7. 9p $ 9x = 4r
1. m = mv = 0.2 # 5 3600 . 10 -30 m.
6.62 # 10 -34
2.
m1
= V2 =
V 200 2 ⇒ 9x = = 5.27 # 10 -30 m.
m2 1 50 = 1 . 4 3.14 # 1 # 10 -5
#

3. r1 = 0.529 A c my mx vx my m vx 16
8. = & 1 = (0.25m )x (0.75v = 3 .
r3 = 0.529 # (3) 2 Ac = 9x mx my vy x x)

2rr 2r (9x)
So, m = n = 3 = 6rx. h
9. m = mv
o
12.3
4. m = A gives the De-Broglie wavelength of an mp m v
V = map vap m = 4mp
ma
electron accelerated by V volts.
mp 4m v 1 va vp 8
n n n 2 4 = m pav pa 2 = 4 # vp va = 1
5. m \ Z ` Z1 = Z2 or 3 = 6 (n = 4 of C5+ ion) ma
1 2

h 10. l = v
6. For a charged particle m = h h
2mqV
then m = mv or m2 = m
1

\ m\ . h
V
So, m= m
INTEXT EXERCISE: 7
h
\ No. of unpaired electron = 4.
1. Orbital angular momentum = , (, + 1) 2r = 0.
X26 : 1s22s22p63s23p63d84s2.

\ l = 0 (s orbital).
To get 4 unpaired electrons, outermost configuration
2. Cu:1s 2 2s 2 2p6 3s 2 3p6 3d10 4s1 . will be 3d6.
\ Cu 2+:1s 2 2s 2 2p6 3s 2 3p6 3d9 or 6Ar@ 3d9 .

\ No. of electrons lost = 2 (from 4s2).
3. Magnetic moment = n (n + 2) = 24 B.M.
\ n = 2.
2.4 Chemistry
4. Zn2+ : [Ar]3d10 (0 unpaired electrons). h
8. Orbital angular momentum = , (, + 1) 2r
Fe 2+
: [Ar]3d (4 unpaired electrons) maximum.
6
Now for 2s orbitals, l = 0
Ni 3+
: [Ar]3d7 (3 unpaired electrons). \ Orbital angular momentum = 0
Cu+ : [Ar]3d10 (0 unpaired electrons). 9. 17
Cl- : [Ne]3s23p6
5. d7 : 3 unpaired electrons Last electron enters 3p orbital.

n
\ Total spin = ! 2 = ! 2 .
3
\ , = 1 and m = 1, 0, - 1.
10. n = 3 can have l= 0, 1, 2
6. X23 : 1s22s22p63s23p63d34s2.
if l = 0 ⇒ orbital angular momentum = 0
No. of electron with , = 2 are 3 (3d3) .
2h
if l= 1 ⇒ orbital angular momentum = 2r
7. Cr (Zn = 24)
electronic configuration is :1s 2 2s 2 2p6 3s 2 3p6 4s1 3d5 6h
if l= 2 ⇒ orbital angular momentum = 2r
So, no. of electron in , = 1 i.e., p subshell is 12 and
no. of electron in , = 2 i.e., d subshell is 5.
INTEXT EXERCISE: 8
1. For 6g orbitals, n = 6 and l = 4. 6. Spherical node = n -l- 1

\ Radial nodes = n -l- 1 = 6 - 4 - 1 = 1 and non spherical =l.
Angular nodes =l= 4 7. The dual nature of electrons is the premise on which

\ Total 5 nodes. this model is based.
2. The radius of maximum probability of 1s orbital of 8. i) Electron density in the XY plane in 3d x - y 2 2

H atom has a value of 0.529Å. orbital is not zero

ii) Electron density in the XY plane in 3d z orbital
3. A has 0 radial nodes and B has 1 radial node
2

is not zero

\ A is 1s and B is 2s iii) 2s orbital has one nodal surface-it is a spherical
4. Dumbell lies at 45o to x & y axis represents dxy node.
orbital. iv) For 2pz orbital, XY is the nodal plane
5. XZ is the nodal plane of px orbital 9. Factual.
10. n, l and m, are obtained from y.

EXERCISE - 1

1. Cathode rays are made up of electrons and have 8. Urea - CO (NH2)2

mass and charge both
Atomic no. of C=6
2. It is average isotropic weight
O=8
3. Isotones ⇒ same value of A - Z i.e. same number of
2N = 7 x 2 = 14
neutrons.
2H2 = 4

\ 6 + 8 + 14 + 4 = 32
4. A nucleus has a much smaller volume than that of an
atom, hence only small fraction get deflected. 9. In the equation E = hu, E is the energy of a packet
which discribes the particle nature and u is the
5. The fraction of a particles undergoing a deviation
frequency of the EM wave and describes the wave
of ~ 180o is proportional to the ratio of volume of
nature.
nucleus to that of the atom.
10. Photons denote the particle nature of light.
6. They are deflected more as the experience stronger
electrostatic forces of repulsion. 11. Red has the largest wavelength and hence smallest
frequency among the visible light.
7. Nucleon are the particles in the nucleus.
12. A packet of energy of EM waves is called quantum.
 Atomic Structure 2.5
13. l = 600 nm 29. Lyman series emits highest energy lines and 2nd line
c of any series has more energy than 1st line. So 3 → 1
u=
m has highest energy and lowest l.
3 ×108
\ =u 30. Max number of spectral lines
600 ×10 -9
\ u = 5 × 1014 Hz
(n 2 - n1) (n 2 - n1 + 1) ]4 - 1g]4g
= 2 = 2 =6
n2 31. As atom is excited so energy of electron >
14. rn = 0.529 × Z ⇒ rn \ n2
\ r1 : r2 : r3 = 1 : 4 : 9 - 13.6 eV

1 32. When electron is in excited state in H atom,

15. K.E. = - T.E. = - 2 P.E.
E > - 3.4 eV. Thus, I.E. < 3.4 eV
For n = 3 and Z = 1
33. A quantum having energy equal to the difference
T.E. = - 1.51 eV of energy between the two energy levels is emitted

\ K.E. = + 1.51 eV and P.E. = - 3.02 eV during de-excitation.

16. E = - 13.6
Z2 34. E in L shell (n = 2) = - 3.4 ev. If it loses 10.2 eV then
n2 new energy = (- 3.4 - 10.2) eV
1
= - 13.6 × 9 = - 1.51 eV
= 13.6 eV.
17. rn = ron2 ⇒ r3 = 9ro 35.
E1 = - 54.4 eV
18. E at n = 1 = - 13.6 eV - 54.4

\ E2 = 4 =- 13.6 eV

E at n = 2 = - 3.4 eV
\ I. E. = 13.6 eV.
E at n 3 = - 1.51 eV 36. Theoretical.
1st excitation potential = 10.2 37. Free electron at rest has 0 energy. When the electron
2nd excitation potential = 12.09 eV is bound to an atom, its energy is therefore negative.
19. I.E. from n = 3 is 1.51 eV 38. Shortest line of Paschen ⇒ n = 3 $ n = 3

\ E3 = - 1.51 eV = R3 2 : 9 - 3 D
1 1 1
m
\ E1 = - 1.51 × 9 = - 13.6 eV
1
13.6 \ m= R
\ E2 = - 4 = - 3.4 eV
39. Z = 2
20. L shell —→ n = 2
n = 2 → n = 1 (Longest l for lowest DE)
E = -3.4 eV
= R × 4 :1 - 4 D = R × 4 : 4 D
1 1 3
P.E. = 2 × -3.4 eV = -6.8 eV m
- P.E.
1
\ m = 3R (b)
21. K.E. =- T.E. = 2
22. K.E. decreases and total energy increases. 40. Z = 1
23. The charge on the nucleus of atomic no. Z n = 3 $ n = 1 (Shortest l for highest DE)
1
24. -3.4 eV (When n = 2). The possible value of energy = R x 1[1] = R
m
13.6
are - 2 eV where ‘n’ is an integer.
1
\ m= R =x
n
25. Bohr said that electrons neither gain nor lose energy Z = 2
when they are at particular orbits and hence these
n = 3 → n = 2
orbits are called stationary states.
= x × 4 : 4 - 9 D
1 1 1 1
26. Emission spectra is diserect, unequally spaced with m
varying intensity.
= x × 4 : 36 D
1 1 5
m
27. Electrons reaching n = 4 from any state emits EM
1 1 5
waves in the Brackett series. = x × 9
m
28. The energy of the orbits keep changing with Z. Thus m= 5
9x
line spectra of no two elements are indetical.
2.6 Chemistry
41. Photo electric emission occurs when u > uo 61. spin Q.N. is not obtained from y.
42. Refer Q. 41 above 62. Theoretical

43.
V I B G Y OR 63. Degenerate orbitals are such orbitals which have
decreasing frequency same energy.

So only V, I, B, G can cause emission
64. For m = 2, l can be from - 2 to 2.
44. It establishes the existence of quantum of energy,
65. If n = 2 then l can be 0 or 1.
which is a packet of energy.
66. An orbital can have maximum 2 electrons.
45. h (u1 - u0) = K.E.

u1 then K.E. - 67. For principal Q.N. n, number of possible values of
l is also n. Hence, number of subshells = n.
1242
46. E =
m (nm) 68. No. of subshells = n = 3.

4=
1242
No. of orbitals in 3s, 3p, 3d = 1 + 3 + 5 = 9.
m (nm) o
69. d x - y orbitals are double dumbbell shaped with its
2 2

l (nm) = 310.5 nm = 3105 A
two lobes lying along the x and y axes.
47. KE1 = E1 - f 70. Since m = -1, therefore it cannot be s-orbital
KE2 = 2E1 - f becuase for s-orbital m = 0.

\ K.E.2 = 2K.E.1 + f 71. Orbital anular momentum = 2r , ], + 1g
h
48. Changing frequency changes the K.E. of the 72. m can be -lto +l. So no. of values possible
photoelectron, not the number of electrons ejected.
= 2l+ 1.
Photocurrent increases on increasing intensity to
and attains a maximum value. 73. 19th electron goes to 4s orbital in Chromium.
49. Theoretical. 74. Fe3+ : [Ar] 3d5
50. Definition of work function. Co2+ : [Ar] 3d7

51. For an a particle (a), m =

0.101 c
A.
n = n (n + 2)
V
52. r1 = x ⇒ r4 = 42 x = 16 x. n Fe3+
5×7

\ nCo =
2+
3×5

\ circumference = 2p (16x) = 4 l3

\ l3 = 8 p x = 35 : 15

53. According to Heisenberg it is impossible to 75. For g-subshell l = 4

determine both the position as well as the momentum \ No. of e- = 2 ( 2 × 4 + 1) = 18.
of a particle simultaneously and accurately. 76. Two spectral lines of an element cannot have the
54. d orbital has l = 2 same wave number.

55. For 2 electrons in same orbital, n, l, m are same but 77. s-orbital is spherically symmetric. The probability
s is different. distribution is uniform in all directions independent
of the angle.
56. No. of orbitals in a subshell = 2l+ 1. Since each
orbital can have a maximum of 2 electrons therefore, 78. No. of radial nodes = 4 - 2 - 1 = 1.
maximum no. of electrons = 2 (2l+ 1). Thus (d) is correct.
57. n = 2 may have one s and 3 p orbitals with a maximum 79. Radial nodes = 5 - 3 - 1 = 1
of 8 electrons. Node at infinity = 1
58. n = 2 l= 1 for 2 p orbitals. Angular node = 3
\ Total number of nodes = 5
59. f-subshell has 7 orbitals and can accomodate a
maximum of 14 electrons. 80. No. of radial nodes for 5d orbitals = 5 - 2 - 1 = 2
60. p orbital is dumbell shaped. Number of maxima in graph = no. of nodes + 1 = 3.
 Atomic Structure 2.7
EXERCISE - 2
1. Isotopes have same atomic number and different 13. N shell ⇒ n = 4
mass number. - 13.6 × 4 2
\ E4 (Bi3+) = = - 13.6 eV
2. The a-particles (He2+ ions) get deflected by the 42
Now, P.E. = 2 T.E. = - 27.2 eV
positively charged nucleus due to the electrostatic
forces of repulsion. - 13.6
14. P.E. of n = 2 of H atom = 2 × 4 = - 6.8 eV
3. Rutherford discoverd the nucleus by this experiment.
6.8
Now, for He+ if K.E. = 2 , T.E. = - 3.4
4. rnu = 1.25 × 10-13 × 641/3 cm = 5 × 10-13 cm \
- 13.6×4
=- 3.4 & n = 4
n2
ratom = 10-8 cm
0.529×9 o 0.529×16 o
V r3 125×10 -39 15. r3 (He+) = A and r4 (He+) = A
\ V Nu = 3nu = = 1.25 × 10-13 2 2
atom r atom 10 -24 0.529 o o
\ r4 - r3 = 2 (16 - 9) A = 1.851 A
5. A = 39 ⇒ n + p = 39 and n = (p+1)
\ (p + 1) + p = 39 ⇒ p = 19 16. r4 (Z) =
0.529×16
Z
Also, e = 19 r1 (H) = 0.529
Now, n = (p + 1) = 20.

\ Z = 25 from the given options
6. In an element ZA X ,
Note: for Z = 16 it overlaps and doesn’t fit inside
no. of protons = electrons = Z the 1st Bohr orbit of H.
no. of neutrons = A - Z ro $ 4 RZ
28×60 + 29×30 + 30×20 17. r2 = R =
Z & ro - 4
Mavg = 110 RZ 9 9
= 28.64.
\ r3 = 4 × Z = 4 R = 2.25R
7. Atomic electrically neutral as well as nucleus n2
18. 0.529× 4 = 0.529
contains neutrons but neutrality is due toequal
number of protons and electrons. n ` =2
hc
8. Eabsorbed =
Ereleased =
hc hc
+ 19. rn \ n 2
^n 22 - n12h
m m1 m 2

\ Spacing is maximum when is
hc hc hc mm maximum.

\ = + &m= 1 2
m m1 m 2 m1 + m 2
1
9. Let no. of quanta absorbed and emitted be na and ne 20. E n \
n2
respectively.
\ spacing in terms of energy is maximum when
n a hc n e hc 1 1
\ Ea = 4500 and Ee = 5000 - is maximum.
n12 n 22
ne 1 na
Now, Ee = 0.5 Ea & 5000 = 2 × 4500 21. r \ n 2 and time period \ n3
n 5000 5
\ nea = 9000 = 9 = 0.55 r 1 T
since r12 = 4 & T1 = 8
1
2
10. Rest mass is not property of photons as they never 22. Bohr’s model cannot explain emission spectrum of
come to rest. multi electron species.
hc
11. E of quanta able to ionize Na = 23. More the energy of an electron, for ther a way from
2414 ×10 -10
the nucleus it can stay.
hc 6.02×10 23

\ I. E. of Na = -10 × kJ/mol
2414×10 103 24. E3 - E1 = 12.09 eV for H atoms.
. 497 kJ/mol.

Other permissible values of energy absorbed may be:
12. P.E. = 2 T.E. ⇒ T.E. = - 3.4 eV. Z = 2 : 12.09 × 4 = 48.36 eV
- 13.6
\ - 3.4 = Z = 3 : 12.09 × 9 = 108.81 eV
n2
⇒ n=2 and so on.

\ 1 excited state.
st
2.8 Chemistry
25. From III to I,
= R ×1 : 9 - 16 D
1 1 1

\
hc m
2 E - E = E = .....(1)
m 1 7R
From II to I =
m 144
4E E hc 144
3 - E = 3 = m1 ..... (2) m = 7R

Divide (1) and (2) 36. I.E. = x ⇒ E1 = - x

m
x x
\ E2 = - 4 and E3 = - 9
3= 1
m
` m1 = 3m (d)
x x 5x
\ E2 → 3 = 4 - 9 = 36
Z
26. v \ n & lower the value of n, higher the speed. 37. K.E.1 = hui = f and K.E.2 = 2hui - f
27. Since EC→B + EB→A = E C→A
\ K.E.2 = 2 K.E.1 + f
hc hc hc
Now, as intensity is doubled, the number of ejected

\ + =
m1 m 2 m3 electrons increases and hence photo current increases
m1 + m 2 1 but it might not necessarily double.
=
m1 m 2 m3 38. As l is doubled, Eincident is halved and hence stopping

mm
\ m3 = 1 2 potential decreases provided Eincident > f.
m1 + m 2
39. The number of photons emitted per unit time
28. I.E. of He+ = (I.E. of H) Z2 = 13.6 × 4 eV increases incident as higher intensity EM waves of

= 54. 4 eV suitable l are
29. He(g) —→ H(g)+ + e- 24.5 eV (given) 40. As intensity remains constant, the number of
electrons emitted are constant. But as l increases
He (g)
+
—→ H(g)2+ + e- 54.4 eV (Bohr’s Model)
Ein decreases and hence K.E.max and magnitude of

\ He(g) —→ H +
(g)
+ 2e-
78.9 eV (Adding the 2) stopping potential decrease.
30. Second series is Balmer series (n1 = 2) fourth line 41. ui = 4 × 1015 Hz
would be 6 → 2.
K.E. = 10.33 eV
31. 4 excited state is n = 5. So, 5 → 3 and 4 → 3 i.e., 2
th

\ f = hui - K.E.
lines in the Paschen sereis.
4×1015 ×6.6×10 -34
f = - 10.33 +
32. n2 - n1 = 2 and n2 + n1 = 4 1.6×10 -19

\ f = 6.2 eV

\ n2 = 3 and n1 = 1
hc
42. = f + 3qVo
= R4 ; 2 - 2 E & = 9
1 1 1 1 32R

\ m
m 1 3 m hc
= f + qVo
= RZ 2 ; 2 - ]n + 1g2 E
1 1 1 2m
33. hc
m n Solving we get f = ⇒ lo = 4l
4m
or y = = RcZ 2 c 2 ]n + 1g2 m
c 2n + 1
hc
m n 43. E = mc2 and also E =
m

if n > > 1 then (n + 1) . n and (2n + 1) . 2n \ mc =2 hc
m
2RcZ 2

\ y= h
n3 \ m=
cm
= R ;1 - 2 E & 2 = 1 -
1 1 1 1 1
34. 44. m =
h
& more the charge lesser the l
m n n Rm
2mqV
=b l& n =
1 Rm - 1 Rm

or, m1 q2 2
n2 Rm Rm - 1
\ = q = 1
m2 1
nh 2h
35. 2r = r ⇒ n = 4
\ l1 = 1.414 l2
n = 4
\ n2 = 4 → n1 = 3 is the transition that takes place.
 Atomic Structure 2.9
h
45. Dp = Dx ⇒ (Dp)2 = 4r n
` 2 + 2n - 15 = 0 ⇒ n = 3 or - 5
n
` =3
h

\ Dp = m Dv =
Mn +x = 1s 2 2s 2 2p6 3s 2 3p6 3d3
2 r
1 h x = + 4.

\ Dv = 2m r
56. By Aufbau Principle. {(n+l) rule}
h
46. 0.1 × 10 = p
-9
1 1
57. The quantum numbers + 2 and - 2 for electron
6.6 ×10 -34 spin represent two quantum mechanical spin states
` p =
0.1×10 -9 which have no classical analogue.
p = 6.6 × 10-24 kg m/s 58. By convention d z and pz are the orbitals which are
2

h assigned m = 0 for l= 2 and l= 1 respectively.

47. m = p . Hence l is inversely proportional to p.
59. No. of values of m is the no. of orbtials present in a
48. Eabsorbed = 1.5 × 13.6 eV given subshell. Thus, m = 2l+1 ⇒ l = 2
m-1

\ K.E. of emitted electron = 6.8 eV
60. Hund’s Rule says that pairing of electrons in an
6.625×10 -34
\l=

h
= atomic orbital will take place only after all the
2m ]K.E.g 2×9.1×10 -31 ×6.8×1.6×10 -19 degenerate orbitals have been singly occupied.
o

\ l = 4.70 A 61. Greater the atomic number smaller is the size of the
species. Thus, the graph with the highest probability
h 6.625 × 10 -34 at the least value of ‘r’ represents Li­2+ while the one
49. m = =
2m ]KEg 2×9.1×10 -31 ×4.55×10 -25 with the peak at largest value of ‘r’ represents H.

= 7.28 × 10 -7 m 62. For radial node, y = 0
50. E A - WA = K A ` E A - 2 = 4K B 6 ! 36 - 24
E A - 2 = K A E A + 0.5 - 4 = K B
⇒ 6 - 6s + s2 = 0 ⇒ s = 2
E B - 4 = K B E A - 3.5 = K B 2rZ
2

\ s = 3 ! 3 ⇒ 3a = 3 ! 3
o
h
m a2 = 2m × K ` E A - 2 = 4E A - 14
\ r1 = 2Z 3 + 3 h and r2 = 2Zo ^3 - 3 h
3a o ^ 3a
A

h2
4m 2A = m 2B = 2m × K ` E A = 4eV 3 3 ao
3a

\ (r1 - r2) = 2Zo ×2 3 = Z
B
EB = 4.5eV
1 K
` 4 = K B ` K A = 4K B \ VA = 2V, VB 0.5 V 1 :ZD 2
3
v
A 63. } = [(v - 1) (v 2 - 8v + 12)] e - 2
16 4 a o
h
51. Orbital angular momentum = , (, + 1) 2r
\ (v - 1) (v 2 - 8v + 12) = 0

for d-orbital l= 2
\ v = 1 or v 2 - v6 - 2v + 12 = 0
h
⇒ Orbital angular momentum = 2r 6
v (v - 6) - 2 (v - 6) = 0
52. No. of electrons for given value of l= 2 (2l+ 1) \ v = 2 or v = 6

and l can very from 0 to (n-1)
\ smin = 1
, = n-1

\ No. of electrons = / 2 (2, + 1)
a
& rmin = 2Z0
,=0

53. All p orbitals and d x - y , d z are located along the

2 2 2 and smax = 6
axis. 3a
& rmax = Z0
54. s-orbitals are spherically symmetric hence the
probability of finding an electron at a given distance 64. Since y becomes 0 for one value of r, therefore
is same in all directions. number of radial nodes = 1.
65. The probability of finding on electron in nucleus is
55. Mn = 1s 2 2s 2 2p6 3s 2 3p6 4s 2 3d5
zero.

3.873 = n (n + 2)
15 = n2 + 2n
2.10 Chemistry
EXERCISE - 3
h k h k 11. 13.6 Z2 = 217.6 ⇒ Z2 = 16
1. m1 = = , m2 = =9
2mq (100) 10 2mq (81)
\ Z=4
h k
m3 = =7 12. No. of radial nodes = (n -l- 1) = 1
2mq (49)
m3 - m 2 ]1/7g - ]9/4g 10 × 2 13. Any d subshell can have maximum of 5 unpaired
\
m1
= ]1/10g = 63 = 0.317
electrons. so the maximum spin possible
1240 1
2. 3 E = 1282.5 eV = 0.96 eV = 5× ! 2 = S
This is the difference in levels 5 and 3 of atomic
5
\ Spin multiplicity = 2 | s | + 1 = 2 × 2 + 1 = 6.
hydrogen.
14. Longest wavelength of any series is the lowest
Now series limit means 3 → ∞ energy line of that series and has the lowest energy.
= (912) -1 c 2 m & = (912) -1 b 9 l & m For Lyman it is 2 → 1.
1 1 1 1
m n1 m
= 8208 Å 15. P.E. = - 6.8 eV = 2 × T.E.
- 13.6
3. 5th excited state ⇒ n2 = 6 \ T.E. = - 3.4 eV ⇒ = 3.4 ⇒ n2 = 4.
n2
For H-atoms: if n1 = 1 ⇒ UV region \ n = 2 ⇒ 1st excited sate.
if n1 = 2 ⇒ visible region h
16. 3 x = 3 p & 3 p =
if n1 = 3, 4, 5..... ⇒ IR region 4r
⇒ Maximum possible transitions are :- ` m 3 v =
h
4r
6→5, 6→4, 6→3, 5→4, 5→3, 4→3
1 6.625×10 -34
⇒ Ans = 6. ` 3 v = -31 × 4×3.14
9.1×10

= 7.98 × 10+12 ⇒ y = 8
4. m1 = d n & V1 = 150
150 1/2
V1 m12 17. No. of volues of m = 2l+ 1
150 150 7-1
]1.41g2 ]1.73g2
V
& 1= = 75 V and V2 = = 50 V ⇒ l= 2 = 3

Hence, potential should be dropped by 25 V. 18. 3 → 2, 2 → 1, 3 → 1.
19. 5.27 = f + 0.27 ⇒ f = 5
12400
5. 3 E = 6200 = 2 eV.
20. Cu+ has 3d10 configuration. So, no. of electrons with
bs =- 1 l = 5
2 × 1.6 × 10-19 × n = 40 × 60 × 0.5
n = 3.75 × 1021 2
6. n = 2, l= 0, 1, 2, 3 = 1.09677×10 +7 ×Z 2 :1 - 4 D
1 1 1
21.
\ for last subshell, n = 2 and l= 3. 3×10 -8
4
\ n + l= 5
or, Z 2 =
3×3×1.09677×10 -8 ×10 +7
7. 2nd line of Balmer series ⇒ 4 → 2
⇒ Z2 = 4 ⇒ Z =2
1st line of Lyman series ⇒ 2 → 1 22. Maximum m = + 3 ⇒ l= 3 ⇒ n = 4
= R : 4 - 16 D & m B = 3R \ No. of waves by the electron in the fourth orbit = 4
1 1 1 16
\
mB
23. n l = 2 p r ⇒ n 200 = 400 ⇒ n =2
= R :1 - 4 D & m L = 3R
1 1 4

mL R R
24. E =- 9H =- 2H & n = 3 .
m n
\ B =4
mL No. of electrons that can be accomodated in n = 3 is
8. 4d orbital has 1 radial node, as (n -l- 1) = 1 2 × 32 = 18.

\ No. of peaks = no. of radial nodes + 1 = 2
\ No. of orbitals presents in n = 3 is 9. All these
9. b ro n4 l = ]rogH & n = 2
2 orbitals are degenerate as of single electron
Be 3+ species E3s = E3p = E3d and so on.
]5 - 2g]5 - 2 - 1g 25. 4→2 in He+ has same energy difference as 2→1 of H.
10. No. of lines = 2 =6
 Atomic Structure 2.11
EXERCISE - 4

10.
1. For He+ : x = R $ 4 ; - E & Rc 2 - 2 m = x
1 1 1 1 C contains 6 protons and 6 neutrons and weighs
12
6
n12 n 22 n1 n 2 4 12 amu. Mass of electron is negligible.

Now, if mass of neutrons is halved, total mass = 9
= R $ 16 ; 2 - 2 E & = 16× 4 = 4x
1 1 1 1 x

For Be3+ : amu.
m n1 n 2 m
3
2. Note that the question states that there is a single H
\ reduction in mass = 3 amu ⇒ 12 × 100 = 25%
atom. Thus, the maximum number of spectral lines 11. Factual.
emitted may be 3 for the transitions 5 → 4, 4 → 3
and 3 → 2. 12. Red end ⇒ we are talking about visible region of the
EM spectrum ⇒ Balmer series in an H atom.
3. The lowest value of ‘n’ for which d-orbital exists Thus, 3rd line from red end ⇒ 5 → 2
is 3.
13. Balmer series lines of only H atoms lies in the visible

\ Electronic configuration given is 3d6 4s2
region of the spectrum

⇒ 4th period.
14. K shell ⇒ n = 1
4. DE2→1=13.6 b 1 - 4 l eV and DE3→2=13.6× b 4 - 9 l eV
1 1 1 1
L shell ⇒ n = 2
3 5 M shell ⇒ n = 3 and so on
= 13.6× 4 eV = 13.6× 36 eV

Since electron makes transition into n = 2, therefore
\
3 E2 " 1 3/4 27 it is Balmer series.
3 E3 " 2 = 5/36 = 5
15. Lyman series is in UV region and Balmer series is in
5. Factural : K. E.max = huincident - f visible region in H atom.
6. Orbital angular momentum = 2r , ], + 1g = 2r
The 4 lines in uv region are : 2→1, 3→1, 4→1, 5→1
h 6h

3h
In the visible region we will have: 3→2, 4→2, 5→2
angular momentum in 3rd Bohr orbit = 2r

\ In the IR region: 5→4, 5→3, 4→3 ⇒ 3 lines
6 2
\ ratio = 3 = 3 16. Rest mass of photon =
h
cm
7. m = 1.73 = 3 = n ]n + 2g
= R :1 - 4 D = 4 R
1 1 3

For 1st line of Lyman:
\ no. of unpaired electrons, n = 1. m1
Now Vanadium has Z = 23 and E.C. of [Ar] 4s23d3 = R : 4 - 9 D = 36 R
1 1 1 5

For 1st line of Balmer:
m2
\ for 1 unpaired electron V must be in + 4 state
m m m 3 36 27
with an E.C. of [Ar] 3d1.
\ m12 = 2 & m12 = 4 × 5 = 5
m1
1
8. 4.25 - fA = TA and lA \ 17. Series limit of any series & 3 $ n1
TA
= R :1 - 3 D &
1 1 1 1
1 \ For Lyman series: =R

4.2 - fB = TB and lB \ mL mL
= R; 2 - 3 E &
TB 1 1 1 1 R

a lB = 2 lA For Balmer series: = 4
mB 2 mB

\ TA = 4 TB \ lL = 4 lB
Now, TB = TA - 1.5 ⇒ 3 TB = 1.5 18. Theoretical

⇒ TB = 0.5 eV and TA = 2 eV
19. Spectral lines of an element are unique and no 2

\ fA = 4.25 - TA = 2.25 eV lines of an element can have the same wavelength.
and fB = 4.2 - TB = 3.7 eV 20. y2 is the probability density i.e. probability per unit
9. 2.1 g ion of Cl- ⇒ 2.1 mole Cl- ions volume.
Now, 1 Cl- ion has 18 electrons 21. In H atom

\ 2.1 mole Cl contains 18 × 2.1 × NA electrons
- Elyman > Ebalmer > Epaschen ..........

\ llyman < lbalmer < lpaschen

\ l1 < l2 < l3 < l4
2.12 Chemistry
o o 36. In H atom, or any other H like species, the energy of
22. For n = 10, r = 0.529 × 102 A = 52.9 A
the electron depends only on the value of ‘n’. Thus,

\ 2 × p × 52.9 = 10 l we can say that for such species (E4s = E4p = E4d =

\ l = 33.22 A
o
E4f) > E3s = E3p = E3d and so on.

23. I : 1 radial node ⇒ 3p 37. Factual.

II : Zero radial nodes ⇒ 3d 38. Heisenberg’s uncertainty principle is valid for both
microscopic and macroscopic moving objects.

III : 2 radial nodes ⇒ 3s
However, it is significant only for microscopic
24. Longest wavelength of Paschen series ⇒ 4 → 3 objects.
\ R $ 1 2 $ : 9 - 16 D = R $ 2 2 $ ; 2 - 2 E
1 1 1 1
h
39. 3 x $3 p $ 4r : by definition.
n1 n 2
⇒ n1 = 6 and n2 = 8. dp
Now, dt = dF⇒ dp = dF.dt
25. y2 is maximum at nucleus and 3p r2 y2 is zero.
dE
hc And dF $ dx = dE & dx = dF
26. E = hu = = hc y
m dE
hc \ dp $ dx = dF $ dt $ dF

Also, Etot = n , but energy of 1 quantum is
m
= dE $ dt
independent of n.
27. \ -particles being much heavier than the electrons \ Heisenberg’s principle may be written as
have a high penetrating power, which is also because h
3 E $3 t $ 4r
of the high speed of \ -particles.
40. In option (c) if l= 2 then ‘m’ cannot be - 3.
28. All 4 statements were conclusions drawn from the
\ -scattering experiment. 41. a) s-orbital is spherically symmetric
29. Bohr’s model doesn’t talk about probabilities. Hence b) shape of orbital is given by ‘l’ not ‘m’.
statement (a) is incorrect. c) Since l= 0, therefore orbital angular momentum
30. Bohr derived that = 0 for 1s, 2s, 3s etc.
n2 Z Z2 d) Electrons in different orbitals have different
rn \ Z , v n \ n and freq \ 3 velocities.
n
Kq q KZe 2 Z3 42. For 3d orbitals: n = 3, l= 2, m = -2, -1, 0, 1, 2 and
]n /Zg
Now, Force = 2 & F \ 2 2 F \ 4
1 2

r n s = + 0.5
31. 2nd I.E. of He = 13.6 × 22 eV 43. If ‘l’ of two orbitals are same, then their orbital

3rd I.E. of Li = 13.6 × 32 eV angular momentum is also same.

1st I.E. of H = 13.6 eV 44. b) if n = 3 then l can be 0 or 1 or 2.

o o

9
r3 of H = 0.529 × 9 A and r3 of Li2+ = 0.529 × 3 A d) if l= 2 then m can be 0 or + 1 or + 2.

n,land m are the 3 Q.N. which can define an orbital. 45. All 3 statements are correct.
Z2 Z2 46. All 3 statements are correct.
32. K.E. K.E. \2 , P.E. \ - 2 , rn = ro $ n
2
n n
47. electron cloud density of px orbital

Energy levels or orbits are not equally spaced. The
distance between the orbits keeps increasing with n. 48. (2) is the excited state electronic configuration of (1)
33. Bohr’s model can be applied to all single electron 49. (b) For n = 2 we can have 4 orbitals but they all do
species. not have different energy levels. Out of the 4, 3
are degenerate and have different energy from
n3
34. Frequency \ the 4th.
Z2
(c) M ⇒ n = 3 and can accommodate a maximum
35. For electronic transition from n → 1, no. of spectral
n ]n - 1g
of 18 electrons.
lines is given by 1 + 2 + 3 + ..... (n - 1) = 2
 Atomic Structure 2.13
50. Quantum numbers are derived by solving the 2.178×10 -11
Schroedinger’s wave equation. For a multi-electron 62. T.E.3 = - 9 ergs
species, the energy of an electron can be estimated a P.E. = 2 T.E.
by the value of (n +l). Azimuthal Q.N. does not 2
\ P.E. = - 9 ×2.18×10 -11 ergs
tell us anything about the motion of the electron.
= - 4.84 × 10-12 erts
51. Shape of orbital is given by azimuthal Q.N. Also,
h
electrons in an atom move with different velocities. 63. m =
2m ]K.E.g
52. Due to the presence of electrons in them the orbital
h
ex-orient themselves to minimize the energy in a 64. m =
2m qV
magnetic field.
6.625×10 -34
53. Spin of electron does not affect the angular 5 × 10-12 =
2×1.67×10 -27 ×1.6×10 -19 ×V
momentum and can be represented as clockwise and
Solving we get V = 32.8 V
anticlockwise.
65. m = 75 × 10-3 kg
54. Factual.
h 6.625×10 -34
55. E4s < E3d. If n = 2 the there are 4 orbitals which can Dx = l = mv ⇒ Dx =
75×10 -3 ×4
accommodate a maximum of 8 electrons. If n = 5 h 6.625×10 -34
then maximum no. of electrons in it is 2n2 = 50. \ Dv = 4rm 3 x =
4×3.14×75×10 -3 ×2.2×10 -33
56. If n = 4 and m = - 2, it may be a 4d or 4f orbital. The = 0.32 m/s
1 1
spin of the electron may be + 2 or - 2 . 66. No. of electrons possible in an orbital = 4
57.
l= 1 to (n + 1)
, ,
electron p q x y z
m = - 2 to + 2
n 3 3 3 3 3
For n = 1 ⇒ l= 1, 2 i.e. 1A, 1B
m 0 0 1 1 1
For n = 2 ⇒ l= 1, 2, 3 i.e. 2A, 2B, 2C
l 0 0 -1 +1 0
For n = 3 ⇒ l= 1, 2, 3, 4 i.e. 3A, 3B, 3C, 3D
s - 1
2 + 1
2 + 1
2 + 1
2 + 1
2
Now, for l= 1 : m = - 1 2 , + 1 2
58. K.E. of photo electron varies with u and l but is
l= 2 : m = -1, 0, 1
constant with intensity.

3 1 1 3
l= 3 : m = - 2 , - 2 , + 2 , + 2
59. A1 = p r12 An = p rn2 and rn = r1n2
l= 4 : m = -2, -1, 0, 1, 2

\ A subshell has 2 orbitals
\ b An l = & ln b An l = 4ln (n)
2 4
A rr n A
1
1 rr12 1 B subshell has 3 orbitals
\ Graph of ln b An l vs ln (n) is a straight line with
A
\ E.C. would be as follows:

1
1A 8
1B 12
2A 8
2B 12
3A 8
.....
slope = 4 and this line passes through the origin. 14444244443 14444244443
1st period 2 nd period

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

Therefore, number of elements in 2nd period are 20.
of multi electron species while in case of single
67. No. of elements in 1st period = 20
electron species, it depends on only the value of ‘n’.

\ 2nd period beings with Z = 21. i.e. Sc.

P.E.1 = - 2 K.E.1 and P.E.2 = - 2 K.E.2 ⇒ |DP.E.|=
2|DK.E.| 68. For Z = 100

It is practically not possible for an electron in an
excited state to absorb another photon.
o
61. r3 = 0.529 × 9 A

\ 2 p r3 = 3 l ⇒ 2 p 0.529 × 9 = 3 l

\ l = 9.96 × 10-10 m . = 10-9 m


\ outermost configuration = 4A8 3C16
2.14 Chemistry
69. For E subshell, l= 5 Z2
74. T.E. =- 13.6eV ⇒ as n increases, T.E. increases.
\ No. of orbitals = 6 n2
Z2
\ No. of electrons = 24 (as 4 electrons in each
Frequency of revolution \ 3 ⇒ as n increases,
n
orbital) frequency decreases.
70. Shortest wavelength is emitted for highest energy. Z3

Acceleration \ 4 ⇒ as n increases, acceleration
n
\ D has shortest wave length emitted decreases
71. Ionization potential is the amount of energy absorbed
n2
radius \ Z ⇒ as n increases, radius increases
by a gaseous atom to lose the aoutermost electron.
75. a) 3 radial nodes ⇒ 4s / 5p but for p orbitals, y < 0
72. B = 4 → 3 ⇒ EB = 13.6 b 9 - 16 l eV
1 1
at r = 0
= 13.6 × 9×16 eV
7
b) 3 radial node ⇒ 4s and 5 p
c) For p and d orbitals angular probability
= 1.983 eV
distribution depends on q and f
\ EB = 1.983 × 1.6 × 10-19 J
d) S - orbital is spherically symmetric, hence no
= 3.18 × 10-19 J energy is released.
planar node.
22
73. a) U1, 2 = - 2 × 13.6 × = - 13.6 × 8 eV 76. a) Orbital angular momentum
12
K1, 1 = + 13.6 eV = 2r , ], + 1g = 2 2r & , = 1 & p orbital.
h h
\ Ratio = - 8 : 1 b) Bohr’s model ⇒ classical mechanical

22
b) r2, 1 = ro 1 = 4 ro
12 r
and r1, 2 = ro 2 = 2o c) 5 degenerate orbitals ⇒ d-orbitals

\ Ratio = 8 : 1 d) N shell ⇒ n = 4 ⇒ no. of waves made by the

electron = 4
2 4

c) V1, 2 = Vo 1 = 2 Vo and V2, 4 = Vo 2 = 2 Vo
\ Ratio = 1 : 1
n3 T1,2 1

d) T \ 2 & T = 8
z 2,2

1. E = W + K.Emax 4. (i) Higher the value of ‘l’, lesser is the penetration

K.Emax = E - W of the orbital. Thus, given statement is correct.
hc (ii) Size of orbit depends on value of ‘n’. Thus,
= - 4.41 × 10-19
m incorrect.
6.63 ×10 -34 × 3 ×108
= - 4.41 × 10-19 nh
(iii) Angular momentum = 2r and for ground state
300 ×10 -9
= 222 × 10-21 joule n=1. Thus, given statement is correct.
(iv) As value of azimuthal Q.N. increases, y vs r
2. 2 pr = nl
graph has more number of peaks and hence
n2 shift towards lower value of r. Thus, incorrect

2p × a = nl
Z 0 statement.
42 5. The graph given in option (c) is incorrect as the

2p × a = 4l
1 0 emission will begin only once n > n0. The graph for
l = 8pa0 (c) should look similar to that of (a) because E a n.

3. v = R H c- m cm -1 6. Radius of nth Bohr orbit in H-atom = 0.53 n2 Ao

1 1
82 n2
Radius of 2nd Bohr orbit = 0.53 × (2)2 = 2.12 Ao
R R R R
= 64H - 2H = 2H + 64H 13.6
n n 7. \ E n =- eV Where, n = 1, 2, 3...
Comparing it with general straight line equation, n2
- 13.6
y = mx + c, we get In excited states, E2 = 4 =- 3.4 eV
R
Slope (m) = – RH, Intercept (c) = 64H
 Atomic Structure 2.15
8. Given, atomic number of Rb, Z = 37 10. According to Bohr’s model,
Thus, its electronic configuration is [Kr]5s1. nh n2 h2
mvr = 2r & (mv) 2 =
So, the quantum numbers are n = 5, l = 0, (for 4r 2 r 2
s-orbital) m = 0 ( a m = + l to - l), s = + 1 /2 or - 1/2. 1 n2 h2
⇒ KE = 2 mv2 = ...(i)
8r 2 r 2 m
9. Given, in the question E = - 2.178 × 10-18 J ; 2 E
Z2
n Also, Bohr’s radius for H-atom is, r = n2a0
For hydrogen Z = 1, Substituting ‘r’ in Eq. (i) gives
So, E1 = - 2.178 ×10-18 J ; 2 E
1
h2 h2
1 KE = When n = 2, KE =
8r n a 0 m
2 2 2
32r 2 a02 m
E2 = - 2.178 ×10-18 J ; 2 E
1
2 11. The number of radial nodes is given by expression
Now, E1 - E2 (n - l - 1).
i.e. DE = 2.178 ×10-18 c 2 - 2 m =
1 1 hc For 3s, number of nodes = 3 - 0 - 1 = 2
1 2 m
For 2p, number of nodes = 2 - 1 - 1 = 0
2.178 ×10-18 c 2 - 2 m =
1 1 6.62×10 -34 ×3.0×108
1 2 m r n2
12. Expression for Bohr’s orbit is , rn = 0Z
m . 1.21×10 m -7

When n = 2, Z = 4.

- 13.6 # 22 5. n = 3 can accommodate maximum of 18 electrons of

1. - 3.4 = 1 1
n2 which 9 will have s = + 2 and 9 will have s = - 2
n = 4
 = 2 6. S1 is spherically symmetrical state, i.e. it correspond
to s-orbital. Also, it has one radial node.
Subshell = 4d
Number or radial nodes = n - l - 1.
Angular nodes =  = 2
Radial nodes = n -- 1 = 4 - 2 -1 = 1
⇒ n-0-1=1

⇒ n = 2 i.e. S1 = 2s-orbital.
2. At d = d0, nucleus-nucleus & electron-electron
repulsion is absent. - 13.6×3 2
7. E2s (Li2+) =
Hence potential energy will be calculated for 2 H 22
atoms. (P.E. due to attraction of proton & electron)
9
\ E2s (Li2+) = 4 E1 (H)
- Kq1 q 2 ]9 ×109g^1.6 ×10 -19h2
= 2.25
P.E. = =
r
]Bohr radiusg
0.529 ×10 -10
8. S2 is 3 p orbital, for energy of nth shell of Li2+
= -4.355 × 10-21 kJ
= -13.6 eV, n = 3

For 1 mol = -4.355 × 10-21 × 6.023 × 1023
As it has 1 radial node hence it is 3p
= -2623.249 kJ/mol
h

For 2 H atoms = -5246.49 kJ/mol \ orbital angular momentum = 2 2r
13.6Z2
9, 10. KE = eV / atom
r r n2
  2 # 13.6Z2
e e PE = eV / atom
n2
do
n2 %
Radius 0.529 Z A
3. n = 4 nh
Angular momentum of electron (mvr) = 2r
| m | = 1 ⇒ l may be 1 | 2 | 3 and m = +1 or -1
\ 2 electrons each of 4p, 4d and 4f can have
1 11. 2s orbital has are radial node and y(r) = 0 once
n = 4, | m | = 1 and s = + 2
12. 1s has zero radial nodes
⇒ Total 6 electrons
1242 13. E2 → 4 = 2.55 eV E2 → 6 = 3.022
4. z = 300 = 4.14 eV
\ Li, Na, K, Mg show P.E.E.
2.16 Chemistry