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

λ =−

91.18 nm1

2

12 2n

⇒ =−

=λ 91.18 nm410.3 nm1

2

1

62 2

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

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

λ =−

91.18 nm1

2

12 2n

n = 3, 4, 5, 6, …

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

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

λ = =−

⇒ − = ⇒ =389.0 nm 91.18 nm

1

2

11

4

10 2344 8

2 2

2

nn

n.

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

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

cos

coscos cos cos cos .

θθ

θ θ2

12

11

12 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 mmcosθ λ= . For m = 1 and for two different wavelengths,

2 11 1d cosθ λ= ( ) 2 11 1d cos ′ = ( ) ′θ λ

Dividing these two equations,

cos

cos

cos

coscos . .

′ = ′ ⇒ ′°

= ⇒ ′ = ( ) = °−θθ

λλ

θ θ1

1

1

1

11

1

540 4408 63 8

0.15 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 cosθ λ θ λ= ( ) = ( )Dividing these two equations,

coscos cos

coscos

.θ θ θ12

11

2

45

2

45

269 3= = ° ⇒ = °

= °−

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

22 2

4 2d m md= ⇒ = = ( )

( ) =λλ

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 hc

photon Js m / s

m J= = = ×( ) ×

×

= ×−−

−

λ6 63 10

3 0 10

500 103 98 1034

8

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 hc= =

= ×( ) ×

×

= ×−−

−

λ6 63 10

3 0 10

1 0 101 99 1034

8

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 λ1 is E hf hc1 1 1= = λ . Similarly, E hc2 2= λ . Since E2 be equalto 2E1,

hc hc

λ λ2 1

2= ⇒ λ λ2

1

2= = =600 nm

2300 nm

Assess: A photon with λ = 300 nm has twice the energy of a photon with λ = 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 hc

photon

Js m / s

m J= =

=

×( ) ×( )×

= ×−

−−

λ6 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

λ = = × ≈ ×−

−h

p

6 63 109 10

3436. Js

70 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

λ = = × −h

p0 20 10 9. m ⇒ × = ×

×( ) ⇒ = ×−−

−0 20 109 11 10

3 6 10934

31

6..

. 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

λ = = × ⇒ × = ××( ) ⇒ = ×− −

−

−

h

p vv0 20 10 0 20 10

6 63 10

1 67 102 0 109 9

34

27

3. ..

.. 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 . Usingλ = h p , for an electron we have

Eh

m= =

×( )×( ) ×( )

= ×−

− −

−2

2

34 2

31 9 219

2

6 63 10

2 9 11 10 1 0 102 4 10

λ.

. ..

Js

kg m J

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

λ = = × = ×−

−h

p

6 63 101 1 10

3434.

. Js

6.0 kg m / s m

(b) Using λ = =h p h mv , we have

vh

m= = ×

( ) ×( ) = ×−

−−

λ6 63 10

0 20 101 7 10

34

9

23.

..

Js

0.200 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 mvh

LL

h

mv1

34

312 2

6 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

Eh

mLnn =

2

22

8 n = 1, 2, 3, 4, …

The minimum energy is

Eh

mLL

h

mE1

2

21

34

31 18

10

8 8

6 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 proton’s energy is restricted to the discrete values

Eh

mLn

nnn = =

×( )×( ) ×( )

= ×( )−

− −

−2

22

34 2 2

27 15 212 2

8

6 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

λ =−

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

λ λ1

2

2

2

91 18

11

21

1

3

=−

= =−

=. m121.6 nm

91.18 nm102.6 nm

For n = 4 and n = 5, λ λ3 4 95 0= =97.3 nm and nm. .(b) The Lyman series converges when n → ∞ . This means 1 02n → → and 91.18 nm.λ(c) For a diffraction grating, the condition for bright (constructive interference) fringes is d ppsinθ λ= , where p =1, 2, 3, … For first-order diffraction, this equation simplifies to d sinθ λ= . For the first and second members ofthe Lyman series, the above condition is d dsin sinθ λ θ λ1 1 2 2= = = =121.6 nm and 102.6 nm . Dividing these twoequations yields

sin sin . sinθ θ θ2 1 10 84375=

= ( )102.6 nm

121.6 nm

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

tan . sin . sin . .θ θ θ θ11

1 2 214 072 0 84375 14 072 11 84= = ⇒ = ° ⇒ = ( ) °( ) ⇒ = °y

L

0.376 m

1.5 m

Applying the position formula once again,

y L2 2 11 84= = ( ) °( ) = =tan tan .θ 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

λ =−

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

λ =−

=−

91.18 nm 820.62 2

1

3

1 92 2

2

n

n

n

The first four members are λ λ λ λ1 2 3 4= = = =1876 nm, 1282 nm, 1094 nm, and 1005 nm.(b) The Paschen series converges when n → ∞ . This means

102 1

32n

→ ⇒ →( )

=λ 91.18 nm821 nm

(c) For a diffraction grating, the condition for bright (constructive interference) fringes is d ppsinθ λ= , where p =1, 2, 3, … For first-order diffraction, this equation simplifies to d sinθ λ= . For the first and second members ofthe Paschen series, the condition is d sinθ λ1 1= and d sinθ λ2 2= . Dividing these two equations yields

sin sin sin . sinθ θ λλ

θ θ2 12

11 10 6834=

=

= ( )1282 nm

1876 nm.

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

tan .θ11 0 4047= = =y

L

0.607 m

1.5 m⇒ θ1 22 03= °. ⇒ sin . sin . .θ θ2 20 6834 22 03 14 85= ( ) ° ⇒ = °

Applying the position formula once again,

y L2 2 14 85= = ( ) ° = =tan tan .θ 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 hc

gamma

Js m / s

J m= =

⇒ =

×( ) ×( )×

= ×−

−−

λλ

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 hc

visible

Js m / s

m J=

=

×( ) ×( )×

= ×−

−−

λ6 63 10 3 0 10

500 103 978 10

34 8

919

. ..

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

nE

E= = ×

×= ×

−

−gamma

visible

J

J

1 0 10

3 978 102 51 10

13

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 hc

photon Js m / s

m J= =

= ×( ) ×

×

= ×−−

−

λ6 63 10

3 10

633 103 14 1034

8

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 4πr2. The number of photons per second per unit area at the location of your retina is

3 10

4 10 102 387 10

18 1

3 29 1 2×

×( )= ×

−− − s

m s m

π.

The number of photons that enter your pupil per second is

2 387 10 3 0 10 6 75 109 1 2 3 2 4 1. . .× × ×( ) = ×− − − − s m m sπ

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 mmcosθ λ= . To determine the anglesof incidence θm, we need to first calculate λ. The wavelength is related to the photon’s energy as E hc= λ . Thus,

λ = =×( ) ×( )

×= ×

−

−−hc

E

6 63 10 3 0 10

1 50 101 326 10

34 8

1510

. .

..

Js m / s

J m

From the Bragg condition,

θ λ θm

m

d

mm=

=×( )

×( )

= ( ) ⇒ = ( ) = °− −−

−− −cos cos

.

.cos . cos . .1 1

10

9

11

1

2

1 326 10

2 0 21 100 3157 0 3157 71 6

m

m

Likewise, θ21 0 3157 2 50 8= ×( ) = °−cos . . and θ3 18 7= °. . Note that θ4

1 0 3157 4= ×( )−cos . is not allowed becausethe cosθ 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 mmcosθ λ= . We have

2 45 6 2 21 0 1d m d mcos . cos .° = ° = +( )λ λ

Notice that θm decreases as m increases, so 21.6° corresponds to the larger value of m. Dividing these two equations,

cos .

cos ..

45 6

21 0 10 7494 3

°°

=+

= ⇒ =m

mm

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

d = × ×°

= × =−

−3 0 070 10

2 45 61 50 10

910.

cos ..

m m 0.150 nm

24.27. Model: The x-ray diffraction angles satisfy the Bragg condition.Solve: (a) The Bragg condition ( 2d mmcosθ λ= ) for normal incidence, θm = 0°, simplifies to 2d = mλ. For a thinfilm of a material on a substrate where nair < nmaterial < nsubstrate, constructive interference between the two reflectedwaves occurs when 2d = mλ, where λ 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

λ1

2

1= d

λ2

2

2= d

This means that λ1 2= ( ) =1.2 nm 2.4 nm and λ2 = 1.2 nm .

24.28. Solve: A small fraction of the light wave of an appropriate wavelength is reflected from each little“bump” 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 (θ = 0).

2d mm

n= =λ λ

glassglass

⇒ = =×( )( )

=−

λ2 2 0 45 10 1 50

11 35

6dn

mglass m. .

. µm

24.29. Model: Particles have a de Broglie wavelength given by λ = 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 neutron’s momentum is

p m vm

mm v

m

mm v

m

mpn n n

n

ee n

n

ee e

n

ee= = = =

where mn and me are the neutron’s and electron’s masses. The de Broglie wavelength for the neutron is

λ λnn e

e

ne

e

n

= = =h

p

h

p

p

p

m

m

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

∆ ∆ ∆y ym

myn

n

ee

e

ne

kg

kg m m 0.818 m= = = ×

×

×( ) = × =−

−− −λ

λµ9 11 10

1 67 101 5 10 8 18 10

31

273 7.

.. .

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

∆ ∆y yh

m v

h

m vv v

m

me n e ne e n n

n ee

n

m / s kg

kg m / s= ⇒ = ⇒ = ⇒ = = ×( ) ×

×

= ×−

−λ λ 2 0 109 11 10

1 67 101 09 106

31

273.

.

..

24.30. Model: Electrons have a de Broglie wavelength given by λ = 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

Lh

ap

Lh

amv= = = =

( ) ×( )×( ) ×( ) ×( ) = × =

−

− −−2 2 2 2 6 63 10

1 0 10 9 11 10 1 5 109 70 10

34

6 31 6

4λ 1.0 m Js

m kg m / s m 0.970 mm

.

. . ..

24.31. Model: Neutrons have a de Broglie wavelength given by λ = 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 µm. 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 µm.

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

λ = ⇒ = = ⇒ = =×( )( )

×( ) ×( ) ×( ) =−

− − −

∆ ∆∆

y d

L

h

p

h

mv

y d

Lv

hL

y md

6 63 10

70 10 1 67 10 0 10 10

34

6 27 3

.

. .

Js 3.0 m

m kg m170 m / s

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

simplifies to Dsin2θ λ= . Using λ = h mv , we have

Dh

mv= = ×

×( ) ×( ) °( )= × =

−

−−

sin

.

. . sin.

2

6 63 10

9 11 10 4 30 10 601 95 10

34

31 6

10

θ 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

Eh

mLn nn = =

×( )×( ) ×( )

= ×( )−

− −

−2

22

34 2

31 9 219 2

8

6 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 E3181 11 10= × −. J.

(b) The energy is E E2 119 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 1193 69 10− = × −. J . That is,

E E E hfhc hc

E E2 12 1

34 8

19

6 63 10 3 0 10

3 69 10539− = = = ⇒ =

−=

×( ) ×( )×

=−

−photon

Js m / s

J nm

λλ

. .

.

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

Eh

mLn = =×( )

×( )( )= ×

−

−−

2

2

34 2

3 264

8

6 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 10

10 103 32 10= × = = ×( ) ⇒ =

×( )×

= ×− −−

−−.

..J kg

J

kg m / s1

2

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 Eh

mL2 1

2

22 2

82 1− = −( ) ⇒ × =

×( )×( ) ( ) = ×−

−

−

−

1 0 108 9 11 10

31 809 1019

2

31 2

37

2..

. J

6.63 10 Js

kg

Jm34 2

L L

⇒ = ××

= × =−

−−L

1 809 10

1 0 101 35 10

37

199.

..

Jm

J m 1.35 nm

2

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

Eh

mLnn =

2

22

8 E

h

mLnn+ = +( )1

2

2

2

81

Taking the ratio,

E

E

n

nn

n

+ = +

121

Thus

n

n

+ = = =1 6 4

3 61 333

4

3

.

..

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

EL

n = × =×( )×( ) ( ) ⇒−

−

−3 6 108 9 11 10

319

34 2

31 2

2..

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,

Eh

mLn

p

mmv v

h

mLn nn n n= = = ⇒ = =

2

22

212

2

8 2 21 2 3, , , K

(b) The first allowed velocity is

v1

34

31 9

6 63 10

2 9 11 10 0 20 10= ×

×( ) ×( ) = ×−

− −

.

. .

Js

kg m1.819 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

Eh

mLn mnm = +( )

2

22 2

8

Solve: (a) The minimum energy for a particle is for n = m = 1:

E Eh

mL

h

mLmin = = +( ) =11

2

22 2

2

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 mmcosθ λ= . The plane spacing is dA = 0.20 nm and the x-raywavelength is λ = 0 12. nm. Thus

cos.

.. cos . .θ λ θm

m

d

mm= =

×( )×( ) = ( ) ⇒ = ( ) = °

−

−−

2

0 12 10

2 0 20 100 3 0 3 72 5

9

9 11

AA

m

m

Likewise for m = 2 and m = 3, θA2 = ( ) = °−cos . .1 0 6 53 1 and θA3 = ( ) = °−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 .θ λm

m

dm= =

20 4243

B

For m = 1, θB1 = ( ) = °−cos . .1 0 4243 64 9 . For m = 2, θB2 = ( ) = °−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, θ1 64 9 45= ° − ° =.19 9. °.θ = ° + ° = °64 9 45 109 9. . also, but we can’t see beyond 90°. For m = 2, θ2 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 mmcosθ λ= , 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 10 100 10 10

2 71 31 56 109

910d 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

Dd=

°=

°=

sin60

0.156 nm

sin600.180 nm

24.41. Model: Electrons have a de Broglie wavelength given by λ = 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 λ = 2L , L, 2

3 L , and 12 L . This follows

from the general relation for closed-closed tubes: λ = 2L n/ . Thus, λ = 10.0 nm, 5.0 nm, 3.33 nm, and 2.50 nm.(b) We have

p mvh

vh

m= = ⇒ = = ×

×( ) = ×−

−

−

λ λ λ λ6 63 10

9 11 10

0 7278 1034

31

3.

.

. Js

kg

m / s2

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

v1

3

940 7278 10

10 0 1010= ×

×= ×

−

−

.

.

m / s

m7.28 m / s

2

v2510= ×1.46 m / s , v3

510= ×2.18 m / s, and v4510= ×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

λ = = = ××( )( )

= ×−

−−h

p

h

mv

6 63 10

3 84 103 45 10

34

26

10.

..

Js

kg 50 m / s m

The slit spacing of the “diffraction grating” is d = = =12

12λlaser 600 nm 300 nm. Using the diffraction grating

equation with m = 1, we have

dd

sin sin .θ λ θ λ= ( ) ⇒ = = × −1 1 151 10 3 rad

In the small-angle approximation, sin tanθ θ≅ = y L . We get

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