I. (a-r), (b-q), (c-p), (d-s) 2. (a-<j), (IK), (e-p, q. r), (d-p, q)Linked Comprehension Type Questions:

1. (b) 2. (e) 3. (b)Integer Answer Type Questions:

I. (9) 2. (4)

SECTION-IStraight Objective Type (Single Correct Answer)

1. (a) 2. (e) 3. (b) 4. (d) 5. (a)6. (d) 7. Cd) 8. (d) 9. (d) 10. (d)II. (a) 12. (b) 13. (e) 14. (c) 15. (a)16. (d) 17. (a) 18. (a) 19. (a) 20. (a)21. Cd) 22. (d) 23. (c) 24. (c) 25. (e)26. Ca) 27. (c) 28. (d) 29. Ca) 30. (b)31- (a) 32. (c) 33. (a) 34. Cd) 35. (d)36. (h) 37. (d) 38. (d) 39. (a) 40. (a)41. (d) 42. (a) 43. (c) 44. (e) 45. (a)46. (b) 47. (d) 48. (a) 49. (e) 50. (c)51. (c) 52. (a) 53. (e) 54. (h) 55. (a)56. Cb) 57. (b) 58. (c) 59. (b) 60. (c)61. Ca) 62. (d) 63. (a) 64. (c) 65. (c)66. (b) 67. (a) 68. (c) 69. (b) 70. (a)71. (b) 72. (a) 73. (c) 74. (d) 75. (c)76. (b) 77. (e) 78. Ca) 79. (a) BO. (d)81. (b) 82. (a) 83. Cc) 84. Cd) 85. (a)86. (d) 87. (c) 88. (b) 89. (d) 90. (d)91. (a) 92. (d) 93. (a) 94. (b)

4. (a)9. (b)14. (d)19. (a)24. (b)29. (a)34. (e)

5. 210. 5

5, (d)10. (d)15, (a)

2. )7. 412. 2

1. 56. 311. 7

SECTION-VIILinked Comprehension Type

SECTION-VMatrix Match Type

2 Ca-p q) (Irq, r), (c-r), (d-q, s)1. (a-q, r), (b-p), (c-s), (~~: 4: (a-r), (b-r), (c-s), (d-<j)3.. (a-<j), (b-p. q), (c-p, s), () 6. (a-s), (b-p), (c-q), (d-r)5. (a-<j), (b-p), (c-s), (d-r) B. (a-r). (b-p), (c-q, s), Cd-r)7, (a-q), (b-s), (c-p), (d:;r\d_r) 10. (a-q. r, s). (b-r), (c-p), (d-q, s)9, (a-q, s), (b-p), (e-q,' , s 12, (a-<j), (b-5), (c-T), (d-p)11, (a-q, 5), Cb-p, 5), (e-r, 5), (d-p,) 14. (a-p, q), (b-q, r), (c-r), (d-q, s)13. (a-q), (b-p), (o--r), (d-5)

SECTION-VIQuestions with Single Digit Answer

3. 0 4. 3B. 4 9. 5

Paragraph-l3. (b) 4. (e)

1. (c) 2. (d)

Paragraph.23. (c) 4, (b)1. (b) 2. (a)

Paragraph-33. (d) 4. (d)

I. (b) 2. (b)

Paragraph-4(d) 4. (b) 5. (b)

1. (a) 2. (b) 3.

Paragraph-54. (d) 5. (d)

2. (c) 3. (d)1. (a)Paragraph-6

1. (d) 2. (c)

1, (a) 2, (a)'6, (c) 7. (a)11, (a) 12, (a)16, (a) 17, (a) IV

SECTION-. I d Olagrammatlc AptitudeObjective Questions on Graphlca an ( ) 5, (d)

3 (e) 4, a d)1, (c) 2, (d) 8' a 9, (b) 10. (a,6, (c) 7, Ca) 13' ~a~ 14. (d) IS. (a)II. (e) 12. (d) , b) 19. (c) 20. (a)16, (a) 17, (a) IB, (a,

SECTION-IIM e than One Correct Answer)Multiple Correct Answer Type ( or ) 5. (a.b)

3 (c d) 4. (b,c d)1. (b, c) 2. (a. d) B' (a' c d) 9. (b. c) 10. (b.6. (a. c) 7. (a, d) ., c' d 14. (a, b) 15. (a, c)II. (a, d) 12. (a. c) 13. (b, 'i) 19. (a, b, c. d) 20. (a. b, c, d)

d) 17 (a b c) 18. (b,c, )16. (b, c, '. d)' 23 (a b d) 24. (a, b, c21. (a. b, c) 22. (a, . "

SECTION-IIIAssertion-Reason Type

3. (c) 4. (c)8. (b) 9, (a)13, (d) 14, (a)

•

•

5. (a)10. (b) ,15. (a)20. (c)25. (c)30. (b)35. (d)

5. (a)

PANORAMA (liT ROUND-UP)Multiple Choice Questions with One Correct Answer:

1. (d) 2. (c) 3. (d)6. (d) 7. (a) 8. (d)II. (b) 12. Cb) 13. (c)16. (e) 17. (b) 18. (c)21. (c) 22. (d) 23. (e)26. (a) 27. (a) 28. (a)31. (c) 32. (d) 33. (d)36. (a) 37. (e)

--- _..--Answea'sheet

Multiple Choice Questions with One or More than One Corl"eerAnswer:I. (b, d) 2. (a, c) 3. (a. c) 4. (d)6. (a, b, c,) 7. (a. d)

Matrix .l\fatching Problems:

ANSWERS & SOLUTIONS

Panorama (liT Round-up of Thirty Three Years)

Angular momenrum,

For 2s-orbital,l = 0 h

Orbital angular momentum = Ji (l + I) 2n = 0

For d-e!ectrons, I = 2 h h r; h~(I 1)_=.J2x3-=oJ6-Orbital angular momentum = Vi \l + " 2n 2l't 2n

40Ca [Even number of protons, even number ofneutronsl· ]30AI (Odd number of protons, odd number of neutronstJ f odd mber of protons and neutrons.~gAl is less stable due to presence o. nu of subshel1. When (n + l) values a~eSmaller is the value of (n + i~ lesser IS th~ ene.r~ pal quantum number' n'; smaller Willsame then energy of subshell depends on e poncibe the value of'II' lesser is the eneT~.~fs~b;h~l.: 2 < (i) n = 4 1= 1(') =3 1=1«;;)n=4.1=0«III)n- • - •lvn, I' ' planeorbital has only one noda pane, I.e., yz-. .

PX2 22 6 3 '3 63d~ 4s1 is ground state electronic configuratlOn ofCr24'1s,2sp;sP,h 6.626 x 10-34

1.=--mu 0.2 x 5/3600 d' th R therford's gold leafex.periment.

33. (d) Beamofa~partic1es(nuc1eiofheliu~)areuse I,n tre ~ with opposite spin and no two. mmodate max.lmum two e ec on

34. (c) An orbital can aceo I es of all four quantum numbers.electrons in the atom can have same va u

n'35. (d) r = - x0.529 AZ

For BeJ+, n = 2, Z = 4

:, r=0.529ANumber of cadi a! nodes = (n-i-I)

36. <a) Number ofcadial nodes for 3$ = (3 -0-1) = 2Number afcadial nodes for 2p = (2-1-1) = 0

37. (e) Nuclear attrnction on electron EO Centrifugal forceKZl- m,}-7=-r-

,}-=wmr

Comparing eqns. (i) and (ii)

KZl- n'h'--;;;;:- = 41t2m2,)

n'[ h' 1 "on'r=z: 4n'mKl- ~zhZ

v =---" 2nmaon

r

24. (b)

r 26. (a)

I, 27. (a)

28.. (a)

29. (a)30, (b)

31. (c)

Electrons revolve around the nucleus.If 1= 2, then m, = -2,- 1,0, + I, + 2, thus, m{ = -3 is impossible.£1 = helAI = I., = 4000 = 2£, hell., AI 2000

I~C, I~N, I~F;these elements have 8 neutrons hence, they are isotone to each other.heM=(£'-£I)=_A

A a:...l-!>£

2s should be filled before 2p because according to Aufbau principle the subshelJ withlower value of(n + I) is filled first.Fluorine is the most electronegative element.

Fg ~ Lr2, 2s22p5

Cr24 ~ 1s2, 2<;22p6, 3s23p6, 3d~, 4s1

Cl17 -+ 1s2, 2s22p6, 3s23p~

Unpaired electron is present in 3p-subshell.Thus. n = 3,[ = 1,m, = I

X-rays do not contain any charged particle hence, they are not deflected in electric andmagnetic. field.3p has one spherical and one non-spherical node.N~!mber of spherical or radial node = n -[- 1

=3-1-1=1Number of angular or non-spherical nodes = I

=1

3. (dJ4. (a)

5. (a)

6, (d)7. (a)

9. (b)

I I. (b)

12, (b)13. (c)

14, (d)

15. (a)

16. (c)

17. (b)

18. (c)

19. (a)

20. (c)

23. (c)

21. (c)

lJfll!tipie Choice Questions with One Correct Answer:

1. (d) In the zinc atom d8Zn); number of neutrons will be (70 _ 30 = 40).

2. (e) Scattering of a-particles takes place due to coulombic repulsion between positivelycharged a-particles and the nucleus.

Any orbital can accommodate two electrons with opposite spins.Size of orbitals increases with incre.1se in the value ofprincipa! quantum number.

1s<2s<3s<4s I " f rh' I-------->1 ncrea."lngSIze0 0 Itas.et m ratio lies in the sequence n < a < p < e.Rb)7 -+ Ii, 2r22p6, 3s23p63dlO, 4s24p6, 5s1

For outermost electron (5sl). n = 5,1 = O.m = O. s = +1/2

8. (d) The electron in ground state (lsi) can absorb a photon to be excited hut cannot emit aphoton because the ground state is the lowest energy state of the atom.Bohr model is valid for single electron species.X - ray < Ultraviolet < Infrared < Radio waves

Increasing wavelength

h'= 321t1maJ

.Jlultiple Choice Questions with One or Jlore than 0 C A71 78 ne orrect nSK'er:

I. (b,d) 33As, 34Se have same' number of ncutron!'i as th t' 76G2 . am32e.. (a,c) Atomic mass is calculated by weight mean method.

Average atomic mass =L% abundance of isotope .100 )(IsotopIC mass

4. (d) Number of neutrons + Number of protons'" 3Z'

5. (a) E = -2" x 13.6 eVn

Energy of second shell of hydrogenI'

E = -"4 x 13.6 cV= -3.4 cV

6. (a,h,c) Cr24 -. [At J 3dS 4sl. It has extra stability of half filled d.subsh 11 .number'm'- (/ 0 I) h c magnetic quantum

- - - - + ,t us negative value i••possibleA~7 has 23 electrons and 24 electrons have clockwise and ;nticlock' .

7. (a,d) N7 -+ Is2 152 2p3 "'1se SplD.

-> rn rn ITIIITJ} Both these fillings

-> rn rn ~ are cnrrect

2. (c)

Z2 = n2

Z=n=3 for Li2+

Thus, S2 will be 3pbecause it has one radial node.For 3p.n = 3,1 = 1Number of radial node = n-I-I = 3 -1-1 = 1

Integer Answer TJpe Q"e.'~tions :t. (9) Total number of electrons with (n = 3) = 2n2 = 18

Out of these 18 electrons, 9 electrons will have anticlockwise spin (ms = -~).2. (4) Photoelectrons are ejected only when the energy of absorbed quantum is greater than the

threshold energy or work function (~) of metal.

Ab hdE Ehe 6.626xl0-34)(3x108

sor energy, =-= 9). 300 x te-

=6.626x10-19J 6.626 x 10-19

eV=4.14eVl6x10-19

Thus. four metals i.e., Li, Na, K, Mg will show photoelectron emission.

SECTION-I: Straight Objective Type (Single Correct Answer)1. (a) For m{ = + 2 (maximwn), 1 = 2 and n = 3

Number of waves in a shell = n = 3E -313.6

2. (c) En = -t = -- = -78.4 kcaVmoln 4

3, (h) ..t.=R(..!..._..!...J').. nt n~For Balmer series. nl = 2

3. (h) Es, = Ell

Z'-2" x 13.6 eV = -13.6 cVn

(0-<1); (1)-.); (c-p, q, ,); (d-p, q)

.•• Orbital angular momentum = .j'i(i+I) l!-2.i.e, it depends on azimuthal quantum number.

.•• Shape, size and orientation of hydrogen like atomic orbitals is detennined byprincipal. azimuthal and magnetic quantum numbers .

.•• Probability density of electron at the nucleus is detennined by the principal andazimuthal quantum number.

Li"k~d Compreh~"sionTypeQuestions:1. (b) 2s orbital is spherically symmetrical and it has one radial node., Number of radial node of2sorbital = n-I -I = 2- o-} = 1

Z, 3' 9Es = _~ x13.6 eV= --, x13.6 eV= -- x13.6 eV

I n 2 . 4),

EH =--, x13.6eV1 '.

Es, = 2. = 2.25EH 4

2,

(for/=O)

KZe'2r

K n = Kinetic energy

In lowest orbit (Is), the angular momentum = ~ 1 (/ + I) ..!!... = 02n

n'r•• = - x0.529 A

Z

..!.. oczY I- i.e, oc Z i.e, y= Ir,

•••-

(0-') (I>-q) (c-p) (d-s)

Ell (Total energy) = -KZe'- .2, 'Z'

=-2x2.l8xlO-18Jn

VII = Potential energy = _ KZt?r,

r•• = Radius of orbit = !!..... x 0 529 AZ .•.• V,-=-2

K,r•• oc £11-1

Answer ofJUatrix ftfatching Problems:

I.

_I

...(2)

...(1)

.fiI

ITIIIJJn 3 3 3I I I 1ml -) + I 0

:En+l+mt:25r= /ri("I)1f'; 'b = IO-I'm.!=R["!"_"!"]=R[~-~] = 3:A. nrnj 12

3-+2, 4-t2, 5-+)1st line lndline Jrdlme

he = he +.!.mv2~ ~o 2

v= 2hC[~0-~]m ),,).0

c.. K=-~l.o

'6U [ 1]' ,_ 6U33. (a) P.E.= J ---;:> dr = 6Ze' -; ~' r

35. (d) .!=R~[L~]=R['.'-,4]A 22 n 4n

4 .'"'=-x-,-R n -4

.'Given: A.= K )(-,-• -4

32. (e)

31. (a)

30. (b)

27. (c)

28. (d)29. (a)

r...(1)

... (2)

I1xl1v~_h_4"",

6XX[lOxo.l] ~ 6.626)(}0-34100. 4 x3.14 x0..2

Orbital angular momentum = "1(/ + I) l!.. =.J6 xl!..2rr 2ft

Magnetic moment = In{n+ 2) B.M.

E. -E, = -0.85 -(-1.5) = + 0.65 eVThis energy is compatible with the infrared region of the specrrum.(E,-EI) > (E.-E,)

E=EL. ,•

.' =!l= 13.6=4Ell 3.4

n=2

Angular momentum (nwr) = nh =!!. = 2.11 )(10-34 Jsec.2. •

.hAngular momentum •• _2..'r=- x0.529AZ

r rx;.r;.:. Angular momc:ntwn a::..[;.

.!=R[L..!..J =R[1._..!..]1 "J2 n? 12 n2

.=[~]I/2~-I

For first excited state, n = 2

A, _ "",' _!'l. _ 16A;- "'" - .t-Tlr=; x0.529A)

T' a: r'T, (r)'/2 IT4r = 4r = 8'

4. (d)

6. (d)

9. (d)

10.. (d)

13. (e)

15. (a)

16. (d)

17. (a)

18. (a)

19. (a)

20.. (a>

21. (d)

---------------,...---::-h ------For fourth line •• , = 4 + 2 = 6 22. (d) ~ = .hEmFor fifth line,., = 5 + 2 = 7 ~ h/.,fif;,Ionization energy ofli

2+ = Z2 x/H = 9)(13.6 eV ),,2: = hi J2E(2m)

:::=1" 23. (c) ~=:;:kT.=2- li Balmer ~= (f1=~1200 =2

(visible) ),,2 vT; 300n= 1- Lyman

(UV) AI = 2'-.2 = 2)" 'th th edium25. (c) Speed of electromagnetic radiations changes W1 em.

h26. (a) ~ - .hEm

h'E=-2m-~-'

_ h _ 6.626x 10.-" = 9.99 x I0.-28= I0.-27 kg mise<P-I- 6630.x10.-IO . . IIn none of the given pairs, probability of finding the electron IS zero In xy pane.CI17 (second excited state)

= 3p'3d'

IJJJTITI, 3, 2+2 +1

36. (b)

37. (d)

38. (d)

I 2 3 4

ITIIIIIJIJ

Comparing Eqos. (I) and (2), we have

K=iR

Change in P.E. = -2 x2:-(_2y) = _2: + 2y= +!. y4 2 4

2p cannot be filled before 2.1'.

A="-ph h v

p=- =- =h-A elv c

Orbital angular momentum = ~l(l+ 1)..!!-2.

m=-/--O-- +/Valucsof'm'= 2/+1

1= m-l2

(n + l)s

(n + 1)(n + 1)

np

(n + 1)

np will be filled before (n + l)s.d:2 has a ring around ii, hence, electron will be along all the three axes.

BS ---io 1$2,2s22plElectron in p-subshell will revolve in elliptical path.After emission of visible radiation, the electron will be in 2nd shell, hence other possible

transition will be :n2 = 2 --+ I!J = 1 (Lyman series)

Distance of Closest approach

K 'IJ;' = (KE.).

_1_2ex2e = (KE.).4ltf'-o r

_1_ 2x 29.' = (KE.).41t£o r

I 29.'r=--X---

2."0 (KE.).

. ~ h ~( 1) h J3 hSpmangularmomentum::::vs(s+l)-:::: - 1+- _::::_x_21t 2 22n 221t

hmvr= n- n= 1,2,3.....2.1.7h 1 .. 'bl:. ~ va ue IS Impossl e.

At nodal surface \j1 :::: 0.. 2-2r=0

r=~=~=IA2 2

A hE"= ~; = Kinetic energyv2Em

for same de Broglie wavelength the K.E. of particles will lie in following sequence:E1 (electron) > E3(proton) > E2(a - particle)

Number of radial or spherical node:::: n -'1- 1::::1- 0 -1 ::::0Absorbed energy:::: 13.6 eVxt5:::: 2Q4 eVKinetic energy of electron:::: 2004 -13.6:::: 6.8 eV

h2nr=n-

mu2w = nA

hAngular momentum (mur) = n x-2.n and 'It are dimensionless hence, dimensions of angular momennun and Planck's constant

are same.When (n + I) values are same then orbital with lower value of' n' is filled first.

62. (d)63. (a)

60. (e)

59. (b)

58. (e)

57. (b)

51. (e)

49. (e)

'.' 53. (e)54. (b)

55. (a)

56. (b)

.•

...{I)

...(2)

1-4

(AI = A. given)

Possible exchange of electrons are:

1-2. 1-3.2-3. 2-43-4

Radii nodes =n-I-I = 3-1-) = I

Spin multiplicity = lIs + I = 2(+.!._..!. _.!] +:) = 02 2 2

(E, -E,» (E. -E,) > (E, -E.) > (E, -EdGreater IS the energy lesser is the .•••.avelength. Thus,

f :{~: Sr~[;~l~].!._5R. 36A -"36 .. A= 5R

M:, =(2E-£)=E= heA,

M:,=(~E-E)=f=~:

Fcom Eqns. (I) and (2)he he3l., = A,

A, = 3A, = 3Ahmur=n-2.

39. (a)

40. (a)

41. (d)

42. (a)

43. (e)

44. (e)

45. (a)

47. (d)

48. (a)

74. (d)

73. (e)86. (d) Na --Jo ls2. 2!122p6. 3s1

3s1 orbital has n ""3,1- 0

Number of radial node •••n -1-1=3-0-1=2

n=3. nh 3xh

Orbital angular momentum :;z - "" -21t 21t

'l'" = 0,2_.!l!.= 0 I.e, 'Il = 200

ao

v = E. = eRZ' [.!.. __ 1_]A .' (n+ I)'

• CRZ,[(n+ I)' -n'] =CRZ'[ 2n+1 ]1I2(n+l)2 n2(n+l)2

When n »> I, then (n+ l):IlI nand (2n+ I)", 2n

v "" cRZ1 [2"] == 2cRZ2

.. n4 n3

In multi electron system. second and third electrons will be ejected by increasing thephoton energy. Hence, electron current will increase in step ladder type parent.Path of electron is at closed loop on spherical surface and looks like an ellipse.

lLi2+=~_!!.IHe

+ zi - 4Iu2+ = ~XI9.6XIO-18

"" 4.41 x10-17 J lltom-I

Energy in ground state ofU2+ = -4.41 xlO-17 J atom-I

n'r=-xO.529AZ

n'4.768 = - xO.5291

n2.9

hAu - ~2m.2eVA::.P.. ~.rs~2.fiA.A- h _123 A- :}2eVm-"JV

F;+ has 5 unpaired electrons and eo)+ has 4 unpaired electrons

~F'" =~n(n+2)=.J5x7=-f3S~Co" =.In(n+ 2) =.n;;s =f1SNearer is the electron to the nucleus, greater is its shielding effect.For radial node,

85. (0)

83. (e)

lW. (d)

81. (b)

80. (d)

78. (0)79. (s)

76. (b)

77. (e)

'.'

64. (e) r" = n2 x'I

~ =4~

n'Til = - x l5 xlO-16secZ,T, 8 2=!.1j=' or T

28

65. (e) v, =CR(.2.._...!-) = cR}2 co2

Vz=CR(~- ;2) = 3:Rv3 = CR("'!'--CO) ~ cR22 4 .. VI -v2 = v)

66. (b) v=~=eR[2,--_I_] =eR[(n-I)'-n']", -,A n (n_l)2 n2{n_l)2 n

h67. (s) A= ..J2Em

E = kT (Kinetic energy of material particle)

:. A>:;, h:}2mkT

68. (e) ~ = I!i= /200A, V7\ 300

A,=&2z,; ,;

69.(b) K.E.=-=- (forH-atom,Z-I)2r 2r

70. (a) Number of waves = 2nr = 21tr'A h/mv

2rr 2rc nh=-xmur=-X-=nh h 2.

71. (b) T, = nl x Z? = 2' x 2' _ 32T, Z,' ni I' 3' - 27

72. (0) K.E. = eV1 ,'2mtl = eV

v= J2::jj =i=RZ'(2,-2,) =R x2' (.2.._..!..) = 4R x~ = 32R

11] 112 12 32 9 9h

A.p =----~2mpeV

A: h . 6.626x 10-'4.J2Em - J2x6.8xL6xlO-19x9.lXIO-31

= 4.70 xlO-1om = 4.7 A

8 dE f . 242 x 10'on nergy 0 smgle bond =----J = 4.017)( 10-19 J6.023 x 1023

E=h.E..A

4.017xIO-19 = 6.626)(10-34 x3)(108

AA.= 494 x IO-9m = 494 nm

Shortest wavelength corresponds to series limit

.!. = RH x...1.-= 109678A. nr 1

A.= 9.117 x lO-6cm = 91 1.7 x 1O-IOm = 911.7 A

v =.!. = Rz' [...1.-_...1.-] = 109737 xl' [L~]A 1lJ.2 n1 22 002

= 27434.25 cm-.' (S . I.' .11enes Imlt WI be of highest energy)

Initial Energy of photon'" Energy ofphotan after scattering +K.E. of scattered electron

hVj hV2 + !mv22

v = '/14571428

= 3.8)(103 l::l4 x 103 ms-I

Absorbed Energy = Emitted Energyhe he he-=-+-A A, A,I I I-=-+-A A, A,I I I

-=-+-355 680 A,

.!. = Rz' [L A~]= 743 lUll

A "J.2 ni(Hydrogen) _I_=RXI'['!'_-'-]

6561 4 9

(Helium He+) .!.= R x2' [.!._1-]A 4 16

Dividing eqn. (I) by (2)i. 5/36 5 4 5

--=---=-x-=-6561 4 x3/16 36 3 27

A= 5 x 6561 = 1215 A27

D<B<C<A

Increasing energy

£3=£(+£2

hv] =' hv} + Jrv2'0'3='0'1+'0'2

c c c-=-+-)..3 AI A2

I I I-=-+-A, A, A,

A -~3 - Al + 1..2

L8- 2rt2mz2e4.- + 2 2

n hWhen mass becomes double, the ionization energy will he Jouhlc.

2188)(108 m/v=----c secn

r = tl2 xQ529 x IO-l!cm

218 xlO-18E = - n2

These relations show thai vrand!::.will be proportional 10the principal qL:ant\lffi number 11.E

in option (n)' r cnnnot be equal to' n' because values of'/, lie between O. 1,2,3 •...... (11-1)

. In option (e) nil cannot be -3 for I = 2 becauseml =-1,--0--+/

Scattering is due to coulombie forces and it follows hyperbolic palh.

n' ,r=-xO.529A (:.rocT! )7.

mvr= n(-"-) i,e., int •..~. ..' multiple of..!!....2n 2n

7. (a,d)

8. (a,c,d)

6. (a,c)

94. (b) n / (n+ I)

(A) 4 1 5(8) 4 0 4(C) 3 2 5(D) 3 I 4• Orbital with lower (n + I) value has lesser energy.• When (n+ /) values are same then the orbital with lower value of 'n' has lesser

energy .

4. (b,c)

3. (c,d)

2. (a,d)

SECTION-II: Multiple Correct Answer Type (More than One Correct

Answer). h

I. (b,c) ChangeIn angularmomenlum = (n2 -"1)- 2n

.. hh3h (h )POSSible values will be _, -, -,.... h and 4n are not possible1t 2n 21t

...(1)

...(2)

2)(6.63)(10-34 (l011_0.9xI01I)

9.1 x 10-3 1

I ,'2mv = h (VI -'0'2)

v2 = llre"'l -'0'2)m

90. (d)

89. (d)

88. (b)

91. (a)

92. (d)

93. (a)

10.(b,d) l. =..!!...-mv

For same speed, "'.of < A~ because m. > m2 n B'

11. (a.d) r =~ xo.S29AZ

Here, ;~iusis inversely proportional to atomic number Z.E = -2 x2.18 xlO-18J

nE ccZ2. here -ve signifies the release of ene bnucleus. rgy w en the electron comes nearer to the

13. (b,t,d) Species having single electro . .104.(a.b) 'c' and 'd' options are not comn8Cll:e_~mes~trum, similar to that of hydrogen ato.m.

, , VC\;8use,m cannot be + I for I = 0 and' r cannot be equal to '2' ti - 2Valuesofm=-I-_O __ +l orn-Values of / = O. I. 2•......(n -I)

l$. (a,c) r = n2 x529 pmFor first shell, r = 529 pmFor,secondshell, r = 52.9x4 = 21L6 pm

17. (a,b,e) OphOn (d) is incorrect becau th k' ,different. se emetiC energy and potential energy of electrons are

lL (b,c,d) Option (a) is not possible becaus b'number' n' • rand' , e an or ltal can be defined only by the three m"."tu, m. "I-.m19. (a,b,e,d) Cr24~ tr2, 2.r22i, 3i3p63dS, 4s1

Total spin = Number of unpaired electrons x.2. = 6 x.!. = 32 2

Spin multiplicity =(2I.S + I) = 2 x 5 x..!.+ 1= 62

Magnetic moment = ~n(n + 2), where n = number of unpaired electrons

= J6;;8 =.J48 RM.21. (a,b,e) Th .e given expression is valid for single electron ~ies' ~ ' . 2+deuterium. ~I"~" ,t.e., lor tritlum. Li and

21. (a,d) Til = 2wll

v,

12. (a)

Here. 8=(~r''1. = 2n2

23. (a,b,d) Photon Energy = 13.6 (1- s;)"'l: 13 eV = 13 x 1.6 )(10-19J

Momentum of hydrogen atom"" momentum of photonhv

mV=-ehv 13 x1.6><10-19

v = - =- ---~----mc 1.67 x 10-27 >(3)(108

o::4msec-1

24. (a,b,c) Elastic collision of electron with metal surface is not possible.

SECTION-III: Assertion-Reason TypeWork function of a metal varies with the depth from surface, that is why the Kineticenergy of photoelectrons are not uniform,Absorbed energy (hv) =Work function (Eo) + Kinetic energy of photoelectrons

. ' Circwnference 2w 2ftrNumber of waves m a shell = ----- = - =--Wavelength A. hlmu

= (mur) x 2ft = (nh) )(2ft = nh 2. h

Second statement is false because the energy of orbitals of hydrogen depends only onprincipal quantum number' n'.Proton is not a fundamental particle, elm ratio of anode rays varies with the gas filled in

the discharge tube.d '2 orbital is not spherically symmetrical. Its electron density is non~zero in xy-planeb~cause there is a ring around the orbital. ,.Radial part of wave function depends on' n', Angular part of wave function depends on' rand'm'.Number of photoelectrons emitted per unit time increases with the intensity of radiation,therefore, photoelectric current also increases.Orbital angular momentum = ~l(l + 1).!!.. (for d-orbitals) 1= 2

2n

:.Orbital angular momentum = J6'!!"2.nh

Angular momentum of electron (mvr) =-2.For Balmer series,lowerstsre is n == 2and this series fall in visible region of the spectrum.Line spectrum is found only in case of atomic species,ForI-sub,belll = 3:. Total number of electrons = 2(21+ 1)= 2(2 x 3+ 1)= 141 = 0 for both 2, and 3selectrons, therefore. orbital angular momentum of these electronswilt be same,

Orbital angular momentum = ~/(l+ l)..!!.. = 02.

8. (b)

7. (a)

6. (b)

5. (d)

4. (c)

9. (a)10. (d)11. (a)

3. (e)

2. (a)

I. (a)

13.6 xZ1K.E. = eV 27.2Z'n2 P.E ..•• ---- eVn'

Second quantum will be in Lyman series bee frfirst shell or ground state. ause om 2nd shell the electron will fall to

.!. = R[-.!.._-.!..J = 109677[2. _-.!..]A. nf n! 12 22

A.= 122 x 10-7 em = 122 mn

9. (b.e)

y=mx+e ...(ii)Comparison of equation (i) and iii) shows that the slope i.e., tan e will be equal to hie.

hm (Slope) = Ian 0 = _e

J. (e) EK = hv -Eo where. Eo = Work functiony= mx+ c

Comparison shows that the Planck's constant will be the slope of the line.

2. (d) A = ~ ,there is inverse relation between de Broglie wavelength and momentum, hence thep

graph will be rectangular hyperbola.

3. (e) When v = vo, the kinetic energy of photoelectrons is zero, further the K.E. ofphotoelectrons increases with increase in the frequency of the radiation.

4. (a) EK = hv - ~

14. (d)

DD

9. (b) ~Ption (b) is correct plot because only for-2s~~;bital,~here~illb~single node.Number of node = n-l-1

=2-0-1=1. . tensity of incident radiation while it is10. (b,d) Photoelectric current increases With the In

independent of the frequency. . h . 'n frequency of incidentI .ncreases Wit Increase I

11. (e) Kinetic energy of photoe eettoRs ; ABC will be parallel to each other.radiation. The graphs for three meta s •.•. .There are two nodes in the graph, hence, It IS for 3s-orbltal.12. (d)Number of radial nodes = n -[-I

=3-0-1=2

13. (a) u c(.!. thus the plot will be exponential curve like (1).n

~ =aZ-abab = I, a = ta045°= 1';;;=52-1=51

v = 512 = 260ls-1

. fi origin corresponds to 3s-orbital.15. (a) Three peaks in the curve startmg rom . . t t which radial probabilitv function16. (a) There is only one node in the plot. Node IS pam a •

18. (a.h)e::grro::T+v:.EOC_(~:). KE.<:20. (a) A (3p):Number of radial nodes = ,r-l-1

=3-1-1=1B (3d) :Number of radial nodes = n -1-1

=3-2-1=0C(3s) :Number ofmdial nodes = n-l-I

=3-0-1=2SECTION-VI: Questions with Single Digit Answer

The next electron will enter into 5d subshell whose (n + I) will ~ equal to 7 (seven).

11. he 6.626xI0-34x3x108 =6.52xto-18j12. aE=l:= 3.055x108

dE -~x2176xto-18 J= 1.63x10-18 JH - 4 .

liE = li.EH xZ2

0-18, t:.E _ 6.52 xl _ 4

Z = liEH - 1.63 x to-18

2=2

...(i)

eVs=hv-4lV _ hv _1,-

e e

5. (d)

6. (e)

E =!!:.A

Eoc.!.).

Inverse relation shows that there will be reclangular hyperbola, in the plot of E against A..), = h

./2eVm

For the plo! of A against ~V'the slope Will be _h_'" .rz;;;;

Slope oc}; i.e, heavier is the particle, lesser will be the slope.

7. (a) Kinetic energy of photoelectrons is independent of the intensity of radiation.8. (a) 'A' is for Is because Is has no node; 'B' is for 2s because this orbital has single node.

13. (d) Kinetic energy of photoelectron does not depend on the intensity of incident radiation.According to the Einstein's theory of photoelectric effect, one photon can eject only onephotoelectron from the metal !;urface.

14. (a) Rest mass of photon is zero; its momentum can be calculated using.h

P=7"A

15. (a) )'J~l = R[*-31,] = 8:

AJ~' =R[~-~]= ~:A'~l =RU-~]=34R

1 1 1 8R--=--+--=-"3-+1 "]-.2 "'2-+1 9

SECTION-IV: Objective Questions on Graphical and DiagrammaticAptitude