ATOMIC STRUCTURE (ADVANCED)

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

(A) 6  2 He 4  1H 3 3 Li  ?  By law of conservation of mass and change the missing particle in neutron

2.

 n 0 0

1

(D) 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 order   n  P  e m

 

3.

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

4.

(D) Same number of neutrons hence, Isotones.

5.

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

6.

(B) 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.

(B) f

1  f  1 Hz 2 T

8.

(D) wavelength VIBGYOR highest lowest frequency Energy  freq.  (D) red

9.

(C)

10.

(C)

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Wave number   

1  2

 11.

hc E1  2 ,  2  E2 1

(B) Frequency =

13.

velocity 3  108  wavelength 5090  103

(B) E  nhν E n hν =

14.



1000  10 500

(B) E

12.

1 500  109

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

(C) 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.

(A) 50  E emitted out 100 hc 50 hc  n1   n2   absorbed 100  emitted 5 n2 50  50 5000   0.55   emitted   9 n1 100  absorbed 100 4500

E absorbed 

16.

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

17.



2x 4

x 3  2x   x 2 2

(A) TE 

PE , so first excited state 2

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18.

(D)

TE   19.

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

(B)

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

20.

(C)

r r3rd

21.

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

r4 th 

0.529  16  4.232 2

(D)

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 22.

(B) v  2.18  106 v 

23.

Z n

Z n v1 n 2 5   v 2 n1 3

(B)

(D) a0  4 R Z 9R r3   4 r2 

24.

r3 

a0  9 R Z

(B) Ground state of hydrogen atom = 0.529 Å

0.529  n 2 0.529  (n)2 r   0.529 z 4 25.

 n2

(D)

V

2.18  10 6 Z 1 , v  z, v  n n

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26.

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

27.

(C) E

nhC  nhc   n

10 = nhcx 28.

(C)

E 29.

10 hcx

13.6 Z2 n2

=

13.6  1  3. 4 4

(D) nh 2 0.529n 2 r mvr  r Z Angular momentum  r mvr 

30.

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

31.

 He  x

4 x 8

27 T 2

(A)

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

32.

TE Li2 

13.6  9 x 1

(B)

 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



24 n2 1  4 22 25 n2



6n 22  25n 22  25

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33.



19n 22  25



(b)

n 22 

25 1 15

(D)

f  =

KZe 2 r2 KZe 2

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

34.



2



Z3 n4

f Li2 

27 f 16

27 f  16 x

(C)

V2 r (2.18  106 ) 2 Z2 n2 = 0.529n 2 Z 8 64 a1,He  a 2,Be3  1 16 1 a 2,Be3    2 a

35.

Follow the expression n 2  0.529 r Z (d) 

36.

Follow the expression  13.6 Z 2 E n2 (a) 

37.

See theory

38.

2n2 + 3n1 = 18 2n2 – 3n1 = 6 Solve this and we get n1  2, n 2  6 So,

39.



z3 n4

 6  2  6  2  1  10 2

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

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1 1 1  R H  22  2  2   3  1 8 = R H  4  9  1 1 1   R H Z 2  2  2   n2   n1 

40.

 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   2cR H Z2 4  n n3 1  1   32  R  2  0   R  min 3  1   min  R 1 1 1  R H  (2) 2  2  2   max 2  1 1 1 1 1   R H  4     max   max 3R H 1 4  

41.

42.

43.

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

44.

45.

46.

 R  2 n   R  1  E = E1 + E2 hc hc hc    1  2   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 ( n  1)  15 2 n=6  1 1 1   R H Z 2  2  2   n2   n1

 n4

1 1 1  109677 2  2   6  1 = 937.3 Å

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47.

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

48.

1/2 = 1/1 + 1/3 2

3

1

49.

50.

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 h x  mv  4 h x  p  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 

52.

53.

5 m/s 18

h = 6. 6 × 10–34 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

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

54.

55. 56. 57.

58.

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 x   4mv 4  200  0.1  10 1000 100 Follow theory (Follow theory) ν  3.5  10Hz ν 0  1.5  1015 Hz h = 6.6 × 10–34 KE = hν  hν0 KE = 6.6 × 10–34(3.5 × 1015 – 1.5 × 1015)  1.32  1018 J KE = hν  hν0

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

59.

v

2hc  1 1     m   0 

v

2hc   0      m   0 

h mv A v  B B vA When  B  2 A , then VA = 2VB 1 KE  mv 2 2 TA VA2  2 TB VB TA 4  TB 1 Also TA – TB = 1.50  TB = 0.50 TA = TB + 1.5 = 0.50 + 1.50 =2 Also, 4.25 = WA + TA 4.20 = WB + TB WA = 4.25 – 2 = 2.25 

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60.

WB = 4.20 – 0.50 = 3.70 KA  EA  2 KB  EB  4 h h A  , B  2m K A 2m K B

h

61. 62. 63. 64. 65. 66.

2m K B 2m K A 1 4  KB KA E A  2  4E B  16 E A  2  4E A  2  16 3E A  12  E A  4 E B  4.5  See theory h h Orbital angular momentum = l (l  1)  6 2 2 h  mV See theory See theory h l(l  1) 2

67 to 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88.

h

2

See theory

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 angular part cannot be 0 so no angular node , Hence s orbital. Two radial node means 3s see theory n=5 See the graphs Follow n – l – l increasing Z will decrease radius  3s

1 1     9 3  a0 

32

(6  6  2 )e



 2

; where r 

2r.Z 3a 0

The maximum red all distance of node from nucleus will be r  radial node occurs where probability of finding e– = 0   2  0 or   0 

89.

6  6   2  0 or   3  3 

3 ( 3  3) a0 2 Z

2rZ 3 3 3 r a0 3a 0 2 z

Probability of finding e– is zero implies mat  2  0 or   0 a ( 1)  0 ,   1  r  0 2z 2 (  8  12)  0

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(  6) (  2)  0 6a 3a   6,  r  0  0 2Z Z a0 R = 2,  r  Z

90.

26(Iron) follow electronic configuration

91.

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

92.

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

93.

n  n  2

Fe  3d 6 4s 2

n=5

5 5  2 94.

1 5 s   5  2 2

95.

See configuration.

96.

Same as 92

97.

See Theory

98.

  n n  2 2.83  n  n  2 

99.

Same as 98

100.

  n n  2 1.73  n  n  2  N=1

101.

  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 FOUNDATION BUILDER (SUBJECTIVE)

1.

E

nhc 

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6.626  1034  3  108  2  1022  4 39.759  10 600     6630 nm 600 

2.

hc 19.878  1026 E   4.995  107  3.979 1019 J 3.979  1019 J  1photon 103 J  0.251 10 22 photons 0.25  1022 6.022 10 23 0.0416  101 4.16  103

3.

4.

moles dissociated moles of photons abscrbed 0.01 2 moles of photons absorbed 0.005  6.022  1023 5  103  6.022  1023 30.11 1020 QE 

hc 6.626  1034  3  108   450  1010 19  4.417  10 J Also energy used for breaking up of I2 molecules Energy given to I2 molecule 

240 103  3.948  1019 J 23 6.022 10 Energy used in importing KE to two I atoms = [4.417 – 3.984] 3.984   KE   4.417   1019  Iodine atom  2   0.216  1019 J 



5. (a)

(b)

1  1 1 1 1  R H  2  2   109677     2 3   4 9 0 109677  5   6564 A 36 1 1 1 1 1   R H  2  2   109677     4 5   16 25 

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

(c)

0 109677  9  40523A 400

1 1  1  1 1  R H  2  2   109677      9 10   81 100 

6. Lyman Series 1 1 1   RH  2  2   1 2  1 109678  3   4 0

  1015 A Balmer Series 1  1 1  109678  2  2   2 3  1 1  109678     4 9 109678  5  36 0

  6564 A Paschen Series 1 1 1   109078  2  2   3 4  0 109678  7   18756 A 144 7.

8.

9.

 1 1 1   RH  2  2    n1 n 2  1 1 1  109678  32  2  2   1 3  1 1   109678  9    1 9  0 109678  9  8   113.9 A or 11.39 nm 9  1 1 1   RH  2  2    n1 n 2   1 1 1   1.09677  107  2  2  2170  n1 7  Solve n1  4  1 1 1   RH  2  2    n1 n 2 

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1 1 1   109678  2  2   1 2  109678  3  4 0

  1215 A 10.

6.626  1034  3  108 Energy given to H atom = 1028  1010  1.933  10 18 J  12.07 eV  Energy of the H atom offer excitation = – 13.6 + 12.07 E  E n  21 n  13.6 n2  9 n=3 1.53 Thus electron in H atom is excited to 3rd orbit hc I induced 1  E 3  E1 E1  13.6eV , E3  1.53eV 0 6.026  10 34  3 108 10  1028  10 m  1028 A  1.53  13.6  1.602 1019 hc  II Induced  2  E 2  E1 13.6 E2  eV E1  13.6eV 4 0 6.626 1034  3 108 2   1216  1010 m  1216 A  13.6  19  4  13.6  1.602  10   hc  III Induced  3  E3  E 2 13.6 13.6 eV , E3  eV E1  13.6eV , E 2  4 9 6.626 1034  3 108 3   13.6 13.6  19  9  4   1.662  10  

1 

0

 6568  1010 m  6568 A 11.

For visible line spectrum, i.e Balmer series n1  2 Also for minimum energy transition. n 2  3

 1 1 1   RH  2  2    n1 n 2  7  1 1  1.110  5  1.1107     36 4 9 7   6.55 10 meter

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hc 6.626  1034  3 108   3.03  1019 Joules 7  6.55  10 Energy released = E  N A E

 3.03 1019  6.022 1023  18.25  104 J = 182.5 kJ 12.

1 1 1   RH  2  2   1 6  109677  35  36 0

 937.8 A 13.

Threshold wavelength   0  230nm  230  10 9 m   1 E   13.6 1    12.09 eV  9

K E max  12.09 1.6 10

19

6.626  1034  3  230 109

 1.07  1018 J 14.

U  1.096 107 1  n 2  where n = 2

Maximum wavelength means. Minimum Energy (minimum tranish)



U  1.096 107 1  22  0

Maximum wavelength = 1215 A or 1.216  107 meters Minimum wavelength means maximum energy (max to)



U  1.096 107 1  2 

0.912  107 meter

Series will be ultraviolet region 15.

for He , For H

1 1 1  R H Z2  2  2   4  2  1 1 1   RH  2  2    n1 n 2 

Since  is same 1  1 1  1 z2  2  2    2  2  4   n1 n 2  2 1  1 1   1 12  22    n 2  n 2     1 1   n1  1 and n2  2

16.

shortest wavelength [largest energy] max transition 1 1  1  U  RH  2  2   2  

= 109677/4

 27419 cm1

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17.

18.

2.18  106 Z n 1 2.18 106  1 V V  1.09 106 m sec n 2 2 r T V V

i)

E

13.6 Z2 2

13.6  1  1.51eV 9

n 0 0.529 n 2 r  0.529  9  4.761A z 1 1 1  RH  2  2   1 3  1 8  109677  1   9

ii) iii)

0

 1025 A

19.



iv)

v

r

0.529 n 2 Z

or 1.032 107 m

C 3 108  2.90 15  1.032  107

0

rI  0.529 A 0

rII  0.529  4  2.116 A 0

rIII  0.529  9  4.761A 0 0.529 1 rI He   0.2645A 2 0 0.529  4 rII He   1.058 A 2 0

rIII He  0.529  9  2.38 A 20.

21.

22.

313.6  Z2 313  1 E   19.6 Kcal n2 16

13.6 _____________ 4 16 13.6 _____________ 3 9 13.6 _____________ 2 4 – 13. 6 _____________ 1 E 3  E1  1.51   13.6 

= 12.09 eV

2.17  1012 erg n2 21.7  1012 E2    5.425  1012 erg 4 En 

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For removal of electron E 2 

hc E 2 should be given to remove electron i.e. + ve 

0 6.626 10 27  3 1010 8   3663.6  10 cm  3663.6 A 5.425 10 12 0

So the longest wavelength = 3663  6 A

E1 for H  Z2 13.6  9   30.6eV n2 4 E1 for H  Z2 13.6  16 3 E1 for Be    54.4 eV n2 4

23.

E1 for Li 2 

24.

Energy of one photon =

hc 

6.626  1034  3  108 J 4500  1010  4.42  1019 J 

8 J J    watt   sec 100 sec   8  150  100

Energy emitted by bulb  150 



n  4.42 1019

n  27.2 1018 25.

E3 for H = 2.41 10 12 erg E 2 for H = 5.42  1012 erg  for a jump from III to II shell hc E  E3  E 2   hc 6.626  1027  3  1010   E3  E 2 2.41 10 12  5.42  1012 0

 6602.9  10 8 cm  6603A

2.18  106  Z 2.18  106  n 3 6 = 0.726 10

26.

V

27.

v

28.

E1 for H = – 13 .6 eV 13.6 13.6   3.4eV E 2 for H =  22 4 E1  E 2  3.4   13.6   10.2 eV Difference in two level = 10.2 eV Also for transition of H like atom

V 2.18 106  2r 2  2  3.14  0.529  4  1010  0.0819 1016 108  0.0819 108

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  3  108 m 1  R H  Z2 _______  1 1 1   R H Z2  2  2   1 2  R H  109677 cm 1  109677  10 2 m 1 1 3  109677  102  2 2   8 3  10 4 2  Z 4 Z = 2 He



29.

The no of waves made by a bohr is equal to orbit no.

30.

E  hv 3.97 1019  hv 3.97 1019 v= 6.626  1034

31.

32.

v  0.599 1015

hc  6.626 1034  3  108  19 copper surface   w  4.5eV  0  4.5 1.6 10    0  19.878  1026 7.2 1019   2.76  107 m 0  Similarly for sodium and cerium surface. Energy of photon = work function + KE Energy of photon = work function + eV0 e = electronic charge V0 = slopping potential eV0 = energy required to stop the ejection to electron

hc 6.626  1034  3 108   7.834  10  253.7  109 7.834 10 19 eV  4.89eV  1.602  1019 Work function = 4.65 eV E photon 

33.

Binding energy of electron = 180.69 kJ mol1

180.69  103 J 6.022  1023  30.00 49  1020

Binding energy of one electron =

Binding energy  hv0 30.0049  1020 V0   4.52  1014 sec 1 34 6.626 10 34. 35.

Energy of photon liberated from He during emission of first line of lymon sense.  1 1  1 1  E  13.6 22  2  2   13.6  4  2  2  1 2   h1 h 2 

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13.6  4  3  40.8eV 4 This energy is used in liberating electron from H atom from ground state. Therefore, 1  40.8 eV  E1 of H  KE   mv 2  2  40.8eV  13.6  KE KE = 40.8 – 13.6 = 27. eV  27.2 1.602 1012  43.57 1012 

36.

37.

38.

6.626  1034  3 108 360 1010  5.52  1018 Joules kE = hv – hvo  5.52 1018  7.52 1019  47.68  10 19 Joules 

h mv 6.626  1034   0.6626 10 34 , 6.626 10 35 100 100 1000 1 kE  mv 2  4.55 1025 J 2 4.55 10 25  2 42  9.108 1031 4  103 m sec 1



h 6.629 10 34  mv 9.108  1031 103  7.27 107 meter



12

39.

150     0  V   in A   

40.

 



150 v



150 100



0

 1.227 A

h mv

6.626  1034 3  108 1 9.108 1031  20 11 = 4.899  10 m 

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41.

 x v  v 

h 4 m

h 4m x

6.626 10 34 4  3.14  9.108  1031  1010  5.8  105 m sec 1 

42.

v  3  107  0.02  6 105 6.6 1034 x  4  3.14 1.67  1027  6  105

43.

for an electron 1 2 mv  eV 2 h  mv 1 h2 Thus, m 2 2  eV 2 m 1 h2 V 2 m 2e 1 6.626  10 34  2 2  9.108  1031  1.54  1010   1.602  1019 = 63.3 volt

44.

Due to Hund’s Rule

45.

A d subshell can have maximum 10 electrons

46.

S2  dim agnetic

  magnetic moment  0  1s 2 2s 2 2p 6 3s 2 3p6  Co 3  Paramagnetic

1s 2 2s2 2p6 3s6 4s6 3d 6

  n n  2 4  4  2

= 4.8913 BM 1

47.

1 1 2 r   r 2a0  2s    2   e 2 2  a 0   a 0  AT r  r0 , radial node is formed

 r  For radial node at r  r0 ,  22s  0 this is possible only when  2    0  a0  r r0  2q 0 2 0 a0

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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 

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 9  500 10 9.

(D) 1  9  15200  136800 

10.

(D)

11.

(A) Atomic no.  25

12.

 Mn

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

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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)

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)

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p

6.6 1034  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 2 R R  n  4

32.

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

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36.

37.

38.

39. 40.

e in 1 orbital still  2 Since it has only 2 types of spin (b) nh L 2

(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   4E hc E111  E  3 ' E hc   3 '   '  3 GET EQUIPPED TO ADVANCE

1.

 1 1 1   RHz 2  2  2   n2   n1 1 1  1  1069n  4 2  2     3

  2051Å

2.

3.

5.

0.529n 2 3r z T  2  V 2.18  106 z n 3 n T 2 z  13.6 ___________________4 16  13.6 ___________________3 9  13.6 ___________________2 4 13.6 ___________________4 E4 – E2 = 2.55eV h  2 KE.m 

6.626  1034 2  6.8  1.6  10 19  9.106  10 31

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Total is 20.4 out of which 136 goes for ionization. So rest is 6.8 which goes for KE  4.7 Å 6.

x  P 

P  = = 7.

8.

9.

10.

11.

12.

13. 14. 15 16.

h 6.626  10 34  4x 4  3.14  1  10 9 0.527 × 10–25 5.2  1026

 1 1 1  1  R H Z 2  2  2  ,  109677   n2    n1

1  1 4 2    2

  911Å h 6.628  10 34   mc 3  108 m By looking at wavelength increasing for can say it belongs to visible range 1242 Ep   2.55 eV  4th orbit to 2nd orbit 486.4 0.529  4 r Z 0.529  4 =  0.705Å 3 WF = Ep – Kmax (6.626  10 34 ) 2 = 4  10  20  2  9.11  10 31  (59  10 10 ) = 3.313 × 10–20J hc hc hc    8208 22800   12825Å 150 …(1) Å 100 150 …(2) 2  Å 81 150 …(3) 3  Å 49 From (1), (2) and (3)  3   2 20  1 03 No. of node = 1 90 = 0 is electron is in nucleus e–r/90 = 0 1 

(a) P.E. = - 2 K.E. 

17.

h 4

P.E.  mv2

(b) 

h 1 , K.E.  mv 2 mv 2

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

h 2m K.E. 6.6 1034



2  9.110

31

 4.55  10

25

 7.27 107 m

18 19.

(c) No. of orbitals  2 l  1

20.

(a) E1H

E 2Be 21.

22.

23.

24.

33.



11 2

 4   2

2



4  1: 4 16

1 1  3R H 1  RH 2  2    2  4 1 1 1  R 43 1  R H  4 2  2   H  RH3  2  4 1 1 1  R  16  3 1  R H  16 2  2  H  R H  12  2  4 1 (D) (Be3+) 1  1 U  R H  2  2   152000 3  2 1  R 95  1 U  R H  9 2  2   H 3  364 2 = 137096.25 KE1  hν1  hν 0

KE2  hν 2  hν 0 1 ν1  ν 0  k ν 2  ν0 kν  ν 2 ν0  1 k 1 th –(n ) + (n + 1)th = (n – 1)th (n + 1)2 – (n)2 = (n – 1)2 2n + 1 = n2 – 2n + 1 N2 – 4n = 0  n = 4 (a) U  2  Total Energy

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