ATOMIC STRUCTURE (MAIN)

ATOMIC STRUCTURE (MAIN) FOUNDATION BUILDER (OBJECTIVE) 1.

3

Li6  ?   2 He 4  1H 3

By law of conservation of mass and change the missing particle in neutron 2.

 n 0 0

1

e ratio lies in the sequence n    p  l M Particle Change Mass  +2 +4 n 0 +1 p +1 +1 e –1  11837

 e m  order  n  P  e 3.

Atomic Number = No. of protons in atom By equation of change 1 56  1 x  2 x = 54 

4.

Same number of neutrons hence, Isotones.

5.

Cathode Ray are made of electrons hence, same change/mass ratio as of  particle.

6.

From Muliken’s oil drop experiment, it was found that change on oil droplets is qualified. Hence, q = ne . where e  1.6  10 19 , n = 1, 2, 3…  (B)

7.

f

8.

wavelength VIBGYOR highest lowest frequency

1  f  1 Hz 2 T

Energy  freq.  (D) red

9.

(c)

10.

Wave number   

1  2



1 500  109



1000  10 500

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 (c)

hc E1  2 ,  2  E2 1

11.

E

12.

velocity 3  108 Frequency =  wavelength 5090  103

13.

E  nhν

n

= 14.

E hν

103  1.72 1030  34 3 6.626  10  880  10

E photon 

12400 12400   1.393eV 0   inA  8900

1.393  1.6  1019  x  3.15  1014 3.15 x  105 x  1.41 105 1.393  1.6

 c



15.

50  E emitted out 100 50 hc  n1   n2  100  emitted

E absorbed 

hc  absorbed

5 n2 50  50 5000   0.55   emitted   9 n1 100  absorbed 100 4500 16.

As PE = - 2 KE PE will change from - 2x to  =

17.

TE 

18.

TE  

19.



2x 4

x 3  2x   x 2 2

PE , so first excited state 2

 13.6 Z2  13.6  16 PE    13.6 and TE  2 n 16 2 PE = - 27.2 eV

 13.6 Z 2  13.6  1    1.511 n2 9 PE TE   PE  3.02eV TE = –KE  KE  1.51eV 2 TE 

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20.

21.

22.

0.529n 2 0 r A Z 0.529  9 r3rd   2.3805 A 0 2

v  2.18  106 Z n

Z n v1 n 2 5   v 2 n1 3

(B)



24.

0.529  16  4.232 2

0.529 n 2 rx  ,n=4 Z 0.529n 2 rH  , n = 1, z = 1 Z 0.529 16  0.529  Z  16  Z Rx < rH

v

23.

r4 th 

a0  4 R Z 9R r3   4 r2 

r3 

a0  9 R Z

Ground state of hydrogen atom = 0.529 Å

r

0.529  n 2 0.529  (n)2   0.529 z 4

 n2

2.18  10 6 Z 1 , v  z, v  n n

25.

V

26.

1 2.18  10 6  V 2    8.13  1014 s 1 10 2r 2 4  0.529  10

27.

E

nhC  nhc  

10 = nhcx

13.6 Z2 n2

28.

E

29.

mvr 

n

=

10 hcx

13.6  1  3. 4 4

nh 2

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r

0.529n 2 Z

mvr 

r

Angular momentum  r 30.

31.

32.

V Z2  3 2 r n 1 H  T 27 2 T  27 x =B 

 He  x

4 x 8

27 T 2

13.6Z2 eV n2 13.6 T E 4,H  eV   K E   E 16 1 E  144 x X = –144 E TE 

TE Li2 

13.6  9 x 1

 1 1   1 1  R h 12  2  2   R h  22  12  12   n1 n 2   n1 n 2  

1 1  1 1  1     4    2   1 25   1 n2  6



33.

24 n2 1  4 22 25 n2



19n 22  25



(b)

f 

KZe 2 r2 KZe 2

Z3 =  4 2 n  0.529n 2    Z   1 fH   x 1 X = 16f/27

34.

V2 r (2.18  106 ) 2 Z2 n2 = 0.529n 2 Z

f Li2 



6n 22  25n 22  25



n 22 

25 1 15

27 f 16

27 f  16 x

a



z3 n4

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a1,He 

8 1

a 2,Be3 

64 16

1 a 2,Be3   2 Follow the expression n 2  0.529 r Z (d)  Follow the expression  13.6 Z 2 E n2 (a)  See theory 2n2 + 3n1 = 18 2n2 – 3n1 = 6 Solve this and we get n1  2, n 2  6 

35.

36.

37. 38.

So, 39.

 6  2  6  2  1  10 2

n1 + n2 = 4 n2 – n1 = 2  n 2  3, n1  1 

1 1 1  R H  22  2  2   3  1

8 = R H  4  9

40.

 1 1 1   R H Z 2  2  2   n2   n1 

 1 c 1   cR H Z2  2  2   n2   n1

 1   2n  1  c 1  cR H Z2  2   cR H Z2  2 2  2   (n  1)  n  n (n  1)  When n >>> 1 then (n +1)  n and (2n + 1)  2n 

n 2c R H Z2  n4 n3 1  1   32  R  2  0   R  min 3 

  2cR H Z2 41.

  min  42.

1  max

1 R

1 1  R H  (2) 2  2  2  2  1

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1  max

1 1   R H  4   1 4 

  max 

43.

 1 1 1  1  1  R  2  2   R  2  2   n2  n  1  n1

44.

 R  2 n   R  1  E = E1 + E2 hc hc hc    1  2

1 3R H

1



45.

46.

1 2 1   2

 1 1 1   R H Z 2  2  2   n2   n1 1 1  1  RH 2  2  9 2170  10 7  n

 n4

n ( n  1)  15 2 n=6

 1 1 1   R H Z 2  2  2   n2   n1 1 1 1  109677 2  2   6  1 = 937.3 Å 47.

n ( n  1) 6 2 n = 4, so excited state is 3rd

48.

1/2 = 1/1 + 1/3 2

3

1

49.

2 = 13/ (1 + 3)

1 1  1 1  RH 2  2    RH L   x 1 1 1  1  R H  4 2  2  B 3  2 1 1 5   B x 9

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50.

x  mv  x  p 

h 4

h 4

x  p

( p) 2 

mv 

h , p  4

h 4

h 4

1 h 2m  mass = 100 × 103 kg V 

51.

V = 23.76 km s/hr = 23.76 

5 m/s 18

h = 6. 6 × 10–34

52.

h 6.626  1034    1039 m mV 100  103  23.76  5 18 h   2   1 1  h  mv  KE  mv 2  m     h  2 2  m   v   m  

1 mh 2 1 h2  2 m 22 2 m2 1 KE  m 2r  n

=

53.



54.

2r n

 

2 32 x  6x 3

m  200g  0.1 V   10 1  100 v  10 ms  h x  mV  4 h 6.626  1034  4mv 4  200  0.1  10 1000 100 Follow theory (Follow theory) ν  3.5  10Hz x 

55. 56. 57.

ν 0  1.5  1015 Hz

h = 6.6 × 10–34 KE = hν  hν0

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KE = 6.6 × 10–34(3.5 × 1015 – 1.5 × 1015)  1.32  1018 J 58.

KE = hν  hν0

1 1 1   mv 2  hc  2   0  v2 

2hc  1 1     m   0 

v

2hc  1 1     m   0 

v

2hc   0      m   0 

59.

See theory

60.

Orbital angular momentum =

61. 62. 63. 64.

h h  6 2 2

h mV See theory See theory h l(l  1) 2 

65 to 74. 75.

l (l  1)

See theory

76. 77. 78. 79.

n = 3, l = 3, m = 0, s = –1/2 Not possible Follow n + l rule Follow theory Follow n + l rule A g subshell will have 9 orbitals so there will be 18 electrons

80.

26(Iron) follow electronic configuration

81.

(D) is not possible because ‘P’ sub shell cannot have more than 7 electrons.

82.

Mn  3d5 4s2 Ti  3d 2 4s2 V  3d 3 4s2 Al  3s2 3p1

83.

n  n  2

Fe  3d 6 4s 2

n=5

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5 5  2

84.

1 5 s   5  2 2

85.

See configuration.

86.

Same as 92

87.

See Theory

88.

  n n  2 2.83  n  n  2 

89.

Same as 98

90.

  n n  2 1.73  n  n  2  N=1

91.

  n n  2 Write the electric configuration for both fe and Co and after removal of 3 electron from cobalt the unpaired in Fe 3  5 and Co 3  4 GET EQUIPPED TO MAINS

1.

(C) 9 9 E1,He   19.6  1018 4 4 17  4.41 10 J E1,Li2 

2.

(D) n : 1s 2 2s 2 2p 6 3s2 3p1  outermost e  : n  3,   1

3.

(A) E

4.

hc E  1 2  E2

(B) 1s 2 2s 2 2p 6 3s2 3p 6 4s1 3d10   1  p subshell  12e    2  d subshell  10e 

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5.

(D) Orbital angular momentum      1  same  value has same orbital angular momentum.

6.

(B) By  n    rule

7.

(B)

r3He   8.

n2 32 a 0  a 0  4.5a 0 Z 2

(C)



C 3 108   6  1014 Hz  500 109

9.

(D) 1  9  15200  136800 

10.

(D)

11.

(A) Atomic no.  25

 Mn

12.

(C) 2nd series  Balmer 4th Line in Balmer  6  2

13.

14.

(A) Paschal Lines : 5  3 43 (B)

15.

(A) E

1240 1 1.6 10 19  6.022  10 23  242 1000

16.

(C) m cannot be greater then 

17.

(A)

18.

(D)

19.

(A) 1  1 1  5R  R      4 9  36

20.

(D)

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r

n2 a0 Z

21.

(A) 1s 2 2s 2 2p 4 No. of unpaired electron  2  total spin  1 Magnetic moment  2  4  8

22.

(B) No. of angular nodes  2

23.

(A)  1 1  5x E  x     4 9  36

24.

(B)

25.

(C) Orbital angular momentum  2  3  6

 2

h 2

26.

(B) No. of radical nodes  n    1  2 1  1  0

27.

(B)

6.6 1034 p  66 1025 9 0.1 10 28. 29.

(D) (D) nh 2h  n4 2  1 1 1   R     9 16  144  7R

30.

(B) Min.   Max.E

31.

(C) 1 1 1   R  2   4 n  4n 2 4   R  R R  n2  4

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32.

(a) ECA  ECB  EBA 1240 1240  eV 364.6 121.5  3.4  10.2  13.6 eV 

= 13.6 

1240 

   91.17 nm

33.

(a) Minimum  1 4 1 Maximum  4 4  3  2 1 & 4 1 n  n  1  only if sufficiently large number of atoms are present 2

34.

(d) Shortest wavelength implies maximum energy n  n  1   15 2 1 1  2 1   R H 1     61  1 36  1 35R 36     36 35R

35.

(c) Total orbitals  3  1  3 2  1 7 e in 1 orbital still  2 Since it has only 2 types of spin (b) nh L 2

36.

37.

38.

39. 40.

(b) 235  1  196  x  3  x  90  3  87 (c) Radial  1  spherical Angular 3  1  1  1 (a) S  spherical (non-directional) hc E1111  2E  E  

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4E hc E  3 ' E hc   3 '   '  3

E111 

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