SUMANDEEP CLASSES

TOPIC NAME:- Atomic Physics

Subject : - Physics Date :
Time:- 60 Min (Solution) Marks :- 180
1. (a) 1 1 7
But    n1  3 and n2 = 4
Sol. For n=1, maximum number of states  2n2  2 and for n 32 4 2 144
= 2, 3, 4, maximum number of states would be 8, 18, 32 (Paschen series)
respectively, Hence number of possible elements 16. (b)
= 2 + 8 + 18 + 32 = 60. Sol. Potential energy of electron in n th orbit of radius r in H-
 n 2 h2 e2
2. (d)Sol. Bohr radius r  0 2 ;  r  n2 . atom U   (in CGS)
Zme r
3. (a)Sol. 1 e2
∵ K.E.  | P.E. |  K 
2 2r
17. (b)
Sol. Let the energy in A, B and C state be EA. EB and EC, then
from the figure
E12  3.4  (13.6)  10.2 eV .
1  1 1 
4. (d) Sol.  RZ 2  2  2 
  1 2 
For di-ionised lithium the value of Z is maximum.
5. (c) Sol. Lyman series lies in the UV region. hc hc hc
(EC  EB )  (EB  EA )  (EC  EA ) or  
6. (b) Sol. The size of the atom is of the order of 1Å = 10 –10m. 1 2 3
7. (b) Sol. Balmer series lies in the visible region. 12
8. (c) Sol. Transition A (n =  to 1) :  3  .
1  2
Series limit of Lyman series
18. (c)
Transition B (n = 5 to n = 2) :
Sol. According to Bohr’s second postulate.
Third spectral line of Balmer series
19. (c)
Transition C (n = 5 to n = 3) :
1  1 1  3R 16 16
Second spectral line of Paschen series Sol.  R 2  2       105 cm
9. (a)  2 4  16 3R 3
Sol. D is excitation of electron from 2nd orbit c 3  1010 9
Frequency n     1015 Hz
corresponding to absorption line in Balmer series and E is the  16 5 16
energy released to bring the electron from  to ground state  10
3
i.e. ionisation potential. 20. (d)
10. (b) Sol. Energy required to remove electron in the n = 2 state
Sol. Paschen series lies in the infrared region. 13.6
11. (b)   3.4 eV
(2)2
Sol. Linear momentum
21. (d)
 mv  9.1  10 31  2.2  10 6
Sol. (Eion)Na  Z 2(Eion )H  (11)213.6 eV
 2.0  1024 kg  m / s
22. (d)
12. (c)
hc hc
Sol. r  n2  rn  n2a0 ( r1  a0 ) Sol. 2E  E  E
 
13. (c) 4E hc E hc '
Sol. For the ionization of second He electron. He  will act as E     3   '  3 .
3 ' 3 ' 
hydrogen like atom.
23. (b)
Hence ionization potential
Sol. Because atom is hollow and whole mass of atom is
 Z 2  13.6 volt  (2)2  13.6  54.4 V concentrated in a small centre called nucleus.
14. (c) 24. (d)
Sol. Energy required  n 2 h2 n2
13.6 13.6 Sol. r  0 2 ;  r 
   0.136 eV Zme Z
n2 102 25. (b)
15. (c) r(n 2) 4 9
1 1 1 1 1 1
Sol. r  n2    r(n 3)  R  2.25 R
Sol.  R  2  2   2  2  r(n 3) 9 4
  n1 n2  n1 n2 R 26. (a)
1 7 Sol. In the revolution of electron, coulomb force provides the
 7 10
 0.0486  .
1.097  10  18752  10 144 necessary centripetal force

Address:- 8, Kalabhavan, near Govt Engg College, Vidyut Colony, Ramanand Nagar, Jalgaon, Maharashtra
425001 Sumandeep Classes, Jalgaon, Contact: 09730373737
4
 ze 2 mv 2 ze 2
 2
  mv 2 
r r r
1 ze 2
 K.E.  mv 2  .
2 2r

39. (c)
1  1 1 
Sol. Wave number   R 2  2 
  
 n1 n2 
27. (d) For first Balmer line n1= 2, n2=3
Sol.According to Bohr’s theory  1 1   9  4  5R
 Wave number  R 2
 2   R  .
h 2 3   9  4  36
mvr  n
2 40. (a)
 h  Sol. Energy required to ionise helium atom
 Circumference 2r  n    n . = 24.6 eV
 mv 
41. (b)
28. (c)
Sol. From diagram
kZe 2 kZe 2 K.E. 1
Sol. K.E  and P.E.   ;   .
2r r P.E. 2
29. (d)
Sol.Lyman series lies in the UV region.
30. (d)
Sol.If E is the energy radiated in transition
then ERG  EQS  ERS  EQR  EP Q E1  13.6  (3.4)  10.2 eV
For getting blue line energy radiated should be maximum E2  13.6  (1.51)  12.09 eV
 1 E3  1.51  (0.85)  0.66 eV
 E   . Hence (d) is the correct option.
  E4  3.4  (1.51)  1.89 eV
31. (b) E3 is least i.e. frequency is lowest.
Sol.Energy released 42. (a)
 1 1 
 13.6  2  2   2.55 eV . ke 2 e2 1 e2
 (2) (4)  Sol. P.E.    ; K.E.   (P.E.) 
r 4 0r 2 8 0r
32. (c) nh h
Sol.The absorption lines are obtained when the electron jumps 43. (c) Sol. mvr  , for n =1 it is .
2 2
from ground state (n = 1) to the higher energy states. Thus
only 1, 2 and 3 lines will be obtained. 44. (d) Sol. Minimum energy required to excite from ground state
33. (c) 1 1
 13.6  2  2   10.2 eV
Sol.Wave number  1 2 
1 1 1   1 1  3R 45. (d)
 R 2  2   R   . 2
  n1 n2   4 16  16 2 2k 2e 4 m  1  2 2me 4
Sol. R    
ch3  4  ch3
34. (a)  0 
Sol. K.E. = – (T.E.) 46. (d)
35. (d) Sol. -particles cannot be attracted by the nucleus.
13.6 47. (c)
Sol.Required energy 3 E  2
 1 .51eV
3 1 1
36. (d) Sol. By using   RC  2
 2
Sol.As n increases P.E. also increases.  n1 n2 
37. (a) 1 1
   107  (3  108 )  2  2  = 6.75  1013 Hz.
Sol.When an electron jumps from the orbit of lower energy 4 5 
(n=1) to the orbit of higher energy (n=3), energy is absorbed. 48. (c) Sol. For M shell (n = 3), orbital quantum number l = 0, 1, 2.
38. (a) n(n  1)
Sol.  E1  E2 49. (d) Sol. Number of possible emission lines 
2
 1   2 Where n = 4; Number 
4(4  1)
 6.
i.e. photons of higher frequency will be emitted if transition 2
takes place from n = 2 to 1. 50. (a) Sol. Diameter of nucleus is of the order of 10–14m and radius
of first Bohr orbit of hydrogen atom
r  0.53  10 10 m.

Address: Purohit’s Academy, MJ College Road, PrabhatChowk Opp. Vodaphon Gallery, Mob : 9823118910, 9168687434