24.1. Model: Balmers formula predicts a series of spectral lines in the hydrogen spectrum.Solve: Substituting into the formula for the Balmer series,

=

91.18 nm

12

12 2n

=

= 91.18 nm 410.3 nm12

162 2

where n = 3, 4, 5, 6, and where we have used n = 6. Likewise for n = 8 and n = 10, = 389.0 nm and = 379.9 nm .

24.2. Model: Balmers formula predicts a series of spectral lines in the hydrogen spectrum.Solve: Balmers formula is

=

91.18 nm

12

12 2n

n = 3, 4, 5, 6,

As , Thus, 91.18 nm 364.7 nmn n n = ( ) =1 0 42 . .

24.3. Model: Balmers formula predicts a series of spectral lines in the hydrogen spectrum.Solve: Using Balmers formula,

= =

= =389.0 nm 91.18 nm1

21

14

1 0 2344 82 2

2

n

nn.

24.4. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Braggcondition.Solve: The Bragg condition is 2d mmcos = , where m = 1, 2, 3, For first and second order diffraction,

2 11d cos = ( ) 2 22d cos = ( )Dividing these two equations,

cos

coscos cos cos cos .

21

21

112 2 2 68 41 5= = ( ) = ( ) =

24.5. Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Braggcondition.Solve: The Bragg condition is 2d mmcos = . For m = 1 and for two different wavelengths,

2 11 1d cos = ( ) 2 11 1d cos = ( ) Dividing these two equations,

cos

cos

cos

coscos . .

=

= = ( ) =

11

1

1

11

1

540 4408 63 80.15 nm

0.20 nm

24.6. Model: The angles corresponding to the various orders of diffraction satisfy the Bragg condition.Solve: The Bragg condition for m = 1 and m = 2 gives

2 1 2 21 2d dcos cos = ( ) = ( )Dividing these two equations,

coscos cos

coscos

. 1 2 1 1245

245

269 3= = = =

24.7. Model: The angles corresponding to the various diffraction orders satisfy the Bragg condition.Solve: The Bragg condition is 2d mmcos = , where m = 1, 2, 3, The maximum possible value of m is thenumber of possible diffraction orders. The maximum value of cosm is 1. Thus,

2 2 2 4 2d m m d= = = ( )( ) = 0.180 nm

0.085 nm.

We can observe up to the fourth diffraction order.

24.8. Model: Use the photon model of light.Solve: The energy of the photon is

E hf h cphoton Js m / s

m J= = = ( )

= 6 63 10

3 0 10500 10

3 98 10348

919

.

.

.

Assess: The energy of a single photon in the visible light region is extremely small.

24.9. Model: Use the photon model of light.Solve: The energy of the x-ray photon is

E hf h c= = = ( )

= 6 63 10

3 0 101 0 10

1 99 10348

916

.

.

.

. Js m / s m

J

Assess: This is a very small amount of energy, but it is larger than the energy of a photon in the visiblewavelength region.

24.10. Model: Use the photon model of light.Solve: The energy of a photon with wavelength 1 is E hf hc1 1 1= = . Similarly, E hc2 2= . Since E2 be equalto 2E1,

hc hc 2 1

2= 2 12= = =600 nm

2300 nm

Assess: A photon with = 300 nm has twice the energy of a photon with = 600 nm. This is an expected result,because energy is inversely proportional to the wavelength.

24.11. Model: Use the photon model of light.Solve: The energy of the single photon is

E hf h cphoton Js m / s

m J= = =

( ) ( )

=

6 63 10 3 0 10

1 0 101 99 10

34 8

619. .

.

.

= = ( ) ( ) = E N Emol A photon J J6 023 10 1 99 10 1 2 1023 19 5. . .Assess: Although the energy of a single photon is very small, a mole of photons has a significant amount of energy.

24.12. Solve: Your mass is, say, m 70 kg and your velocity is 1 m/s. Thus, your momentum is p = mv (70 kg)(1 m/s) = 70 kg m/s. Your de Broglie wavelength is

= =

hp

6 63 10 9 1034

36. Js70 kg m / s

m

24.13. Solve: (a) For an electron, the momentum p mv v= = ( )9 11 10 31. kg . The de Broglie wavelength is = = h

p0 20 10 9. m =

( ) =

0 20 109 11 10

3 6 10934

316

.

.

. m 6.63 10 Js

kg m / s

vv

(b) For a proton, p mv v= = ( )1 67 10 9. kg . The de Broglie wavelength is = = =

( ) =

hp v

v0 20 10 0 20 10 6 63 101 67 10

2 0 109 934

273

. .

.

.

. m m Js

kg m / s

24.14. Solve: The kinetic energy of a particle of mass m is related to its momentum by E p m= 2 2 . Using = h p , for an electron we have

E hm

= =

( )( ) ( ) =

2

2

34 2

31 9 219

26 63 10

2 9 11 10 1 0 102 4 10

.

. .

.

Js

kg m J

24.15. Solve: (a) The baseballs momentum is p = mv = (0.200 kg)(30 m/s) = 6.0 kg m/s. The baseballs deBroglie wavelength is

= = =

hp

6 63 10 1 1 1034

34..

Js6.0 kg m / s

m

(b) Using = =h p h mv , we have

vh

m= =

( ) ( ) =

6 63 10

0 20 101 7 10

34

923.

.

.

Js0.200 kg m

m / s

24.16. Model: The momentum of a wave-like particle has discrete values given by p n h Ln = ( )2 where n =1, 2, 3, .Solve: Because we want the smallest box and the momentum of the electron can not exceed a given value, n mustbe minimum. Thus,

p mv hL

L hmv

1

34

312 26 63 10

2 9 11 100 036= = = =

( )( ) =

.

.

.

Js kg 10 m / s

mm

24.17. Model: A confined particle can have only discrete values of energy.Solve: From Equation 24.14, the energy of a confined electron is

E hmL

nn =2

22

8 n = 1, 2, 3, 4,

The minimum energy is

E hmL

L hmE1

2

21

34

31 18

10

8 86 63 10

8 9 11 10 1 5 102 0 10= = =

( ) ( ) =

.

. .

.

Js

kg J m = 0.20 nm

24.18. Model: Model the 5.0-fm-diameter nucleus as a box of length L = 5.0 fm.Solve: The protons energy is restricted to the discrete values

E hmL

nn

nn = =( )

( ) ( ) = ( )

2

22

34 2 2

27 15 212 2

86 63 10

8 1 67 10 5 0 101 316 10

.

. .

.

Js

kg m J

where n = 1, 2, 3, For n = 1, E1121 316 10= . J , for n = 2, E2

12 121 316 10 4 5 26 10= ( ) = . . J J , and for n =3, E E3 1

119 1 18 10= = . J .

24.19. Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum.Solve: (a) The generalized formula of Balmer

=

91.18 m1 1

2 2m n

with m = 1 and n > 1 accounts for a series of spectral lines. This series is called the Lyman series and the first twomembers are

12

2

2

91 18

1 12

1 13

=

= =

=

. m 121.6 nm 91.18 nm 102.6 nm

For n = 4 and n = 5, 3 4 95 0= =97.3 nm and nm. .(b) The Lyman series converges when n . This means 1 02n and 91.18 nm.(c) For a diffraction grating, the condition for bright (constructive interference) fringes is d ppsin = , where p =1, 2, 3, For first-order diffraction, this equation simplifies to d sin = . For the first and second members ofthe Lyman series, the above condition is d dsin sin 1 1 2 2= = = =121.6 nm and 102.6 nm . Dividing these twoequations yields

sin sin . sin 2 1 10 84375= = ( )

102.6 nm121.6 nm

The distance from the center to the first maximum is y L= tan . Thus,

tan . sin . sin . . 1 1 1 2 214 072 0 84375 14 072 11 84= = = = ( ) ( ) = yL0.376 m1.5 m

Applying the position formula once again,y L2 2 11 84= = ( ) ( ) = =tan tan . 1.5 m 0.314 m 31.4 cm

24.20. Model: The generalized formula of Balmer predicts a series of spectral lines in the hydrogen spectrum.Solve: (a) The generalized formula of Balmer

=

91.18 m1 1

2 2m n

with m = 3, and n > 3 accounts for a series of spectral lines. This series is called the Paschen series and thewavelengths are

=

=

91.18 nm 820.62 213

1 92 2

2

n

n

n

The first four members are 1 2 3 4= = = =1876 nm, 1282 nm, 1094 nm, and 1005 nm.(b) The Paschen series converges when n . This means

1 02 13

2n

( ) =91.18 nm 821 nm

(c) For a diffraction grating, the condition for bright (constructive interference) fringes is d ppsin = , where p =1, 2, 3, For first-order diffraction, this equation simplifies to d sin = . For the first and second members ofthe Paschen series, the condition is d sin 1 1= and d sin 2 2= . Dividing these two equations yields

sin sin sin . sin 2 12

11 10 6834=

=

= ( )1282 nm1876 nm

.

The distance from the center to the first maximum is y L= tan . Thus,

tan .1 1 0 4047= = =yL

0.607 m1.5 m

1 22 03= . sin . sin . . 2 20 6834 22 03 14 85= ( ) = Applying the position formula once again,

y L2 2 14 85= = ( ) = =tan tan . 1.5 m 0.398 m 39.8 cm

24.21. Model: Use the photon model of light.Solve: (a) The wavelength is calculated as follows:

E hf h cgamma Js m / s

J m= =

=( ) ( )

=

6 63 10 3 0 10

1 0 102 0 10

34 8

1312. .

.

.

(b) The energy of a visible-light photon of wavelength 500 nm is

E h cvisible Js m / s

m J= =

( ) ( )

=

6 63 10 3 0 10

500 103 978 10

34 8

919. .

.

The number of photons n such that E nEgamma visible= is

nEE

= =

=

gamma

visible

J J

1 0 103 978 10

2 51 1013

195.

.

.

24.22. Model: Use the photon model.Solve: The energy of a 1000 kHz photon is

E hfphoton Js Hz J= = ( ) ( ) = 6 63 10 1000 10 6 63 1034 3 28. .The energy transmitted each second is 20 103 J . The number of photons transmitted each second is 20 103 J /6 63 10 3 02 1028 31. . = J .

24.23. Model: Use the photon model for the laser light.Solve: (a) The energy is

E hf h cphoton Js m / s

m J= = = ( )

= 6 63 10

3 10633 10

3 14 10348

919

. .

(b) The energy emitted each second is 1 0 10 3. J . The number of photons emitted each second is 1 0 10 3. J /3 14 10 19. J = 3 19 1015. .

24.24. Model: Use the photon model for the incandescent light.Solve: The photons travel in all directions. At a distance of r from the light bulb, the photons spread over a sphereof surface area 4r2. The number of photons per second per unit area at the location of your retina is

3 104 10 10

2 387 1018 1

3 29 1 2

( ) =

s

m s m

.

The number of photons that enter your pupil per second is2 387 10 3 0 10 6 75 109 1 2 3 2 4 1. . . ( ) = s m m s

24.25. Model: Use the photon model of light and the Bragg condition for diffraction.Solve: The Bragg condition for the reflection of x-rays from a crystal is 2d mmcos = . To determine the anglesof incidence m, we need to first calculate . The wavelength is related to the photons energy as E hc= . Thus,

= =( ) ( )

=

hcE

6 63 10 3 0 101 50 10

1 326 1034 8

1510. .

.

.

Js m / s J

m

From the Bragg condition,

mm

dm

m= =

( )( )

= ( ) = ( ) =

cos cos.

.

cos . cos . .1 110

91

11

21 326 102 0 21 10

0 3157 0 3157 71 6 m

m

Likewise, 2 1 0 3157 2 50 8= ( ) = cos . . and 3 18 7= . . Note that 4 1 0 3157 4= ( )cos . is not allowed becausethe cos cannot be larger than 1. Thus, the x-rays will be diffracted at angles of incidence equal to 18.7, 50.8, and71.6.

24.26. Model: The angles for which diffraction from parallel planes occurs satisfy the Bragg condition.Solve: We cannot assume that these are the first and second order diffractions. The Bragg condition is2d mmcos = . We have

2 45 6 2 21 0 1d m d mcos . cos . = = +( ) Notice that m decreases as m increases, so 21.6 corresponds to the larger value of m. Dividing these two equations,

cos .

cos ..

45 621 0 1

0 7494 3

=

+= =

m

mm

Thus these are the third and fourth order diffractions. Substituting into the Bragg condition,

d =

= =

3 0 070 102 45 6

1 50 109

10.

cos ..

m m 0.150 nm

24.27. Model: The x-ray diffraction angles satisfy the Bragg condition.Solve: (a) The Bragg condition ( 2d mmcos = ) for normal incidence, m = 0, simplifies to 2d = m. For a thinfilm of a material on a substrate where nair < nmaterial < nsubstrate, constructive interference between the two reflectedwaves occurs when 2d = m, where is the wavelength inside the material.(b) From a thin film with a period of 1.2 nm, that is, with d = 1.2 nm, the two longest x-ray wavelengths that willreflect at normal incidence are

121

=

d 2

22

=

d

This means that 1 2= ( ) =1.2 nm 2.4 nm and 2 = 1.2 nm .

24.28. Solve: A small fraction of the light wave of an appropriate wavelength is reflected from each littlebump in the refractive index. These little bumps act like the atomic planes in a crystal. The light will be stronglyreflected (and hence blocked in transmission) if it satisfies the Bragg condition at normal incidence ( = 0).

2d m mn

= = glassglass

= =( )( )

=

22 0 45 10 1 50

11 35

6dnm

glass m. .. m

24.29. Model: Particles have a de Broglie wavelength given by = h p . The wave nature of the particlescauses an interference pattern in a double-slit apparatus.Solve: (a) Since the speed of the neutron and electron are the same, the neutrons momentum is

p m v mm

m vm

mm v

m

mpn n n n

e

e nn

e

e en

e

e= = = =

where mn and me are the neutrons and electrons masses. The de Broglie wavelength for the neutron is

nn e

e

n

ee

n

= = =

hp

hp

pp

m

m

From Section 22.2 on double-slit interference, the fringe spacing is y L d= / . Thus, the fringe spacing for theelectron and neutron are related by

y y mm

yn ne

ee

n

e

kg kg

m m 0.818 m= = =

( ) = =

9 11 101 67 10

1 5 10 8 18 1031

273 7.

.

. .

(b) If the fringe spacing has to be the same for the neutrons and the electrons,

y y hm v

hm v

v vm

me n e n

e e n n

n ee

n

m / s kg kg

m / s= = = = = ( )

=

2 0 10 9 11 101 67 10

1 09 10631

273

.

.

.

.

24.30. Model: Electrons have a de Broglie wavelength given by = h p . The wave nature of the electronscauses a diffraction pattern.Solve: The width of the central maximum of a single-slit diffraction pattern is given by Equation 22.22:

wL

a

Lhap

Lhamv

= = = =

( ) ( )( ) ( ) ( ) = =

2 2 2 2 6 63 101 0 10 9 11 10 1 5 10

9 70 1034

6 31 64 1.0 m Js

m kg m / s m 0.970 mm

.

. . .

.

24.31. Model: Neutrons have a de Broglie wavelength given by = h p . The wave nature of the neutronscauses a double-slit interference pattern.Solve: Measurements show that the spacing between the m = 1 and m = 1 peaks is 1.4 times as long as the lengthof the reference bar, which gives the real fringe separation y = 70 m. Similarly, the spacing between the m = 2and m = 2 is 2.8 times as long as the length of the reference bar and yields y = 70 m.

The fringe separation in a double-slit experiment is y L d= . Hence,

= = = = =( )( )

( ) ( ) ( ) =

y dL

hp

hmv

y dL

vhLy md

6 63 1070 10 1 67 10 0 10 10

34

6 27 3

.

. .

Js 3.0 m m kg m

170 m / s

24.32. Model: Electrons have a de Broglie wavelength given by = h p .Visualize: Please refer to Figure 24.11. Notice that a scattering angle = 60 corresponds to an angle ofincidence = 30 .Solve: Equation 24.6 describes the Davisson-Germer experiment: D mmsin 2 ( ) = . Assuming m = 1, this equationsimplifies to Dsin2 = . Using = h mv , we have

D hmv

= =

( ) ( ) ( ) = =

sin.

. . sin.

26 63 10

9 11 10 4 30 10 601 95 10

34

31 610

Js

kg m / s m 0.195 nm

24.33. Model: A confined particle can have only discrete values of energy.Solve: (a) Equation 24.14 simplifies to

E hmL

n nn = =( )

( ) ( ) = ( )

2

22

34 2

31 9 219 2

86 63 10

8 9 11 10 0 70 101 231 10

.

. .

.

Js

kg m J

Thus, E119 2 191 231 10 1 1 23 10= ( )( ) = . .J J , E2 19 2 191 231 10 2 4 92 10= ( )( ) = . .J J , and E3 181 11 10= . J.

(b) The energy is E E2 1 19 19 194 92 10 1 23 10 3 69 10 = = . . . J J J .(c) Because energy is conserved, the photon will carry an energy of E E2 1 193 69 10 = . J . That is,

E E E hf hc hcE E2 1 2 1

34 8

19

6 63 10 3 0 103 69 10

539 = = = =

=

( ) ( )

=

photon

Js m / s J

nm . .

.

24.34. Model: A particle confined in a one-dimensional box has discrete energy levels.Solve: (a) Equation 24.14 for the n = 1 state is

E hmLn

= =

( )( )( ) =

2

2

34 2

3 264

86 63 10

8 10 105 50 10

.

.

Js kg 0.10 m

J

The minimum energy of the Ping-Pong ball is E1645 50 10= . J .

(b) The speed is calculated as follows:

E mv v v164 2 1

23 2

64

3315 50 10 10 10

2 5 50 1010 10

3 32 10= = = ( ) = ( )

=

.

.

.J kg J

kg m / s12

24.35. Model: A particle confined in a one-dimensional box has discrete energy levels.Solve: Using Equation 24.14 for n = 1 and 2,

E E hmL2 1

2

22 2

82 1 = ( ) = ( )

( ) ( ) =

1 0 108 9 11 10

3 1 809 10192

31 2

37

2..

.

J 6.63 10 Js

kg Jm34 2

L L

=

= =

L 1 809 101 0 10

1 35 1037

199.

.

.

Jm J

m 1.35 nm2

24.36. Model: A particle confined in a one-dimensional box has discrete energy levels.Solve: From Equation 24.14,

E hmL

nn =2

22

8 E h

mLnn+ = +( )1

2

22

81

Taking the ratio,

EE

n

n

n

n

+=

+ 121

Thus

n

n

+= = =

1 6 43 6

1 333 43

.

.

.

The two allowed energies have n = 3 and n = 4. Using n = 3 in the first energy equation,

ELn

= =( )( ) ( )

3 6 108 9 11 10

31934 2

31 22

.

.

J 6.63 10 Js

kgL = 1.23 nm

24.37. Model: A particle confined in a one-dimensional box of length L has the discrete energy levels given byEquation 24.14.Solve: (a) Since the energy is entirely kinetic energy,

E hmL

npm

mv vhmL

n nn n n= = = = =2

22

212

2

8 2 21 2 3, , , K

(b) The first allowed velocity is

v1

34

31 96 63 10

2 9 11 10 0 20 10=

( ) ( ) =

.

. .

Js kg m

1.819 10 m / s6

For n = 2 and n = 3, v2 = 3.64 106 m/s and v3 = 5.46 106 m/s.

24.38. Model: The allowed energies of a particle of mass m in a two-dimensional square box of side L are

E hmL

n mnm = +( )2 2 2 28Solve: (a) The minimum energy for a particle is for n = m = 1:

E E hmL

hmLmin

= = +( ) =11 2 2 2 22

281 1

4(b) The five lowest allowed energies are Emin , 52 Emin (for n = 1, m = 2 and n = 2, m = 1), 4Emin (for n = 2, m = 2),5Emin (for n = 1, m = 3 and n =3, m = 1), and 132 Emin (for n = 2, m = 3 and n = 3, m = 2).

24.39. Model: Sets of parallel planes in a crystal diffract x-rays.Visualize: Please refer to Figure 24.7.Solve: The Bragg diffraction condition is 2d mmcos = . The plane spacing is dA = 0.20 nm and the x-raywavelength is = 0 12. nm. Thus

cos.

.

. cos . . mm

dm

m= =( )( ) = ( ) = ( ) =

20 12 10

2 0 20 100 3 0 3 72 5

9

9 11

AA

m

m

Likewise for m = 2 and m = 3, A2 = ( ) = cos . .1 0 6 53 1 and A3 = ( ) = cos . .1 0 9 25 8 . These three angles for thex-ray diffraction peaks match the peaks shown in Figure 24.7c.(b) The new interplaner spacing is d dB A= =2 0 141. nm (see Figure 24.7b). The Bragg condition for the tiltedatomic planes becomes

cos . mm

dm= =

20 4243

B

For m = 1, B1 = ( ) = cos . .1 0 4243 64 9 . For m = 2, B2 = ( ) = cos . .1 0 8486 31 9 .(c) The crystal is already tipped by 45 to get the tilted planes (see Figure 24.7b). So, for m = 1, 1 64 9 45= =.19 9. . = + = 64 9 45 109 9. . also, but we cant see beyond 90. For m = 2, 2 31 9 45 76 9= + = . . . These twoangles match the angles in the diffraction peaks of the tilted planes.

24.40. Model: Sets of parallel planes in a crystal diffract x-rays.Visualize: Please refer to Figure CP24.40.Solve: The Bragg diffraction condition is 2d mmcos = , where d is the interplanar separation. Because smallerm values correspond to higher angles of incidence, the diffraction angle of 71.3 in the x-ray intensity plot mustcorrespond to m = 1. This means

2 71 3 1 0 10 10 0 10 102 71 3

1 56 1099

10d dcos . . .cos .

. = ( ) = ( ) =

m m

m

The cosines of the three angles in the X-ray intensity plot are cos .71 3 = 0.321, cos .50 1 = 0.642, and cos .15 8 =0.962. These are in the ratio 1:2:3, which tells us that these are the m =1, 2, and 3 diffraction peaks from a singleset of planes with d = 0 156. nm.

We can see from the figure that the atomic spacing D of this crystal is related to the interplanar separation d by

D d=

=

=

sin600.156 nm

sin600.180 nm

24.41. Model: Electrons have a de Broglie wavelength given by = h p . Trapped electrons in theconfinement layer behave like a de Broglie wave in a closed-closed tube or like a string fixed at both ends.Solve: (a) The four longest standing-wave wavelengths in the layer are = 2L , L, 23 L , and 12 L . This followsfrom the general relation for closed-closed tubes: = 2L n/ . Thus, = 10.0 nm, 5.0 nm, 3.33 nm, and 2.50 nm.(b) We have

p mv h v hm

= = = =

( ) =

6 63 10

9 11 100 7278 1034

31

3.

.

. Js kg

m / s2

Using the above four longest values of we get the four smallest values of v. Thus,

v1

3

940 7278 10

10 0 1010=

=

.

.

m / s m

7.28 m / s2

v2510= 1.46 m / s , v3 510= 2.18 m / s, and v4 510= 2.91 m / s .

24.42. Model: As light is diffracted by matter, matter can also be diffracted by light.Visualize: Please refer to Figure CP24.42.Solve: The de Broglie wavelength of the sodium atoms is

= = = ( )( ) =

hp

hmv

6 63 103 84 10

3 45 1034

2610.

.

.

Js kg 50 m / s

m

The slit spacing of the diffraction grating is d = = =12 12laser 600 nm 300 nm. Using the diffraction gratingequation with m = 1, we have

dd

sin sin . = ( ) = = 1 1 151 10 3 radIn the small-angle approximation, sin tan = y L . We get

y L= = ( ) ( ) =sin . 1.0 m 1.15 mm1 151 10 3

EOC_Solution_24_01.pdfEOC_Solution_24_02.pdfEOC_Solution_24_03.pdfEOC_Solution_24_04.pdfEOC_Solution_24_05.pdfEOC_Solution_24_06.pdfEOC_Solution_24_07.pdfEOC_Solution_24_08.pdfEOC_Solution_24_09.pdfEOC_Solution_24_10.pdfEOC_Solution_24_11.pdfEOC_Solution_24_12.pdfEOC_Solution_24_13.pdfEOC_Solution_24_14.pdfEOC_Solution_24_15.pdfEOC_Solution_24_16.pdfEOC_Solution_24_17.pdfEOC_Solution_24_18.pdfEOC_Solution_24_19.pdfEOC_Solution_24_20.pdfEOC_Solution_24_21.pdfEOC_Solution_24_22.pdfEOC_Solution_24_23.pdfEOC_Solution_24_24.pdfEOC_Solution_24_25.pdfEOC_Solution_24_26.pdfEOC_Solution_24_27.pdfEOC_Solution_24_28.pdfEOC_Solution_24_29.pdfEOC_Solution_24_30.pdfEOC_Solution_24_31.pdfEOC_Solution_24_32.pdfEOC_Solution_24_33.pdfEOC_Solution_24_34.pdfEOC_Solution_24_35.pdfEOC_Solution_24_36.pdfEOC_Solution_24_37.pdfEOC_Solution_24_38.pdfEOC_Solution_24_39.pdfEOC_Solution_24_40.pdfEOC_Solution_24_41.pdfEOC_Solution_24_42.pdf