NUCLEAR CHEMISTRY. Atomic Structure Recall: Atoms – consist of a positively charged nucleus, which…
Atomic Structure & Nuclear Chemistry SOLUTIONS OF ATOMIC STRUCTURE & NUCLEAR CHEMISTRY · 2020. 4....
Transcript of Atomic Structure & Nuclear Chemistry SOLUTIONS OF ATOMIC STRUCTURE & NUCLEAR CHEMISTRY · 2020. 4....
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Atomic Structure & Nuclear Chemistry
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SOLUTIONS OF ATOMIC STRUCTURE & NUCLEAR CHEMISTRY
EXERCISE # 1
PART - I
A-2. Space occupied by the nucleus = 4
3r3 =
4
3 ×
22
7 × (3.5 × 10–15)3 = 1.8 × 10–43 m3
A-3. nucleus
atom
V
V =
31
32
4r
34
r3
=
3–15
–10
1.5 10
0.5 10
= 2.7 × 10–14
A-4. (A) R = R0 A1/3 = 1.3 × 10–15 × (125)1/3 = 6.5 × 10–15 m
(B) r = 2
2
4 KZe
m v
, Z = 47 for Silver atom
r = 2
2
188 Ke
m v
B-1. Total energy = No of photons × Energy of one photon = 100 × –34 8
–10
6.625 10 3 10
2000 10
= 9.937 × 10–17 J = –17
–19
9.937 10
1.6 10
eV = 621.1 eV
B-2. Let number of photoelectrons emitted per second be n
n × Energy of one photon = Total energy emitted
n × (Å)
12400
× 1.6 × 10–19 = 5 × 10–3
n × 12400
6200 × 1.6 × 10–19 = 5 × 10–3
On solving, n = 1.56 × 1016 photoelectrons.
B-3. C =
= c
=
8
3
3 10
1368 10
= 219.3 m
= 1
=
1
219.3 = 4.56 × 10–3 m–1
B-4. E = hc
× NA =
–34 8 23
–10
6.625 10 3 10 6.022 10
5000 10
J/mol = 239.4 KJ/mol
B-5. n × Energy of one photon = Total energy
n × –34 8
–9
6.6 10 3 10
850 10
= 3.15 × 10–14
n = 1.35 × 105 photons
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B-6. Let n be the no. of photons emitted by bulbs per second & W be wattage of yellow bulb. For Red bulb no. of photons emitted per second × Energy of one photon = total energy
n × 12400
8000 × 1.6 × 10–19 = 100 .........(1)
For yellow bulb, n × 12400
4000 × 1.6 × 10–19 = W .........(2)
By (1) and (2), W = 200 watts.
B-7. For frequency 2.5 × 1016 Hz, h = hKEmax
h (2.5 × 1016) = h + KEmax ...........(1) For frequency 4 × 1016 Hz,
h (4 × 1016) = ho + 2 KEmax ..........(2) Multiply eqn (1) by 2 and subtract eqn (2) from it.
2 ho – ho = h (5.0 × 1016 – 4 × 1016)
ho = h (1 × 1016)
threshold frequency, o = 1 × 1016 Hz C-1. Radius of ground state of hydrogen atom = 0.529 Å
So, 0.529 = 0.529 × 2n
Z
0.529 = 0.529 × 2n
4 n = 2
C-2. v3 = v1 × Z
n
v3 = 2.18 × 106 × 1
3
= 7.27 × 105 m/s
C-3. Angular momentum, = mvr = nh
2 n
Radius, r = 0.529 2n
zÅ r n2 For same Z
Velocity, v = 2.18 × 106z
n m/s v
1
n
v r n × n2 × 1
n
But, v r nx (given) n2 = nx x = 2
C-4. 2
He
Li
T
T
=
2
3
2He
3
2Li
n
Z
n
Z
=
3
2
3
2
2
2
4
3
= 9
32
C-5. x : n = 3 to n = 1 (12.09 eV) y : n = 4 to n = 2 (2.55 eV) z : n = 5 to n = 3 (0.967 eV)
Shortest wavelength maximum energy n = 3 to n = 1 (12.09 eV)
C-6. Clearly, E4 – E2 = (– 0.85) – (– 3.4) = 2.55 eV A =2, B = 4
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C-7. (a) 16.52 = (E3 – E2) × Z2
16.52 = 1.89 × Z2 Z2 = 9 or Z = 3 (b) (E3 – E1)Z = 12.09 × Z2 = 12.09 × 9 = 108.81 eV
(c) = 1
12400
E =
2
12400
13.6 3 = 101.3 Å = 1.013 × 10–8 m,
(d) KE1 = 13.6 Z2 eV = 13.6 × 32 = 122.4 eV
C-8. 40.8 = E)21 × Z2
40.8 = 10.2 × Z2
Z2 = 4 or Z = 2 IE = 13.6 Z2 = 13.6 × 4 = 54.4 eV
D-1. 32 = 3 2
12400
E =
12400
1.89 = 6561 Å ;
42 = 4 2
12400
E =
12400
2.55 = 4863 Å
D-2. H
1
= R(1)2
2 2
1 1–
1 2
&
He
1
= R(2)2
2 21 2
1 1–
n n
But H = He
from above 2 equations, n1 = 2 & n2 = 4.
D-3. E62 = h
= –19
–34
3.022 1.6 10
6.625 10
= 7.3 × 1014 Hz
This frequency lies in visible spectrum.
D-4. E = hc
=
12400
300 =
124
3 = 13.6 Z2
124
3 = 13.6Z2
11–
4
Z = 2
D-5.
But, (H2)2 1 = (He+)4 2 = (Li2+)6 3 are lines of same energy and so, will overlap each other.
Total no. of lines observed = (1 + 6 + 15) – 2 = 20 lines. E-1. Energy absorbed = 1.5 × 13.6 = 20.4 eV. Energy used up in escaping = 13.6 eV
Energy left as KE with electron = 20.4 – 13.6 = 6.8 eV
Associated de-Broglie wavelength, = 12.27
6.8 = 4.71 Å.
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E-2. Debroglie wavelength associated with particle of mass (m) moving with velocity (v) is
= h
mv
p = p p
h
m v and e =
e e
h
m v
Given, p = e p p
h
m v =
e e
h
m v mp vp = me ve
e
p
v
v =
p
e
m
m = 1836 ve = 1836 vp
It means when velocity of electron will be 1836 times velocity of proton, then debroglie wavelength associated with electron would be equal to debroglie wavelength associated with proton.
E-3. 1 = h
2mq (100) =
k
10, 2 =
h
2mq(81) =
k
9, 3 =
h
2mq(49) =
k
7
3 2
1
=
k k–
7 9k
10
= 20
63
E-4. Initial kinetic energy, KEi = 2 eV Increase in KE due to acceleration = q × V = e × 2v = 2eV.
Final kinetic energy, KEf = 2 + 2 = 4 eV.
Associated de-Broglie wavelength, = 12.27
4 = 6.15 Å.
E-5. m = h
4 x v =
34
10 24
6.625 10
4 3.14 10 5.27 10
Ans. 100 gm
F-1. Number of radial nodes = n – – 1
number of angular nodes = total nodes = n – 1 G-1. Ni Atomic No : 28
Ni : [Ar] 3d8 4a2 ; Ni2+ [Ar] 3d8 4s0
No. of unpairecd electron = 2 G-2. Atomic No. 56 Electronic configuration : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2.
G-3. (a) n = 3 , = 1 3p (b) n = 5 , = 2 5d (c) n = 4 , = 1 4p
(d) n = 2 , = 0 2s (e) n = 4 , = 2 4d
G-4. Orbital angular momentum = h
( 1)2
For 4s orbital, = 0 Angular momentum = h
0 (0 1)2
= 0.
For 3p orbital, = 1 Angular momentum = h
1(1 1)2
= h
2.
For 4th orbit, Angular momentum = nh
2 =
4h
2 =
2h
.
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G-5. (i) = 0 m = 0 (m 1) (iii) n = 1 = 0 ( 2) (vi) s = + 1/2 or – 1/2 (s 0) G-6. (i) 26Fe3+ : 1s2 2s2 2p6 3s2 3p6 3d5
It contains 5 unpaired electrons n = 5
Total spin = ±n
2 = ±
5
2
Magnetic moment = n(n 2) = 5(5 2) = 35 BM.
(ii) 29Cu+ : 1s2 2s2 2p6 3s2 3p6 3d10
It contains 0 unpaired electron
Total spin = 0.
Spin magnetic Moment = 0.
H-1. E = m × 931.478
m = E
931.478
=
17.25
931.478 = 0.0185 amu
m = 3.07 × 10–26 g. H-5. Nuclear fission is a series reaction 3, 9, 27.......neutron and E, 3E, 9E......energy are emitted. Hence answer is 3n, 3n – 1E.
PART - II A-1. Hydrogen atom contains 1 proton, 1 electron and no neutrons. A-2. It constitute of electron.
A-3.
e
e /m
e /m
= e
e
e /m
2e / 4 1836 m =
3672
1
A-4. Factual.
A-5. Volume fraction = Volume of nucleus
Total vol. of atom =
13 3
8 3
(4 /3) (10 )
(4 /3) (10 )
= 10–15
A-6. R = R0 A1/3 = 1.3 × 641/3 = 5.2 fm
B-1. = c
=
8
6
3 10
400 10
= 0.75 m
B-2. Violet colour has minimum wavelength so maximum energy.
B-3. I.E. of one sodium atom = hC
& I.E. of one mole Na atom = hC
NA =
34 8 23
9
6.62 10 3 10 6.02 10
242 10
= 494.65 kJ.mol.
B-4. Power = nhC
t 40 ×
80
100 =
34 8
9
n 6.62 10 3 10
620 10 20
n = 2 × 1021
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B-5. For photoelectric effect to take place, Elight W hc
0
hc
or 0.
B-6. Photoelectric effect is a random phenomena. So, electron It may come out with a kinetic energy less
than (h – w) as some energy is lost while escaping out.
C-1. r 2n
Z
As Z increases, radius of I orbit decreases.
C-2. Radius = 0.529 2n
ZÅ = 10 × 10–9 m
So, n2 = 189 or, n 14 Ans.
C-3. E1 (H) = – 13.6 × 2
2
1
1 = – 13.6 eV ; E2 (He+) = – 13.6 ×
2
2
2
2 = – 13.6 eV
E3 (Li2+) = – 13.6 × 2
2
3
3 = – 13.6 eV ; E4 (Be3+) = – 13.6 ×
2
2
4
4 = – 13.6 eV
E1(H) = E2(He+) = E3 (Li2+) = E4(Be3+)
C-4. V = 2.188 × 106 Z
n m/s
Now, V Z
n so,
2Li
H
V
V
= – 1 1
2 2
Z /n
Z /n =
3 / 3
1/1 = 1 or, 2LiV = VH
C-5. r1 – r2 = 24 × (r1)H
210.529 n
1
–
220.529 n
1
= 24 × 0.529
2 21 2n – n = 24 So, n1 = 5 and n2 = 1 C-6. (a) Energy of ground state of He+ = – 13.6 × 22 = – 54.4 eV (iv)
(b) Potential energy of orbit of H-atom = – 27.2 × 12 = – 27.2 eV (ii)
(c) Kinetic energy of excited state of He+ = 13.6 × 2
2
2
3 = 6.04 eV (i)
(d) Ionisation potential of He+ = 13.6 × 22 = 54.4 V (iii) C-7. S1 : Potential energy of the two opposite charge system decreases with decrease in distance, S4 : The energy of Ist excited state of He+ ion = – 3.4 Z2 = – 3.4 × 22 = – 13.6 eV. S2 and S3 are correct statement.
D-1. En = E1 2
2
Z
n E5 = – 13.6 ×
2
2
(1)
(5) = – 0.54 eV
D-2. = hc
E
1
E
D-3. Infrared lines = total lines – visible lines – UV lines = 6(6 –1)
2 – 4 – 5 = 15 – 9 = 6.
(Visible lines = 4 62, 52, 42, 32) (UV lines = 5 61, 51, 41, 31, 21)
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D-4. When electron falls from n to 1, total possible number of lines = n – 1.
D-5. Visible lines Balmer series (5 2, 4 2, 3 2). So, 3 lines.
D-6. According to energy, E4 1 > E3 1 > E2 1 > E3 2 . According to energy, Violet > Blue > Green > Red.
Red line 3 2 transition. D-7. For 1st line of Balmer series
1v = RH (3)2
2 2
1 1
(2) (3)
= 9R 5
36
= 5
4R
For last line of Pachen series
2v = RH (3)2
2 2
1 1
(3) ( )
= R so, 1v – 2v = 5
4R – R =
R
4.
E-1. = h
mv =
346.625 10
0.2 5
× 3600 10–30 m.
E-2. 1
2
= 2
1
V
V =
200
50 =
2
1.
E-3. r1 = 0.529 Å r3 = 0.529 × (3)2 Å = 9x
so, = 2 r
n
=
2 (9x)
3
= 6 x.
E-4. For an particle, = 0.101
VÅ.
E-5. n
Z 1
1
n
Z = 2
2
n
Z or
2
3 =
4
6 (n = 4 of C5+ ion)
E-6. For a charged particle = h
2mqV,
1
V.
E-7. p . x = h
4 x =
34
5
6.62 10
4 3.14 1 10
= 5.27 × 10–30 m.
F-1. Factual.
F-2. A has 0 nodes as 2 is not zero anywhere so it's 1 s (n – 1 = 0) B has 1 node so, it is 2s as n – 1 = 1 F-4. Dumbell lies at 45º to x & y axis. F-5. Factual
F-6. Spherical node = n – – 1
Non spherical = F-7. Factual
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F-8. (a) Electron density in the XY plane in 2 2x y3d orbital is not zero
(b) Electron density in the XY plane in 2z3d orbital is not zero.
(c) 2s orbital has one nodal surface (d) For 2pz orbital, XY is the nodal plane. F-9. Factual
F-10. n, and m.
G-1. Orbital angular momentum = ( 1)h
2 = 0. = 0 (s orbital).
G-2. Cu : 1s22s22p63s23p63d104s1.
Cu2+ : 1s22s22p63s23p63d9 or [Ar]3d9.
G-3. Magnetic moment = n (n 2) = 24 B.M.
No. of unpaired electron = 4. X26 : 1s2 2s22p63s23p63d64s2. To get 4 unpaired electrons, outermost configuration will be 3d6.
No. of electrons lost = 2 (from 4s2).
n = 2. G-4. Zn2+ : [Ar] 3d10 (0 unpaired electrons). Fe2+ : [Ar] 3d6 (4 unpaired electrons) maximum. Ni3+ : [Ar] 3d7 (3 unpaired electrons). Cu+ : [Ar] 3d10 (0 unpaired electrons).
G-5. d7 : 3 unpaired electrons. Total spin = ±n
2 = ±
3
2.
G-6. X23 : 1s2 2s2 2p6 3s2 3p6 3d3 4s2.
No. of electron with = 2 are 3 (3d3). G-7. Cr (Zn = 24) Electronic configuration is : 1s2 2s2 2p6 3s2 3p6 4s1 3d5
so, no of electron in = 1 i.e. p subshell is 12 and no of electron in = 2 i.e. d subshell is 5.
G-8. Orbital angular momentum = ( 1)h
2 = 0 (since = 0 for s orbital).
G-9. Cl17– : [Ne] 3s2 3p6. Last electron enters 3p orbital.
= 1 and m = 1, 0, –1.
G-10. Number of radial nodes = n – – 1 = 1, n = 3. = 1.
Orbital angular momentum = ( 1)h
2 = 2
h
2.
H-1. 11
11 056 1C B e
H-2. n
p > 1
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H-3. It is the order of penetrating power.
H-4. Nucleides having n
p > 1 undergoes -emission to decrease
n
p ratio in order to attain belt of stability.
H-5. AZ X
A 1Z X + 10 n
H-6. 23892 U 21482 Pb + m
42He + n
01e
m = 6 and m = 2. Total 8.
PART - III 1. It is factual.
2. fn = n
n
v
2 r, fn
2
3
Z
n, Tn =
n
n
2 r
v
, Tn
3
2
n
Z.
En = – 13.6 2
2
Z
n, En
2
2
Z
n, rn
2n
Z.
3. It is factual.
4. Number of values of = total number of subshells = n.
Value of = 0, 1, 2 ........ (n – 1).
= 2 m = –2, –1, 0, +1, +2 (5 values)
m = – to + through zero.
EXERCISE # 2
PART - I 1. Factual.
2. m
e = 1.5 × 10–8 and e = 1.6 × 10–19
m = 1.5 × 10–8 × 1.6 × 10–19 m = 2.4 × 10–24 g 3. Charge on oil drop = 6.39 × 10–19 C
1.602 × 10–19 C is charge on one electron
6.39 × 10–19 C is charge on = 19
19
6.39 10
1.602 10
= 4 electrons.
5. hc
= 1 + ...(1)
3 × hc
= 4 + ...(2)
from, e.q., (1) and (2) = 0.5 eV
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6. 1
2
T
T =
3132
n
n =
3
3
1
2 =
1
8.
2 r
TV
so, T 3
2
n
Z
7. 1
2
r
r =
2122
n
n =
R
4R
1
2
n
n =
1
2
1
2
T
T =
3132
n
n =
1
8.
8. Angular momentum J = mvr J2 = m2v2r2
or 2J
2 =
2 21 mv mr2
or K.E. = 2
2
J
2mr
9. P.E. = 1 2Kq q
r =
K(–e)( 4e)
r
=
0
1
4 × –
24e
r = –
2
0
e
r .
10. KE = 1
2
2KZe
r =
2
0
3e
8 r .
11. (He+)2 4 = 4 3
2n n(Li )
1
2
Z
Z = 2
4
n
n = 1
3
n
n or
2
3 =
4
2
n =
3
4
n
n4 = 3 and n3 = 6.
Transition in Li2+ ion = 3 6
12. = RC Z2 2 21 2
1 1–
n n
.
1 = RC Z2 2 2
1 1–
1
= RC Z2, 2 = RC Z2 2 2
1 1–
1 2
= 3
4 RC Z2.
3 = RC Z2 2 2
1 1–
2
= 1
4RC Z2. 1 – 2 = 3.
13. Visible lines Balmer series 3 lines. (5 2, 4 2, 3 2).
14. Shortest wave length of Lyman series of H-atom
1
=
1
x = R
2 2
1 1
(1) ( )
so, x = 1
R
For Balmes series
1
= R(1)2
2 2
1 1
(2) (3)
1
=
1
x ×
5
36 so, =
36x
5.
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15. Number of lines in Balmer series = 2. n = 4 (lines will be 4 2, 3 2).
KE of ejected photoelectrons = Ephoton – BEn = 13 – 2
13.6
4 = 13 – 0.85 = 12.15 eV.
16. y
x
= x x
y y
m v
m v.
y
1
= x x
x x
m v
(0.25m ) (0.75 v ) =
16
3.
y = 5.33Å.
17. = h
mV
p
=
p p
m V
m V
m = 4mp
p
=
p p
4m V
m V
1
2 = 4 ×
p
V
V
pV
V =
8
1
18. For an electron accelerated with potential difference V volt, = h
2mqV =
12.3
VÅ.
19. = v
then = h
mV or 2 =
h
m So, =
h
m.
20. x = 2p
x.p = 2
= h
4 2 p.p =
2
2(mV)2 = 2
; (V)2 = 24m
V = 2m
.
21. 2r = n = circumference 22. s orbital is spherical so non-directional.
23. The lobes of 2 2x yd orbital are alligned along X and Y axis. Therefore the probability of finding the
electron is maximum along x and y-axis. 24. Factual.
25. Rb37 : [Kr] 5s2. n = 5, = 0, m = 0, s = ±1
2.
26. Magnetic moment = 2.83 so, no. of unpaired electrons = 2 so, Ni2+ is the answer.
27. For 1s, 3s, 3d and 2p orbital, = 0, 0, 2, 1 respectively.
Orbital angular momentum = ( 1) .
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28. After np orbital, (n + 1) s orbital is filled.
29. Total number of electrons in an orbital = 2 (2 +1).
The value of varies from 0 to n – 1.
Total numbers of electrons in any orbit =
n 1
0
2(2 1)
.
30. Spin quantum number does not comes from Schrodinger equation.
s = +1
2 and –
1
2 have been assigned arbitrarily.
31. Change in mass number = 32 Change in proton = 10
8 means 8 2He4 i.e., 32 mass number, 16 protons.
Now 6 = 6 neutrons changed to 6 proton. So, net change in proton = 10
Answer is 8, 6.
PART - II
1. p(e /m)
(e /m) =
p p
p p
e /m
2e / 4m =
2
1.
2. 1
2
E
E =
1
hc
× 2
hc
= 2
1
=
600
300 = 2.
3. Heat required for melting 1 mole (18 g) of ice = 330 × 18 = 5940 J
Energy of one photon = h = 6.6 × 10–34 × 5 × 1013 = 33 × 10–21 J
Total number of photons required = 21
5940
33 10 1.8 × 1023
4. W0 + K = h
40 × 1.6 × 10–19 = –34 86.62 10 3 10
= 31 nm
5. Energy of the photon = 0.6375 eV = 3
4 × 0.85 eV =
3
4 ×
2
13.6
4 =
2
13.6 11
44
= 13.62 2
1 1
4 8
Thus, this photon coresponds to transition n = 8 to n = 4 (Brakett series)
x = 8 Ans. 6. IP = 13.6Z2 = 16 (given).
1st excitation potential = 13.6 × 3
4 × Z = 16 ×
3
4 = 12 V.
7. Diameter of Hydrogen atom = 16.92Å Radius of an atom = 8.46Å
rn = 20.529n
Z 8.46 =
20.529n
1 n = 4
Maximum number of electron = 2n2 = 2 × (4)2 = 32.
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8. Number of spectral lines = 2 1 2 1(n n )(n n 1)
2
6 = 2 2(n 3)(n 3 1)
2
12 = 22 2n 5n 6
n22 – 5n2 – 6 = 0.
n2 = 6 or – 1. Since n2 = –1 is not possible. Hence n2 = 6.
9. Total energy = 2
2
13.6 Z
n =
2
2
13.6 (Z)
(4) = 3.4 eV
Now K.E. = 3.4 – 1.4 = 2 eV Now, Total energy = 2 + 4 = 6 eV i.e. potential = 6 V
For electron, = 150
V so = 5 Å.
10. For an electron, de-Broglie wavelength = eV
150
KEÅ
2 = eV
150
KE KE =
2
150
=
150
12.016 =
13.6 11
144
eV (Using given relation)
22n 6 LiE = 3 HE + KEelectron
13.6 (3)2 2 2
2
1 1
6 n
= 13.6 (1)2
2 2
1 1
3
+ 13.6 11
144
On solving, we get n2 = 12. 12. (2 – Zr)2 = 0 Zr = 2
r = 2
z =
2
2 = 1Å
13. dxy, dyz, dxz The lobes of dxy orbital are at an angle of 45º with X and Y axis. So along the lobes, angular probability
distribution is maximum similarly for dyz & dxz. 14. Cr : 1s2 2s2 2p6 3s2 3p6 4s1 3d5
n + = 3 so the combinations are 2p, 3s. So 8 electrons.
15. n(n 2) = 4.9
No. of unpaired electrons, n = 4. 25Mn : [Ar]4s23d5 For having 4 unpaired electrons, a Mn atom should lose 3 electrons (2 from 4s and 1 from 3d).
a = +3.
16. 23592 U + 10n
13954 Xe +
9438Sr + x0n
1
Equating mass number 235 + 1 = 139 + 94 + x x = 3
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PART - III
1. No. of neutrons in 7632Ge = A – Z = 76 – 32 = 44.
No. of neutrons in 7733 As = 77 – 33 = 44.
No. of neutrons in 7834 Se = 78 – 34 = 44.
2. Ne contains 10 electrons O2– and F– contain 10 electrons
3. Since most part of atom is empty space, so, when particles are sent towards a thin metal foil, most of them go straight through the foil.
4. From particle scattering experiment, distance of closest approach of particle with nucleus came out to be of the order of 10–14 m.
5. = c
=
8
–9
3 10
600 10
= 5 × 1014 sec–1
E = 12400
6000 = 2.07 eV.
6. Li2+, H and He+ are single electron species.
7. Velocity Z
n; Frequency
2
3
Z
n; Radius
2n
Z; Force
2
4
Z
n.
8. 1st excitation potential = 10.2 Z2 = 24 V Z2 = 24/10.2
IE = 13.6 Z2 = 13.6 24
10.2
= 32 eV.
Binding energy of 3rd excited state = 0.85 Z2 = 0.85 24
10.2
= 2eV.
2nd excitation potential of sample = 12.09 Z2 = 12.09 24
10.2
=
32 8
9
V.
9. In all the given cases, only one quantum of energy is emitted since only one electronic transition
occurs.
10. Transition is taking place from 5 2
n = 3
Hence maximum number of spectral line observed = 3(3 1)
2
= 6.
(C) number of lines belonging to the Balmer series = 3 (5 2,4 2,32) as shown in figure. 5
3
2
n = 1
4
Number of lines belonging to Paschen series = 2 (5 3, 4 3).
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11. Change in angular momentum for 3 2 transition = (3 – 2)h
2 =
h
2.
Change in angular momentum for 4 2 transition = (4 – 2)h
2 =
h
.
12. = h
mv =
h
2mKE =
h
2mqV.
When v, KE and V are same, as m increasing, decreases. e > p > (if v, KE and V are same). 13. n = 4, m = 2
Value of = 0 to (n – 1) but m = 2. = 2 or 3 only Value of s may be +1/2 or – 1/2.
14. n(n 2) = 1.732
Number of unpaired electrons, n = 1.
25X : [Ar] 4s23d5 For having one unpaired electron, 6 electrons are to be removed (2 from 4s & 4 from 3d).
Y = 6.
15. Spin angular momentum S = h
s(s 1)2
.
s = 1
2 S =
3
2 ×
h
2.
16. (A) 24Cr : [Ar]3d54s1 (B) m = – to + through zero. (C) 47Ag : 1s22s22p63s23p64s23d104p65s14d10. Since only one unpaired electron is present.
23 electrons have spin of one type and 24 of the opposite type. 17. 8O : [He] 2s22p4 ; 16S : [Ne] 3s23p4
18. - rays are uncharged. 19. 1 neutron added increases mass number.
PART - IV
1. As the frequency of incident radiations increases, the kinetic energy of emitted photoelectrons increases.
Decreasing order of Violet > Blue > Orange > Red
Decreasing order of KE of photoelectrons Violet > Blue > Orange > Red 2. The interaction between photon and electron is always one to one for ejection of photoelectrons, Frequency of incident radiations > threshold frequency 5.16 x 1015 > 6.15 × 1014
3. The number of photoelectrons emitted depend on the intensity or brightness of incident radiation. 4. Last line of Bracket series for H-atom
1
1
= R
2 2
1 1
(4) ( )
so, 1 = 16
R
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2nd line of Lyman series
2
1
= R
2 2
1 1
(1) (3)
so, 2 = 9
8R
or, 1
128
=
2
9
5. 1. Spectral lines of H atom only belonging to Balmer series are in visible range. 2. In the Balmer series of H-atom, first 4 lines are in visible region and rest all are in ultra violet region.
3. 2nd line of Lyman series of He+ ion has energy = (E31) × 22 = 12.1 × 4 = 48.4 eV.
6. v = R(4)2 2 2
1 1
(3) (4)
= 7R
9.
7. x = h
4 Me ×
1
V V = V ×
0.001
100 = 300 × 10–5 m/s
x = 5.8 10–5 × 5
1
300 10 = 1.92 × 10–2 m
8. The maximum KE of potoelectron is corresponding to maximum stopping = 22 eV. Eincident = Ethresold + KEmaxi = 40 eV + 22eV = 62 eV
incident = 12400 Å
62 = 200 Å
9. Circumference = 2r = n
de-broglie – = 2 r
n
=
3nm
3 = 1 nm = 10Å
= 12.3
VÅ KE =
212.3
10
= 1.51 eV.
KE of electron in third orbit = 1.51 eV binding energy of third orbit in this atom
= of photon required to ionise = 1240 eV Å
1.51 eV = 821 nm
10. Multiply Angular part and Radial part of 1s orbital and square this. 11. For s-orbital probability of finding an electron is same at all angles at specific radius. 12. Two unpaired electrons present in carbon atom are in different orbitals. So they have different magnetic
quantum number. 13. Electronic configuration of Zn2+ ion is 1s2 2s2 2p6 3s2 3p6 3d10 so no electron in 4s orbital.
14. s (s 1)h
2 =
1 11
2 2
h
2 =
3
2
h
2 = 0.866
h
2
15. Cu+ 1s2, 2s2, 2p6, 3s2, 3p6, 3d10
Fe+3 1s2, 2s2, 2p6, 3s2, 3p6, 3d5
Cr+3 1s2, 2s2, 2p6, 3s2, 3p6, 3d3
Co+3 1s2, 2s2, 2p6, 3s2, 3p6, 3d6 Cu+ ion have maximum number of full filled orbital
Number of electrons related to = 2 are 10
Number of electrons related to + m = 0 are 12
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16. Cr+3 1s2, 2s2, 2p6, 3s2, 3p6, 3d3
Number of electrons related to n + = 5 are 3
Magnetic moment u = n(n 2) B.M. = 3 5 = 15 B.M.
17. Co+3 1s2, 2s2, 2p6, 3s2, 3p6, 3d6
Number of unpaired e– = 4
Number of electrons related to n + = 5 are 6.
EXERCISE # 3
PART - I
1. rn = 0.529 2n
ZÅ
For hydrogen, n = 1 and Z = 1 ; rH = 0.529
For Be3+, n = 2 and Z = 4 ; 3Ber = 20.529 2
4
= 0.529
Therefore, (D) is correct option.
2. 22s = probability of finding electron with in 2s orbital
2at node = 0 (probability of finding an electron is zero at node)
For node at r = r0, 2 = 0
So, 2 = 0 = 1
4 2
3
0
1
a
0
0
r2 –
a
× 0 0r / 2a
e
0
0
r2 –
a
= 0 or 2 = 0
r
a
r = 2a0 (b) The wavelength can be calculated with the help of de-Broglie’s formula i.e.,
= h
mv =
–34
3
6.626 10
100 100 10
=
–34
3
6.626 10
10,000 10
= 6.626 × 10–35 m or 6.626 × 10–25 Å
(c) (i) The atomic mass of an element reduces by 4 and atomic number by 2 on emission of an -particle.
(ii) The atomic mass of an element remains unchanged and atomic number increses by 1 on emission
of a -particle.
Thus change in atomic mass on emission of 8–particles will be 8 × 4 = 32 New atomic mass = old atomic mass – 32 = 238 – 32 = 206
Similarly change in atomic number on emission of 8-particle will be : 8 × 2 = 16 i.e., New atomic number = old atomic number – 16 = 92 – 16 = 76
On emission of 6-particles the atomic mass remains unchanged thus, atomic mass of the new element will be 206.
The atomic number increases by 6 unit thus new atomic nubmer will be 76 + 6 = 82
Thus, the equation looks like : 92X238 –8
–6
82Y206
3. (a) For hydrogen atom, Z = 1, n = 1
v = 2.18 × 106 × Z
nms –1 = 2.18 × 106 ms–1
De-broglie wavelength, = h
mv =
–34
–31 6
6.626 10
9.1 10 2.18 10
= 3.32 × 10–10 m = 3.3 Å
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(b) For 2p, = 1
Orbital angular momentum = ( 1)h
2 = 2 .
h
2
.
4.
2
n
2
n
2
n
KZeK
2r
KZeV
r
KZeE
2r
so, n
n
V
K = – 2 and En
1
r.
5.
6. For lower state (S1)
No. of radial node = 1 = n – – 1
Put n = 2 and = 0 (as higher state S2 has n = 3) So, it would be 2s (for S1 state)
7. Energy of state S1 = – 13.6 2
2
3
2
eV/atom
= 9
4 (energy of H-atom in ground state)
= 2.25 (energy of H-atom in ground state). 8. For state S2
No. of radial node = 1 = n – – 1 ....... (eq.-1) Energy of S2 state = energy of e– in lowest state of H-atom = – 13.6 eV/atom
= – 13.62
2
3
n
eV/atom
n = 3.
put in equation (1) = 1 so, orbital 3p (for S2 state).
9. Ephoton = 12400
3000 = 4.13 eV
Photoelectric effect can take place only if Ephoton Thus, Li, Na, K, Mg can show photoectric effect.
10.
So, electrons with spin quantum number = –1
2 will be 1 + 3 + 5 = 9.
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11.
12. mv (4a0) = h
so, v = 0
h
4m a
so, KE = 1
2mv2 =
1
2m.
2
2 2 20
h
16 m a =
2
2 20
h
32 m a
13. 63 1 1 4 129 1 0 2 1Cu H 6 n 2 H X
64 = 6 + 4 + 2 + A A = 52
29 + 1 = 30 = 0 + 2 + 2 + z z = 26 Element X should be iron in group 8.
14. 9 84 4Be X Be Y
If X is then Y is 0n1 If X is 1P1 then Y is 1D2
15. n = 4, m = 1, –1
Hence can be = 3, 2, 1 i.e. Hf ; 2 orbitals Hd ; 2 orbitals Hp ; 2 orbitals
Hence total of 6 orbitals, and we want ms = –1
2, that is only one kind of spin. So, 6 electrons.
16. Energy order of orbitals of H is decided by only principle quantum number (n) while energy order of H–
is decided by (n + ) rule :
Electronic configuration of ‘H–’ is - 1s2 its Energy order is decided by n+ rule. H– = 1s22s02p0 Its 2nd excited state is 2p and degenery 2p is ‘3’
17. For 1s electron in H-atom, plot of radial probability function (4r2R2) V/s r is as shown :
4r2R
2
r
18. s-orbital is non directional so wave function will be independent of cos
19. For 2s orbital no. of radial nodes = n –– 1 = 1
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n,
,m (r)
0
r
a
20. For 1s orbital should be independent of , also it does not contain any radial node.
4 2
6 2
E – E
E – E =
1 1
1 1
E E–
16 4E E
–36 4
=
1
1
3E–
168E
–36
= 3 36
8 16
=
27
32
21. x1 =
X2 =
X3 =
X4 =
22. HeE = –13.6 ×
2
2
n
)2( = –3.4 =
4
6.13–
n2 = 16 so n = 4 quantum number are
n = 4, = 2, m = 0 so subshell is = d.
angular node = = 2
Radial node = [n––1] = 4 – 2 – 1 = 1
24. rn = 0.529 ÅZ
n2
rn n2
Angular momentum () =
2
nh n1
K.E. = 2mv2
1 =
26
n
Z1018.2m
2
1
K.E.
2
2
n
Z K.E. n–2
P.E. = – 2K.E. P.E. 2
2
n
Z P.E. n–2
PART - II 8. Following Aufbau principle for filling electrons.
9. De-broglie wavelength (for particles) = h
2m KE
As temperature is same, KE is same. So, 1
m .
Hence db (electron) > db (neutron)
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10. n = 5 Possible subshell are
5s, 5p, 5d, 5f, 5g
Total number of orbital = 1 + 3 + 5 + 7 + 9 = 25 11. NaF: Na+ = 1s22s22p6 F– = 1s22s22p6
12. For shortest '' of hydrogen
n1 = 1 & n2 =
1
= Rz2
2 21 2
1 1
n n
1
A= R(1)2
2 2
1 1
1
R = 1
A
for longest '' of He+ n1 = 3 n2 = 4
2
2 2
1 1 1 1 1 72
A A 363 4
or =
36A
7
13. rn = 52.92n
1
pm = 211.6 pm (for H-atom)
n = 2
Higher orbit to n = 2 Balmer series
14. = 250 nm = 2500 Å
E = hc
=
12400
2500 = 4.96 eV
KE = stopping potential = 0.5 eV E = W0 + K.E. 4.96 = W + 0.5
W0 = 4.46 4.5 eV
15. 2r = n = 2 r
n
=
1
Å529.02
16. When temperature is increased, black body emit high energy radiation, from higher wavelength to lower
wavelength. 17.
22
21
2H
n
1–
n
1ZR
2f
2H
8
1–
n
1)1(R
2
64
R–
n
R H
f
H
2
Slope for graph of & 2f
n
1is + RH
–RH
64
1
nf2
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18. mvr = 2
nh
According to wave mechanics, the ground state angular momentum is equal to 2
h.
19.
K.E. of Electron
Frequency of Light
KEmax = h–h0
tan= h
Intercept = –h0
20. E = –13.6 2
2
Z
n
= –13.6 2
2
2
3
= –6.04 eV
21. 2 21 2
1 1 1R
n n
72
1 1 110
(3)
= 9 × 10–7 m
= 900 nm
22. 201
h h mv2
201
h( – ) mv2
1/ 2
02h( – )vm
00
h h m h
mv m m( – )2h( – )
1/ 2
0
1
( – )
23. 2hc 1
mv2
34 8
10
6.626 10 3 10
4000 10
= +
1
2× 9 × 10–31 × (6 × 105)2
= (4.97 – 1.62) × 10–19 J
= 3.35 × 10–19 J 2.1 eV
24. 2r = n 2
0
n2 a n
Z
2
0 0
n2 a n1.5 a
Z
n 1.5 3
Z 2 4 = 0.75
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25. From n + rule the increasing order of energies of electrons will be IV < II < III < I 26. As kinetic energy is much higher than work function so
E = E0 + KE KE
As E =
hc & kE =
m2
P2
So 1
2
=
2
2
1
P
P
2
2
5.1
1
2 = 9
4
27. Balmer)(
Lyman)(
=
9
1
4
1
4
1
4
111
= 4
9
4
14
1
28. (1) Total energy of electron is minimum in first orbit i.e. at a0 distance from nucleus.
(2) P.E. = – r
eZK 2
K.E. = r
eZK
2
1 2 |P.E.| = 2|K.E.|
(3)
2
r
1s
2 is maximum at nucleus
(4) Electron can be found at any distance from nucleus.
P
r
1s
P Probability function.
29. By the graph since 2 is not zero at r = 0 it must be s orbital
also n – –1 = 1
n = 2 ( = 0) it is 2s orbital
30. Shortest for lyman series :
1
1
1R
12
= R ; = R
1
Shortest for paschen series :
1
3
1R
'
12
= 9
R; ’ =
R
9
1
R
R
9'
= 9
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31. (r) is probability density of an electron and it is maximum at a & c. 32. Energies of the orbitals in the same subshell decrease with increase in the atomic number E2s(H) > E2s(Li) > E2s(Na) > E2s(K)SS 34. Theory based.
35. 11H
21H(D)
31H(T)
Number of neutrons 0 + 1 + 2 = 3
36. r = Z
na 20
For Li2+ r = 3
)2(a 20 = 3
a4 0
37. 2 r = n
2 × 2n
Za0 = n
2 × 24
1a0 = 4
= 8 a0
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