atomic structure & nuclear chemistry solutions of atomic structure & nuclear chemistry · 2020. 4....

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

mailto:[email protected]://www.resonance.ac.in/reso/results/jee-main-2014.aspx





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






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.






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

.






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






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)






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






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






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






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.






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) .






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.






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






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).






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






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






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






(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.






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






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)






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






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






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






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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