Chapter 24

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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 xray 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 wavelike 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 5.0fmdiameter 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 firstorder 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 firstorder 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 visiblelight 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 xrays 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 xrays 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 xray 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 xray 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 doubleslit 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 doubleslit 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 singleslit 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 doubleslit 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 doubleslit 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 DavissonGermer 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 onedimensional 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 PingPong 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 onedimensional 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 onedimensional 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 onedimensional 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 twodimensional 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 xrays.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 xraywavelength 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 thexray 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 xrays.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 xray 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 Xray 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 closedclosed tube or like a string fixed at both ends.Solve: (a) The four longest standingwave wavelengths in the layer are = 2L , L, 23 L , and 12 L . This followsfrom the general relation for closedclosed 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 smallangle approximation, sin tan = y L . We get
y L= = ( ) ( ) =sin . 1.0 m 1.15 mm1 151 10 3
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