33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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Telegram @unacademyplusdiscounts114 NEET-AIPMT Chapterwise Topicwise Solutions Physics\ Total energy1= − = − P.E.P.E. P.E.=−34. eV.2 2\ K.E. = + 3.4 eV24. (a)25. (a) : Radius of first orbit, r ∝ 1 ,Zfor doubly ionized lithium Z (= 3) will be maximum,hence for doubly ionized lithium, r will be minimum.2Z26. (a) : E ∝2n27. (a)28. (a) : Centripetal force = force of attraction ofnucleus on electron2 2mv 1 ee= va0 0 a0 2 ⇒ =4πε4πε0a0m29. (c) : Energy of the ground electronic state ofhydrogen atom E = –13.6 eV.We know that energy of the first excited state for secondorbit (where n = 2)13.6 136 .En =− =− =− 34 . eV.2 2( n)() 230. (b) : When a hydrogen atom is in its excited level,then n = 2.Therefore radius of hydrogen atom in its first excitedlevel (r) = n 2 r 0 = (2) 2 r 0 = 4r 0 .31. (d) : According to Bohr’s principle, radius of orbit2 2(r) = 24πε0× nh; r2 24πme∝ nwhere n = principal quantum number.32. (d)33. (a) : As r ∝ n 2 , therefore, radius of 2 nd Bohr’s orbit = 4a 034. (d) : E = E 4 – E 313. 6=− − ⎛ 13.6 ⎞−⎝⎜⎠⎟4 3=− 085 . + 151 . = 066 . eV2 235. (a) : Second excited state corresponds ton = 3Energy needed to ionize,∴ E = 13 . 6 eV = 151 . eV2336. (b) : Bohr used quantisation of angular momentum.nhFor stationary orbits, Angular momentum mvr = 2πwhere n = 1, 2, 3,...etc.37. (d) : E ∝ Z 2 and Z for singly ionised helium is 2 (i.e.,2 protons in the nucleus)\ (E) He = 4 × 13.6 = 54.4 eV38. (c) : When electron jumps from higher orbit tolower orbit then, wavelength of emitted photon is givenby,so,1 1 1λ = ⎛2 − ⎞R⎜ 2 ⎟⎝ n f n i ⎠1 1 1 5λ = R ⎛2 22− ⎞⎝⎜3 ⎠⎟ = R 1 1 1 7and36 ′ = ⎛2 23− ⎞⎝⎜4 ⎠⎟ = RRλ144144 5λ20λ∴ λ′ = × =7 36 739. (c) : Energy of the photon, E = hcλ−34 8663 . × 10 × 3×10E =J−10975 × 10−34 8663 . × 10 × 3×10=eV = 12.75 eV−10 −19975 × 10 × 16 . × 10After absorbing a photon of energy 12.75 eV, the electronwill reach to third excited state of energy –0.85 eV, sinceenergy difference corresponding to n = 1 and n = 4 is12.75 eV.\ Number of spectral lines emitted40. (d)()( n n− 1) ()( 4 4− 1)= = = 62 241. (a) : According to Rydberg formula1 1 1λ = ⎡2 − ⎤R ⎢ 2⎥⎣⎢n f n i ⎦⎥Here, n f = 1, n i = 51 ⎡ 1 1 ⎤∴ = ⎢ − ⎥⎣ ⎦= ⎡1⎣⎢ − 1 ⎤⎦⎥1 5 1 25= 24λ R 2 2R 25RAccording to conservation of linear momentum, we getMomentum of photon = Momentum of atomhh h R hRmv vλ= = mλ= ⎛ 24 ⎞ 24orm ⎝⎜⎠⎟ =25 25 m42. (d)43. (c) :The maximum wavelength emitted here corresponds tothe transition n = 4 → n = 3 (Paschen series 1 st line)
AtomsTelegram @unacademyplusdiscounts11544. (c) : Ionisation potential of hydrogen atom is 13.6 eV.Energy required for exciting the hydrogen atom in theground state to orbit n is given byE = E n – E 1. . .ie .., 12.1 =− 13 6 136 13 6 13 . 62n− ⎛ − ⎞⎝⎜ 2 21 ⎠⎟ =− n+. .or, − . = − 13 6 2 13615 or, n = = 9 or,n=32n15 .Number of spectral lines emittednn− ( 1)3× 2= = = 3.2 245. (b) :(E C – E A ) = (E C – E B ) + (E B – E A )hc hc hc 1 1 1= + or = +λ λ λ λ λ λ3 1 2 3 1 21 λ1+λ2λλ 1 2∴ = or λ3=λ3λλ 1 2 λ1+λ2⎡ 1 1 ⎤46. (a) : υ∝⎢− ⎥⎣n1 2 n2 2 ⎦47. (b) : Jump to second orbit leads to Balmer series.The jump from 4th orbit shall give rise to second line ofBalmer series.48. (d) : Transition of hydrogen atom from orbit n 1 = 4to n 2 = 2.1 ⎡ 1 1 ⎤ ⎡1 1⎤Wave number = = ⎢ − ⎥ = ⎢ − ⎥λ R ⎣⎢⎦⎥⎢⎣( 2)( 4 ) ⎥1 2 2 2 Rn n2 2⎦= R ⎡11 ⎤⎢ − ⎥⎣ ⎦= R ⎡ 4−1⎤⎢ ⎥⎣ ⎦= 3R4 16 16 16 ⇒ l = 16/3R49. (c) : Absorption spectrum involves only excitationof ground level to higher level. Therefore spectral lines 1,2, 3 will occur in the absorption spectrum.50. (b)51. (d) : The circumference of an orbit in an atom interms of wavelength of wave associated with electron isgiven by the relation,Circumference = nl, where n = 1, 2, 3, ...52. (b) : 2dsinf = nl ; (sinf) max = 1i.e., l max = 2d⇒ l max = 2 × 2.8 × 10 –8 = 5.6 × 10 –8 m.53. (b) : hc hc 1= eV or λ = ∝λeV V54. (b)vvv
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Atoms
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115
44. (c) : Ionisation potential of hydrogen atom is 13.6 eV.
Energy required for exciting the hydrogen atom in the
ground state to orbit n is given by
E = E n – E 1
. . .
ie .., 12.
1 =− 13 6 136 13 6 13 . 6
2
n
− ⎛ − ⎞
⎝
⎜ 2 2
1 ⎠
⎟ =− n
+
. .
or, − . = − 13 6 2 136
15 or, n = = 9 or,
n=
3
2
n
15 .
Number of spectral lines emitted
nn− ( 1)
3× 2
= = = 3.
2 2
45. (b) :
(E C – E A ) = (E C – E B ) + (E B – E A )
hc hc hc 1 1 1
= + or = +
λ λ λ λ λ λ
3 1 2 3 1 2
1 λ1+
λ2
λλ 1 2
∴ = or λ3
=
λ3
λλ 1 2 λ1+
λ2
⎡ 1 1 ⎤
46. (a) : υ∝⎢
− ⎥
⎣n1 2 n2 2 ⎦
47. (b) : Jump to second orbit leads to Balmer series.
The jump from 4th orbit shall give rise to second line of
Balmer series.
48. (d) : Transition of hydrogen atom from orbit n 1 = 4
to n 2 = 2.
1 ⎡ 1 1 ⎤ ⎡
1 1
⎤
Wave number = = ⎢ − ⎥ = ⎢ − ⎥
λ R ⎣⎢
⎦⎥
⎢
⎣( 2)
( 4 ) ⎥
1 2 2 2 R
n n
2 2
⎦
= R ⎡1
1 ⎤
⎢ − ⎥
⎣ ⎦
= R ⎡ 4−1⎤
⎢ ⎥
⎣ ⎦
= 3R
4 16 16 16 ⇒ l = 16/3R
49. (c) : Absorption spectrum involves only excitation
of ground level to higher level. Therefore spectral lines 1,
2, 3 will occur in the absorption spectrum.
50. (b)
51. (d) : The circumference of an orbit in an atom in
terms of wavelength of wave associated with electron is
given by the relation,
Circumference = nl, where n = 1, 2, 3, ...
52. (b) : 2dsinf = nl ; (sinf) max = 1
i.e., l max = 2d
⇒ l max = 2 × 2.8 × 10 –8 = 5.6 × 10 –8 m.
53. (b) : hc hc 1
= eV or λ = ∝
λ
eV V
54. (b)
vvv
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