Ch 29 ISM 8e Solutions

CHAPTER 29 PARTICLES AND WAVES

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (a) At the higher temperature, the intensity per unit wavelength is greater, and the aximum

occurs at a shorter wavelength (see Section 29.2). 2. (b) An X-ray photon has a much greater frequency than does a microwave photon (see

Section 24.2). The X-ray photon also has a greater energy E, because E = hf (Equation 29.2), where h is Planck’s constant and f is the frequency. The wavelength λ and frequency of a photon are related by λ = c/f (Equation 16.1), where c is the speed of light in a vacuum. Since the microwave photon has the smaller frequency, it has the greater wavelength.

3. (c) A green photon has a greater frequency than does a red photon (see Section 24.2).

Therefore, the green photon possesses a greater energy E, because E = hf (Equation 29.2), where h is Planck’s constant and f is the frequency.

4. (e) As the wavelength of a photon becomes smaller, its frequency and its energy become

larger (see Section 29.3). Since the energy of an incident photon is equal to the maximum kinetic energy of the photoelectron plus the work function of the metal, increasing the incident energy increases the maximum kinetic energy of the photoelectron. The work function is a characteristic of the metal, and does not depend on the photon.

5. (d) The maximum kinetic energy of the ejected photoelectrons depends on the energy of the

incident photons. Doubling the number of photons per second that strikes the surface does not change the energy of the incident photons, which depends only on the frequency of the photons. Thus, the maximum kinetic energy of the ejected photoelectrons does not change. However, doubling the number of photons per second that strikes the surface means that twice as many photoelectrons per second are ejected from the surface.

6. (b) Whether or not electrons are ejected from the surface of the metal depends on the energy of

the incident photons and the work function of the metal The work function depends on the type of metal (e.g., aluminum or copper) from which the plate is made. (See Section 29.3).

7. W0 = 3.6 × 10−19 J 8. KEmax = 2.7 × 10−20 J 9. (d) According to the discussion in Section 29.4, a photon has a momentum whose

magnitude p is related to its wavelength λ by p = h/λ.

Chapter 29 Answers to Focus on Concepts Questions 1511

10. (b) In the Compton effect, an X-ray photon strikes an electron and, like two billiard balls (particles) colliding on a pool table, the X-ray photon scatters in one direction and the electron recoils in another direction after the collision.

11. (a) In the Compton effect, some of the energy of the incident photon is given to the recoil

electron. Therefore, the energy of the scattered photon is less than that of the incident photon. Since the energy of a photon depends inversely on its wavelength (see Section 29.3), the wavelength of the scattered photon is greater than that of the incident photon.

12. λ′ = 0.35 nm 13. (c) The de Broglie wavelength λ depends inversely on the magnitude p of the momentum;

λ = h/p (Equation 29.8), where h is Planck’s constant. Particle A, having the smaller charge, has the smaller electric potential energy (see Section 19.2). Consequently, after accelerating through the potential difference, particle A has the smaller kinetic energy, and hence, the smaller momentum. Thus, particle A has the longer de Broglie wavelength.

14. (e) The de Broglie wavelength λ depends inversely on the magnitude p of the momentum;

λ = h/p (Equation 29.8), where h is Planck’s constant. Therefore, as the momentum decreases, the wavelength increases, and vice versa. In A, the proton is moving opposite to the direction of the electric field, so the proton is slowing down, and its momentum is decreasing. In B, the proton is accelerating, and its momentum is increasing. In C, the proton moves parallel to the magnetic field. According to the discussion in Section 21.2, the proton does not experience a force, so its momentum remains constant. In D the proton is moving perpendicular to the direction of the magnetic field. According to the discussion in Section 21.3, such a situation does not change the magnitude of the proton’s momentum.

15. λ = 2.1 × 10−14 m 16. (b) According to the Heisenberg uncertainty principle, the uncertainty Δy in a particle’s

position is related to the uncertainty Δpy in its momentum by (see Equation 29.10)

( )/ 4 yy h pπΔ ≥ Δ . If Δpy = 0 kg⋅m/s, then Δy becomes infinitely large.

1512 PARTICLES AND WAVES

CHAPTER 29 PARTICLES AND WAVES PROBLEMS 1. REASONING AND SOLUTION The energy of a single photon is

E = hf = (6.63 × 10–34 J⋅s)(98.1 × 106 Hz) = 6.50 × 10–26 J The number of photons emitted per second is

Power radiatedEnergy per photon

WJ

photons / s=××

= ×−

5 0 106 50 10

7 7 104

2629.

..

2. REASONING The energy of a photon of frequency f is, according to Equation 29.2,

E hf= , where h is Planck's constant. Since the frequency and wavelength are related by f c= / λ (see Equation 16.1), the energy of a photon can be written in terms of the

wavelength as E hc= / λ . These expressions can be solved for both the wavelength and the frequency.

SOLUTION a. The wavelength of the photon is

λ = =× ⋅ ×

×= ×

hcE

(6.63 J s)(3.00 m / s)1.22 J

1.63 m–34 8

–18–710 10

1010

b. Using the answer from part (a), we find that the frequency of the photon is

f c= =

××

= ×λ

3.00 m / s1.63 m

1.84 Hz8

–71510

1010

Alternatively, we could use Equation 29.2 directly to obtain the frequency:

f Eh

= =×

× ⋅= ×

1.22 JJ s

1.84 Hz–18

–341510

6 63 1010

.

c. The wavelength and frequency values shown in Figure 24.9 indicate that this photon

corresponds to electromagnetic radiation in the ultraviolet region of the electromagnetic

spectrum.

Chapter 29 Problems 1513

3. SSM REASONING According to Equation 29.3, the work function W0 is related to the photon energy hf and the maximum kinetic energy KEmax by W hf0 = – maxKE . This expression can be used to find the work function of the metal.

SOLUTION KE max is 6.1 eV. The photon energy (in eV) is, according to Equation 29.2,

hf = × ××

FHG

IKJ =⋅6.63 10 J s 3.00 10 Hz 1 eV

1.6012.4 eV–34 15c hc h

10–19 J

The work function is, therefore,

W hf0 6 3= = =– .maxKE 12.4 eV – 6.1ev eV

4. REASONING According to Equation 29.3, the relation between the photon energy, the

maximum kinetic energy of an ejected electron, and the work function of a metal surface is

max 0 Photon WorkMaximumenergy functionkinetic energy

of ejectedelectron

KE hf W= +

Equation 16.1 relates the frequency f of a photon to its wavelength λ via f = c/λ, where c is

the speed of light in a vacuum. The maximum kinetic energy KEmax is related to the mass m and maximum speed vmax of the ejected electron by KEmax = 1

22mv (Equation 6.2). With

these substitutions, Equation 29.3 becomes

21max 0 max 02KE or hchf W mv W

λ= + = + (1)

SOLUTION Solving Equation (1) for the wavelength gives

( ) ( )( ) ( )

21max 02

34 87

A 231 5 1912

6.63 10 J s 3.00 10 m/s2.75 10 m

9.11 10 kg 7.30 10 m/s 4.80 10 J

hcmv W

λ

λ−

−

− −

=+

× ⋅ ×= = ×

× × + ×


( ) ( )( ) ( )

34 87

B 231 5 1912

6.63 10 J s 3.00 10 m/s3.35 10 m

9.11 10 kg 5.00 10 m/s 4.80 10 Jλ

−−

− −

× ⋅ ×= = ×

× × + ×

5. SSM REASONING The energy of the photon is related to its frequency by Equation

29.2, E hf= . Equation 16.1, v f= λ , relates the frequency and the wavelength for any wave.

SOLUTION Combining Equations 29.2 and 16.1, and noting that the speed of a photon is

c, the speed of light in a vacuum, we have

λ = = = =× ⋅ ×

×= × −c

fc

E hhcE( / )

.(6.63 m = 310 nm10 1010

3 1 10 7–34 8

–19

J s)(3.0 m / s)6.4 J

6. REASONING The photons of this wave must carry at least enough energy to equal the

work function. Then the electrons are ejected with zero kinetic energy. Since the energy of a photon is E = hf according to Equation 29.2, where f is the frequency of the wave, we have that W0 = hf. Equation 16.1 relates the frequency to the wavelength λ according to f = c/λ, where c is the speed of light. Thus, it follows that W0 = hc/λ.

SOLUTION Using Equations 29.2 and 16.1, we find that

W hc0

34 8

919

6 63 10 3 00 10

485 104 10 10= =

× ⋅ ×

×= ×

−

−−

λ

. ..

J s m / s

m J

c hc h

Since 1 eV = 1.60 × 10–19 J, it follows that

W019

194 10 10 110

2 56= ××

FHG

IKJ =−

−. . J eV

1.60 J eVc h

7. REASONING AND SOLUTION Equation 29.3 gives

KE

J s m / s

meV

JeV

J

max = − = −

=× ⋅ ×

×−

×FHG

IKJ = ×

−

−

−−

hf W hc Wo oλ

6 63 10 3 00 10

215 103 68

1 60 101

3 36 1034 8

9

1919

. ..

..

c hc h b g


Converting to electron volts

KE JeV

JeVmax = ×

×FHG

IKJ =−

−3 36 10

11 60 10

2 101919.

..c h

8. REASONING The intensity S = 680 W/m2 of the photons is equal to the total amount of

energy delivered by the photons per second per square meter of surface area. Therefore, the intensity of the photons is equal to the energy E of one photon multiplied by the number N of photons that reach the surface of the earth per second per square meter: S NE= (1) The energy E of each photon is given by E hf= (Equation 29.2), where h = 6.63×10−34 J·s

is Planck’s constant and f is the frequency of the photon. We will use cfλ

=

(Equation 16.1) to determine the frequency f of the photons from their wavelength λ and the speed c of light in a vacuum. SOLUTION Solving Equation (1) for N, we obtain N = S/E. Substituting E hf= (Equation 29.2) into this result yields

S SNE hf

= = (2)

Substituting cfλ

= (Equation 16.1) into Equation (2), we find that

( )( )( )( )

( )

2 9

34 8

21 2

680 W/m 730 10 m

6.63 10 J s 3.00 10 m/s

2.5 10 photons/ s m

S S SNchf hch

λ

λ

−

−

×= = = =

⎛ ⎞ × ⋅ ×⎜ ⎟⎝ ⎠

= × ⋅

9. SSM REASONING AND SOLUTION The number of photons per second, N, entering the

owl's eye is N SA E= / , where S is the intensity of the beam, A is the area of the owl's pupil, and E is the energy of a single photon. Assuming that the owl's pupil is circular, A r d= =π π2 1

22c h , where d is the diameter of the owl's pupil. Combining Equations 29.2

and 16.1, we have E hf hc= = / λ . Therefore,


N SAhc

= =× × ×

× ⋅ ×=

λ π(5.073 photons / s

–1310 8 5 10 10

10 10

12 W / m m (510 m)

(6.63 J s)(3.0 m / s)

2 –3 2 –9

8

) .–34

c h

10. REASONING The wavelength λ of the light shining on the surface is related to the

maximum kinetic energy KEmax of the electrons ejected from the surface by

max 0KEhf W= + (Equation 29.3), where h is Planck’s constant, cfλ

= (Equation 16.1) is

the frequency of the light, and W0 is the work function of the surface. Substituting Equation 16.1 into Equation 29.3, we see that

max 0KEhchf Wλ

= = + (1)

The work function W0 is a property of the metal surface itself, so it remains the same for any wavelength of incident light. When the wavelength of the incident light is λ1 = 221 nm, the maximum kinetic energy of the ejected electrons is KEmax,1 = 3.28×10−19 J, and when the wavelength is λ2, the maximum kinetic energy is twice as great: KEmax,2 = 2KEmax,1. We will use this information, with Equation (1), to determine the unknown wavelength λ2. SOLUTION Solving Equation (1) for the work function W0 yields

0 maxKE hcWλ

= − + (2)

Because the work function W0 isn’t affected by changing the wavelength of the incident light from λ1 to λ2, we have that

0 max, 2 max,1 max, 2 max,12 1 1 2

KE KE or KE KEhc hc hc hcWλ λ λ λ

= − + = − + − + = (3)

Solving Equation (3) for λ2 and substituting KEmax,2 = 2 KEmax,1, we obtain

2max,1

max, 2 max,1 max,1 max,11 1 1

1KE 1KE KE 2KE KE

hc hchc hc

hc

λ

λ λ λ

= = =− + − + +

(4)

In the last step of Equation (4), we have divided both the numerator and the denominator by the product hc. Substituting the given values the into Equation (4), we find that


( )( ) ( )

72 19

934 8

1 1.62 10 m 162 nm3.28 10 J 1

221 10 m6.63 10 J s 3.00 10 m/s

λ −−

−−

= = × =×

+×× ⋅ ×

11. REASONING The wavelength λ of the photon is related to its frequency f by λ = c/f

(Equation 16.1), where c is the speed of light. The frequency of the photon is proportional to its energy E via f = E/h (Equation 29.2), where h is Planck’s constant. Thus, λ = ch/E. The photon energy is equal to the sum of the maximum kinetic energy KEmax of the ejected electron and the work function W0 of the metal; max 0KEE W= + (Equation 29.3). Substituting this expression for E into λ = ch/E gives

max 0KEc h

Wλ =

+ (1)

The maximum kinetic energy is related to the maximum speed vmax by 21

max max2KE mv= (Equation 6.2), where m is the mass of the electron.

SOLUTION Substituting 21

max max2KE mv= into Equation (1), and converting the work function from electron-volts to joules, gives

( )( )

( )( ) ( )

21max 02

8 347

19231 612

3.00 10 m/s 6.63 10 J s 1.9 10 m1.60 10 J9.11 10 kg 1.2 10 m/s 2.3 eV

1 eV

c hmv W

λ

−−

−−

=+

× × ⋅= = ×

⎛ ⎞×× × + ⎜ ⎟

⎝ ⎠

12. REASONING The total energy Q delivered by N photons is NE, where E is the energy

carried by one photon, so that N = Q/E. Equation 29.2 indicates that the photon energy is E = hf, where h is Planck’s constant and f is the frequency. Thus, the number N of photons can be written as

Q QNE h f

= = (1)

Equation 16.1 relates the frequency f of a photon to its wavelength λ according to f = c/λ,

where c is the speed of light in a vacuum. Therefore, Equation (1) can be expressed as


Q QNh f h c

λ= = (2)

According to Equation 12.4, the heat Q required to raise the temperature of a substance by

an amount ΔT is Q = cspecific heatmΔT, where cspecific heat is the specific heat capacity of the substance and m is its mass.

SOLUTION Combining Equation 12.4 with Equation (2), the number of photons required

to raise the temperature by an amount ΔT is

specific heat or c m TQN N

h c hcλλ Δ

= =

Applying this result to each type of photon, we obtain

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

23infrared 8

21blue 8

6.0 10 m 840 J/ kg C 0.50 kg 2.0 C2.5 10

6.63 10 J s 3.00 10 m/s

4.7 10 m 840 J/ kg C 0.50 kg 2.0 C2.0 10

6.63 10 J s 3.00 10 m/s

N

N

−5

−34

−7

−34

× ⋅ ° °⎡ ⎤⎣ ⎦= = ×× ⋅ ×

× ⋅ ° °⎡ ⎤⎣ ⎦= = ×× ⋅ ×

13. SSM WWW REASONING AND SOLUTION a. According to Equation 24.5b, the electric field can be found from E S c= / ( )0ε . The

intensity S of the beam is

( )( )( )

18 34 8

2 93

5 2

Energy perunit time

1.30 10 photons/s 6.63 10 J s 3.00 10 m/s514.5 10 m1.00 10 m

1.60 10 W/m

N h f N h cSA A A λ

π

−

−−

⎛ ⎞= = = ⎜ ⎟⎝ ⎠

× × ⋅ ⎛ ⎞×= ⎜ ⎟⎜ ⎟×⎝ ⎠×

= ×

where N is the number of photons per second emitted. Then,

( )0/E S cε= = 7760 N / C


b. According to Equation 24.3, the average magnetic field is

B = E/c = –52.59 10 T× 14. REASONING The heat required to melt the ice is given by Q mL= f , where m is the mass

of the ice and Lf is the latent heat of fusion for water (see Section 12.8). Since, according to Equation 29.2, each photon carries an energy of E hf= , the energy content of N photons is E NhfTotal = . According to Equation 16.1, f c= / λ , so we have

E NhcTotal =

λ

If we assume that all of the photon energy is used to melt the ice, then, E QTotal = , so that

Nhc mL

EQ

λTotal

f: 4=

This expression may be solved for N to determine the required number of photons. SOLUTION a. We find that

NmL

hcf= =

× ×× ⋅ ×

= ×λ (2.0 kg)(33.5

2.1 photons10 10

10 1010

4 –9

–34 824 J / kg)(620 m)

(6.63 J s)(3.00 m / s)

b. The number ′N of molecules in 2.0-kg of water is

′ =×

FHG

IKJ

×FHG

IKJ ×N (2.0 kg) 1 mol

18 10 kg 6.022 = 6.7–3

10 1023

25 molecules1 mol

molecules

Therefore, on average, the number of water molecules that one photon converts from the ice

phase to the liquid phase is

′=

××

=NN

6.7 32 molecules / photon1010

25

24

molecules 2.1 photons

15. SSM REASONING The angle θ through which the X-rays are scattered is related to the

difference between the wavelength λ′ of the scattered X-rays and the wavelength λ of the incident X-rays by Equation 29.7 as


( )1 coshmc

λ λ θ′ − = −

where h is Planck’s constant, m is the mass of the electron, and c is the speed of light in a

vacuum. We can use this relation directly to find the angle, since all the other variables are known.

SOLUTION Solving Equation 29.7 for the angle θ, we obtain

( )

( )( ) ( )

( )

31 89 9

34

1

cos 1

9.11 10 kg 3.00 10 m/s1 0.2703 10 m 0.2685 10 m 0.26

6.63 10 J s

cos 0.26 75

mch

θ λ λ

θ

−− −

−

−

′= − −

× ×= − × − × =

× ⋅

= = °

16. REASONING The momentum of the photon is related to its wavelength λ and Planck’s

constant h. The momentum (nonrelativistic) of the ball depends on its mass m and speed v. We can set the two momenta equal and solve directly for the speed.

SOLUTION The momentum pphoton of the photon and the momentum pball of the ball are

photon ball (29.6) and (7.2)hp p mvλ

= =

Since pphoton = pball, we have

( )( )34

259 3

Momentum Momentumof photon of ball

6.63 10 J s or 4.2 10 m/s720 10 m 2.2 10 kg

h hmv vmλ λ

−−

− −

× ⋅= = = = ×

× ×

17. REASONING The frequency f of a photon is related to its energy E by f = E/h

(Equation 29.2), where h is Planck’s constant. As discussed in Section 29.4, the energy E is related to the magnitude p of the photon’s momentum by E = pc, where c is the speed of light in a vacuum. By combining these two relations, we see that the frequency can be expressed in terms of p as f = pc/h.

SOLUTION a. Substituting values for p, c, and h, into the relation f = pc/h gives


( )( )9 8

1334

2.3 10 kg m/s 3.00 10 m/s1.0 10 Hz

6.63 10 J spcfh

−2

−× ⋅ ×

= = = ×× ⋅

b. An inspection of Figure 24.9 shows that this frequency lies in the infrared region of the electromagnetic spectrum.

18. REASONING Before the scattering, the electron is at rest and has no momentum. Thus, the

total initial momentum consists only of the photon’s momentum, which points along the +x axis. The total initial momentum has no y component. Since the total momentum is conserved, the total momentum after the scattering must be the same as it was before and, therefore, has no y component.

The total momentum after the scattering is the sum of the momentum of the scattered

photon and that of the scattered electron, and it only has an x component. But the scattered photon is moving along the –y axis, so its momentum has no x component. Therefore, the momentum of the electron must have an x component.

The total momentum after the scattering is the sum of the momentum of the scattered

photon and that of the scattered electron, and it has no y component. But the scattered photon is moving along the –y axis, so its momentum points along the –y axis. Therefore, this contribution to the total final momentum must be cancelled by part of the momentum of the scattered electron, which must have a component along the +y axis.

SOLUTION Since the total momentum is conserved and since the scattered photon has no

momentum in the x direction, the momentum of the scattered electron must have an x component that equals the momentum of the incident photon. According to Equation 29.6, the magnitude p of the momentum of the incident photon is p = h/λ, where h is Planck’s constant and λ is the wavelength. Therefore, the momentum of the scattered electron has a component in the +x direction that is

p hx = =

× ⋅×

= × ⋅−

−−

λ6 63 10

107 37 10

34

1223. . J s

9.00 m kg m / s

The momentum of the scattered electron has a component along the +y direction. This

component cancels the momentum of the scattered photon that points along the –y direction. To find the momentum of the scattered photon, we first need to determine its wavelength λ′, which we can do using Equation 29.7:


( )( )

0

3412 11

31 8

1 cos 90.0

6.63 10 J s9.00 10 m 1.14 10 m9.11 10 kg 3.00 10 m/s

hmc

λ λ=

−− −

−

⎛ ⎞⎜ ⎟′ = + − °⎜ ⎟⎝ ⎠

× ⋅= × + = ×

× ×

Again using Equation 29.6, we find that the momentum of the scattered electron has a

component in the +y direction that is

p hy = =

× ⋅×

= × ⋅−

−−

λ6 63 10

14 105 82 10

34

1123.

.. J s

1 m kg m / s

19. REASONING There are no external forces that act on the system, so the conservation of

linear momentum applies. Since the photon is scattered at θ = °180 , the collision is "head-on," and all motion occurs along the horizontal direction, which we take as the x axis. The incident photon is assumed to be moving along the +x axis. For an initially stationary electron, the conservation of linear momentum states that:

electron

Momentum Momentum Momentum of incident of scattered of recoil

photon photon electron

p p p′= − +

where the momentum of the scattered photon is negative since is moves along the −x

direction (the scattering angle is 180°). Using the relation p = h/λ (Equation 29.6), where h is Planck’s constant and λ is the wavelength of the photon, we can write the expression for the momentum of the electron as

electron1 1 + h hp p p h

λ λ λ λ⎛ ⎞′= = + = +⎜ ⎟′ ′⎝ ⎠

SOLUTION Substituting numerical values into the equation above, we have

–34 –24electron 9 9

1 1(6.626 10 J s) = 4.755 10 kg m/s 0.2750 10 m 0.2825 10 m

p− −

⎛ ⎞= × ⋅ + × ⋅⎜ ⎟× ×⎝ ⎠


20. REASONING The wavelength λ of the incident X-rays is related to the wavelength λ′ of

the scattered X-rays by ( )1 coshmc

λ λ θ′ − = − (Equation 29.7), where h = 6.626×10−34 J·s is

Planck’s constant, c = 2.998×108 m/s is the speed of light in a vacuum, m = 9.109×10−31 kg is the mass of an electron, and θ = 122.0° is the angle at which the X-rays are scattered. The wavelength λ′ of the scattered photon is found from the magnitude p′ of its momentum

via hpλ

′ =′ (Equation 29.6).

SOLUTION Solving ( )1 coshmc

λ λ θ′ − = − (Equation 29.7) for λ, we obtain

( )1 coshmc

λ λ θ′= − − (1)

Solving hpλ

′ =′ (Equation 29.6) for λ′ yields h

pλ′ =

′, which, on substitution into

Equation (1), gives

( ) ( )

( )( )( )( )

3434

24 31 8

10

1 cos 1 cos

6.626 10 J s 1 cos122.06.626 10 J s1.856 10 kg m/s 9.109 10 kg 2.998 10 m/s

3.533 10 m 0.3533 nm

h h hmc p mc

λ λ θ θ

−−

− −

−

′= − − = − −′

× ⋅ −× ⋅= −

× ⋅ × ×

= × =

21. SSM WWW REASONING The change in wavelength that occurs during Compton

scattering is given by Equation 29.7:

( ) ( ) ( )max2– 1– cos or – 1– cos180h h h

mc mc mcλ λ θ λ λ′ ′= = ° =

( )max–λ λ′ is the maximum change in the wavelength, and to calculate it we need a value

for the mass m of a nitrogen molecule. This value can be obtained from the mass per mole M of nitrogen ( 2N ) and Avogadro's number AN , according to A/m M N= (see Section 14.1).

SOLUTION Using a value of 0.0280 kg/molM = , we obtain the following result for the

maximum change in the wavelength:


( )( )

( )

–34

max8

23 –1A

17

2 6.63 10 J s2 2–0.0280 kg/mol 3.00 10 m/s

6.02 10 mol

9.50 10 m

h hmc M cN

λ λ× ⋅

′ = = =⎛ ⎞ ⎛ ⎞

×⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠

= × –

22. REASONING Energy is conserved during the collision. This means that the energy E of

the incident photon must equal the kinetic energy KE of the recoil electron plus the energy E′ the scattered photon:

E = KE + E′ (1)

The energy E of a photon is related to its frequency f by E = hf (Equation 29.2), where h is Planck’s constant. The frequency, in turn, is related to the wavelength λ by f = c/λ (Equation 16.1), where c is the speed of light in a vacuum. Substituting f = c/λ into E = hf gives

E = hc/λ (2) The wavelength λ′ of the scattered photon depends on the wavelength λ of the incident photon according to Equation 29.7, so that we have

( )1 coshmc

λ λ θ′ = + −

Since the photon is scattered straight backward, θ = 180°, and

( ) 21 cos180h hmc mc

λ λ λ′ = + − ° = + (3)

SOLUTION The kinetic energy of the recoil electron is given by 21

2KE = mv (Equation 6.2), where m is its mass and v is its speed. Substituting this expression into Equation (1), we have E = 21

2 mv + E′ . Solving for the speed v of the electron gives

( )2 E Evm

′−= (4)

From Equation (2), we also know that E = hc/λ and /E hc λ′ ′= . Substituting this expression

into Equation (4), we find that the speed of the electron can be written as


2 1 1h cvm λ λ

⎛ ⎞= −⎜ ⎟′⎝ ⎠

Since 2hmc

λ λ′ = + [Equation (3)], the speed of the electron is

( )( )

( )( )( )

34 8

31

9 349

31 8

6

2 1 12

2 6.63 10 J s 3.00 10 m/s9.11 10 kg

1 1 0.45000 10 m 2 6.63 10 J s0.45000 10 m

9.11 10 kg 3.00 10 m/s

3.22 10 m/s

hcvhm

mcλ λ

−

−

− −−

−

⎛ ⎞= −⎜ ⎟+⎜ ⎟

⎝ ⎠

× ⋅ ×=

×

⎛ ⎞−⎜ ⎟× × ⋅⎜ ⎟× +⎜ ⎟× ×⎝ ⎠

= ×

×

23. REASONING AND SOLUTION a. We have

λ = λ' − (h/mc)(1 − cos 163°) = 0.1819 nm b. For the incident photon

E = hf = hc/λ = 1.092 J–15×10 c. For the scattered photon

E ' = hf ' = hc/λ' = 1.064 J–15×10

d. The kinetic energy of the recoil electron is, therefore,

KE = E − E ' = 2.8 J–17×10 24. REASONING

a. Consider one square meter of the sail’s surface. Each of the N photons that strike this square meter in a one-second interval (Δt = 1.0 s) has an initial momentum p that is

determined by its wavelength λ, according to hpλ

= (Equation 29.6), where


h = 6.63×10−34 J·s is Planck’s constant. Each photon is fully reflected, so the final momentum of a photon is equal to –p. The magnitude of the change Δp in a photon’s momentum, then, is equal to Δp = |p − (−p)| = 2p, and the magnitude ΔP of the net momentum change undergone by all N photons is given by ΔP = NΔp = 2Np. In order to cause this momentum change, the sail exerts an impulse of magnitude FΔt = ΔP (Equation 7.4) on the photons, where F is the magnitude of the force exerted on the photons per square meter of the sail. By Newton’s Third Law, that force magnitude is equal to the magnitude of the force exerted by the N photons on one square meter of the sail. From Equation 7.4, then, we have that 2F t P NpΔ = Δ = (1) Lastly, the magnitude ΣF = F of the net force necessary for the sail to attain the desired acceleration of a = 9.8×10−6 m/s2 is given by Newton’s Second Law, F ma= (Equation 4.1), where m = 3.0×10−3 kg is the mass of one square meter of the sail. As instructed, we have ignored all other forces acting on the sail. b. The intensity S of the laser beam depends on the total energy delivered to the sail by

Total energySt A

=Δ

(Equation 24.4), where A is the area of the sail and Δt is the time interval.

The total energy is equal to the number N of photons that strike the area in one second times the energy E of a single photon. Therefore, the intensity of the laser beam is

Total energy NESt A t A

= =Δ Δ

(2)

The energy E of a single photon is given by E hf= (Equation 29.2), where cfλ

=

(Equation 16.1) is the photon’s frequency and c = 3.00×108 m/s is the speed of light in a vacuum. SOLUTION a. Solving Equation (1) for N, we obtain

2F tN

pΔ

= (3)

Substituting hpλ

= (Equation 29.6) and F ma= (Equation 4.1) into Equation (3) yields


( )( )( )( )( )

3 6 2 918

34

2 22

3.0 10 kg 9.8 10 m/s 1.00 s 225 10 m5.0 10

2 6.63 10 J s

F t ma t ma tNhp h

λ

λ

− − −

−

Δ Δ Δ= = =

⎛ ⎞⎜ ⎟⎝ ⎠

× × ×= = ×

× ⋅

b. Substituting cfλ

= (Equation 16.1) into E hf= (Equation 29.2) gives hcE hfλ

= = .

Substituting this result into Equation (2), we obtain

NE NhcSt A t Aλ

= =Δ Δ

(4)

Equation (4) applies to the intensity reaching an area A = 1.0 m2 of the sail in a time Δt = 1.0 s. Therefore, in order for the sail to accelerate at the desired rate, the intensity of the laser must be

( )( )( )( )( )( )

18 34 82

2 95.0 10 6.63 10 J s 3.00 10 m/s 4.4 W/m

1.00 s 1.00 m 225 10 mNhcSt Aλ

−

−

× × ⋅ ×= = =

Δ ×

25. SSM REASONING AND SOLUTION The de Broglie wavelength λ is given by

Equation 29.8 as λ = h p/ , where p is the magnitude of the momentum of the particle. The magnitude of the momentum is p mv= , where m is the mass and v is the speed of the particle. Using this expression in Equation 29.8, we find that λ = h mv/ ( ) , or

v hm

= =× ⋅

× ×= ×

λ6.63 10 J s

1.67 10 kg 1 10 m10 m s

–34

–27 –147

c hc h..

303 05 /

The kinetic energy of the proton is

KE kg)(3.05 m / s) = 7.77 J–27 7 2 –13= = × × ×12

2 12 1 67 10 10 10mv ( .

26. REASONING According to Equation 27.1, the angle θ that locates the first-order bright

fringes (m = 1) is specified by sin θ = λ/d, where λ is the wavelength and d is the separation between the slits. The wavelength of the electron is the de Broglie wavelength, which is given by λ = h/p (Equation 29.8), where h is Planck’s constant and p is the magnitude of the momentum of the electron.


SOLUTION Combining Equations 27.1 and 29.8, we find that the angle locating the first-order bright fringes is specified by

sin θ λ= =

dhpd

Dividing this result for case A by that for case B, we find

( )( )

( ) ( )( )

AA B A AB

B B A B

22 423

B 4

/sin sin or

sin / sin

1.2 10 kg m/s sin 1.6 10 degrees4.8 10 kg m/s

sin 4.0 10 degrees

h p d p pp

h p d p

p

θ θθ θ

− −−

−

= = =

× ⋅ ×= = × ⋅

×

27. REASONING AND SOLUTION The de Broglie wavelength λ is given by Equation 29.8 as

λ = h/p, where p is the magnitude of the momentum of the particle. The magnitude of the momentum is p = mv, where m is the mass and v is the speed of the particle. Using this expression in Equation 29.8, we find that

λλ

= = =× ⋅

× ×= ×

hmv

v hm

or6.63 10 J s

1.67 10 kg 0.282 10 m1 41 10 m s

–34

–27 –93

c hc h . /

28. REASONING AND SOLUTION We know that λ = h/mv. Solving for the mass yields

m hv

= =× ⋅

× ×= ×

−

−−

λ6 63 10

8 4 10 1 2 106 6 10

34

14 627.

. ..

J sm m / s

kgc hc h

29. REASONING AND SOLUTION The average kinetic energy of a helium atom is

KE = (3/2)kT = (3/2)(1.38 × 10–23 J/K)(293 K) = 6.07 × 10–21 J The speed of the atom corresponding to the average kinetic energy is

vm

= =×

×= ×

−

−

2 2 6 07 10

6 65 101 35 10

21

273KE J

kgm / s

b g c h.

..

The de Broglie wavelength is


λ = =× ⋅

× ×= ×

−

−−h

mv6 63 10

6 65 10 1 35 107 38 10

34

27 311.

. ..

J skg m / s

mc hc h

30. REASONING The de Broglie wavelength λ of a particle is inversely proportional to the

magnitude p of its momentum, as we see from hp

λ = (Equation 29.8), where

h = 6.63×10−34 J·s is Planck’s constant. The electron is moving at a speed v = 0.88c, which is close to the speed of light in a vacuum (c = 3.00×108 m/s). Therefore, we will use

2

21

mvpvc

=

−

(Equation 28.3) to determine the magnitude of the electron’s relativistic

momentum, where m = 9.11×10−31 kg is the electron’s mass. SOLUTION Substituting Equation 28.3 into Equation 9.8, we find that

( ) ( ) ( )

2

2

2

2

34

31 8

1

1

6.63 10 J s 0.8819.11 10 kg 0.88 3.00 10 m/s

h h h vmvp mv c

vc

c

λ

−

−

= = = −⎛ ⎞⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

× ⋅= −

× × c

2121.3 10 m−⎛ ⎞

= ×⎜ ⎟⎝ ⎠

31. SSM REASONING The de Broglie wavelength λ is related to Planck’s constant h and the

magnitude p of the particle’s momentum. The magnitude of the momentum can be related to the particle’s kinetic energy. Thus, using the given wavelength and the fact that the kinetic energy doubles, we will be able to obtain the new wavelength.

SOLUTION The de Broglie wavelength is

hp

λ = (29.8)

The kinetic energy and the magnitude of the momentum are

1 22

KE (6.2) (7.2)mv p mv= =


where m and v are the mass and speed of the particle. Substituting Equation 7.2 into Equation 6.2, we can relate the kinetic energy and momentum as follows:

( )2 2 2

1 22

KE or 2 KE2 2

m v pmv p mm m

= = = =

Substituting this result for p into Equation 29.8 gives

( )2 KEh hp m

λ = =

Applying this expression for the final and initial wavelengths λf and λi, we obtain

( ) ( )f if i

and 2 KE 2 KE

h hm m

λ λ= =

Dividing the two equations and rearranging reveals that

( )

( )

( )( )

( )( )

ff i if i

i f f

i

2 KE KE KE or

KE KE2 KE

hm

hm

λλ λ

λ= = =

Using the given value for λi and the fact that ( )f iKE 2 KE= , we find

( )( ) ( ) i10i

f if

KEKE2.7 10 m

KEλ λ −= = ×

i2 KE( )101.9 10 m−= ×

32. REASONING The de Broglie wavelength λ of the woman is found from hpλ

=

(Equation 29.8), where p is the magnitude of her momentum and h = 6.63×10−34 J·s is Planck’s constant. We will use p = mv (Equation 7.2) to determine the magnitude p of the woman’s momentum from her mass m and her speed v at the instant she strikes the water. Once the woman jumps from the cliff, she is in free fall with an initial speed of v0 = 0 m/s, and an acceleration a = −9.8 m/s2. Since we have taken upward to be the positive direction, her displacement during the fall is H = −9.5 m. Her final speed v, then, is given by


2 20 2v v aH= + (2.9)

SOLUTION Solving hpλ

= (Equation 29.8) for λ yields hp

λ = . Substituting p = mv

(Equation 7.2) into this result, we obtain

h hp mv

λ = = (1)

Substituting v0 = 0 m/s into Equation (2.9) and taking the square root of both sides, we find that ( )22 20 m/s 2 or 2 or 2v aH v aH v aH= + = = (2) Substituting Equation (2) into Equation (1), we find the de Broglie wavelength of the woman at the instant she strikes the water to be:

( ) ( )( )

3436

2

6.63 10 J s 1.2 10 m2 41 kg 2 9.8 m/s 9.5 m

h hmv m aH

λ−

−× ⋅= = = = ×

− −

33. REASONING When the electron is at rest, it has electric potential energy, but no kinetic

energy. The electric potential energy EPE is given by EPE = eV (Equation 19.3), where e is the magnitude of the charge on the electron and V is the potential difference. When the electron reaches its maximum speed, it has no potential energy, but its kinetic energy is

212 mv . The conservation of energy states that the final total energy of the electron equals the initial total energy:

212

Initial totalFinal total energyenergy

mv eV=

Solving this equation for the potential difference gives ( )2 / 2V mv e= .

The speed of the electron can be expressed in terms of the magnitude p of its momentum by v = p/m (Equation 7.2). The magnitude of the electron’s momentum is related to its de Broglie wavelength λ by p = h/λ (Equation 29.8), where h is Planck’s constant. Thus, the speed can be written as v = h/(mλ). Substituting this expression for v into ( )2 / 2V mv e= gives ( )2 2/ 2V h meλ= .


SOLUTION The potential difference that accelerates the electron is

( )

( )( )( )

22 344

2 231 9 11

6.63 10 J s 1.86 10 V2 2 9.11 10 kg 1.60 10 C 0.900 10 m

hVmeλ

−

− −1 −

× ⋅= = = ×

× × ×

34. REASONING The linear momentum p of a particle is given by p = mv (Equation 7.2),

where m and v are its mass and velocity. Since particle A is initially at rest, its momentum is zero. The initial momentum of particle B is p0B = mBv0B. This is also the total initial linear momentum of the two-particle system.

After the collision the combined mass of the two particles is mA+ mB, and the common velocity is vf .Thus, the total linear momentum of the system after the collision is pf = (mA+ mB)vf . From Section 7.2, we know that the total linear momentum of an isolated system is conserved. An isolated system is one in which the vector sum of the external forces acting on the system is zero. Since there are no external forces acting on the particles, the two-particle system is an isolated system. Thus, the total linear momentum of the system after the collision equals the total linear momentum before the collision.

The de Broglie wavelength λ is inversely related to the magnitude p of a particle’s momentum by λ = h/p (Equation 29.8), where h is Planck’s constant.

SOLUTION The de Broglie wavelength λf of the object that moves off after the collision is

given by λf = h/pf (Equation 29.8). Since momentum is mass times velocity, the magnitude of the momentum that the object has after the collision is pf = (mA + mB)vf, where vf is the common speed of the two particles. We can evaluate this momentum by using the law of conservation of momentum, which indicates that the total momentum after the collision is the same as it is before the collision. Before the collision only particle B is moving, so that the magnitude of the total momentum at that time has a value of mBv0B, where v0B is the initial speed of particle B. Assuming the particles travel along the +x axis, we write the conservation of linear momentum as follows:

( )A B f B 0B

Total momentumTotal momentumbefore collisionafter collision

+ m m v m v+ + =

Using this result, we find that the desired de Broglie wavelength is

( )ff B 0BA B f

h h hp m vm m v

λ = = =+


But the term on the far right is just the given de Broglie wavelength of the incident particle B. Therefore, we conclude that 34

f 2.0 10 mλ −= × . 35. SSM REASONING The de Broglie wavelength λ of the electron is related to the

magnitude p of its momentum by λ = h/p (Equation 29.8), where h is Planck’s constant. If the speed of the electron is much less than the speed of light, the magnitude of the electron’s momentum is given by p = mv (Equation 7.2). Thus, the de Broglie wavelength can be written as λ = h/(mv).

When the electron is at rest, it has electric potential energy, but no kinetic energy. The

electric potential energy EPE is given by EPE = eV (Equation 19.3), where e is the magnitude of the charge on the electron and V is the potential difference. When the electron reaches its maximum speed, it has no potential energy, but its kinetic energy is 21

2 mv . The conservation of energy states that the final total energy of the electron equals the initial total energy:

212

Initial totalFinal total energyenergy

mv eV=

Solving this equation for the final speed gives 2 /v eV m= . Substituting this expression

for v into λ = h/(mv) gives / 2h meVλ = . SOLUTION After accelerating through the potential difference, the electron has a

de Broglie wavelength of

( )( )( )

3411

31 9

6.63 10 J s 6.01 10 m2 2 9.11 10 kg 1.60 10 C 418 V

hmeV

λ−

−

− −1

× ⋅= = = ×

× ×

36. REASONING AND SOLUTION The energy of the photon is E = hf = hc/λphoton , while the

kinetic energy of the particle is KE = (1/2)mv2 = h2/(2mλ2). Equating the two energies and rearranging the result gives λphoton/λ = (2mc/h)λ. Now the speed of the particle is v = 0.050c, so λ = h/(0.050 mc), and

λphoton /λ = 2/0.050 = 4.0 101×

37. SSM WWW REASONING AND SOLUTION According to the uncertainty principle,

the minimum uncertainty in the momentum can be determined from ( )/ 4yp y h πΔ Δ = .


Since p mvy y= , it follows that Δ Δp m vy y= . Thus, the minimum uncertainty in the velocity of the oxygen molecule is given by

( )( )–34

6–26 –3

6.63 10 J s 8.3 10 m s4 4 5.3 10 kg 0.12 10 my

hvm yπ π

× ⋅Δ = = = ×

Δ × ×– /

38. REASONING When particles pass through the slit the great majority fall on the screen

between the first dark fringes on either side of the central bright fringe. The first dark fringes are located by the angles −θ (below the midpoint) and +θ (above the midpoint), so this is the minimum range of angles over which the particles spread out. As Figure 29.15

shows, the angle θ is found from 1tan y

x

p

pθ −

Δ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠ (Equation 1.4), where Δpy is the

uncertainty in the y component py of a particle’s momentum, and px is the x component of a particle’s momentum. The given de Broglie wavelength λ = 0.200 mm of the particles is the wavelength they possess before passing through the slit, when their momentum has only the x component px. Therefore, the x component of each particle’s momentum is given by

xhpλ

= (Equation 29.8), where h = 6.63×10−34 J·s is Planck’s constant.

We are given that the uncertainty in the position of each particle along the y direction is equal to one-half the width of the slit: 1

2y WΔ = . The minimum uncertainty in the y component Δpy of a particle’s momentum, then, is found from the Heisenberg uncertainty principle:

( )( )4yhp yπ

Δ Δ = (29.10)

SOLUTION Solving Equation 29.10 for Δpy and substituting 1

2y WΔ = , we obtain

( ) ( )1

24 24y

h h hpy WWπ ππ

Δ = = =Δ

(1)

Substituting Equation (1) and xhpλ

= (Equation 29.8) into 1tan y

x

p

pθ −

Δ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠ (Equation 1.6),

we find that


1 1tan tany

x

hp

pθ − −

Δ⎛ ⎞= =⎜ ⎟⎜ ⎟

⎝ ⎠

2 Wh

π( )

91 1

3633 10 mtan tan

2 2 0.200 10 m

0.0289

Wλ

π πλ

−− −

−

⎛ ⎞⎜ ⎟ ⎡ ⎤×⎛ ⎞⎜ ⎟ = = ⎢ ⎥⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎣ ⎦⎜ ⎟⎝ ⎠

=

Therefore, the particles spread out over the range −0.0289° to +0.0289° .

39. SSM WWW REASONING The uncertainty in the electron’s position is Δy = 3.0 ×

10−15 m. The minimum uncertainty Δpy in the y component of the electron’s momentum is given by the Heisenberg uncertainty principle as ( )/ 4yp h yπΔ = Δ (Equation 29.10).

SOLUTION Setting Δy = 3.0 × 10−15 m in the relation ( )/ 4yp h yπΔ = Δ gives

( )34

2015

6.63 10 J s 1.8 10 kg m/s4 4 3.0 10 my

hpyπ π

−−

−× ⋅

Δ = = = × ⋅Δ ×

40. REASONING Suppose the object is moving along the +y axis. The uncertainty in the

object’s position is Δy = 2.5 m. The minimum uncertainty Δpy in the object’s momentum is specified by the Heisenberg uncertainty principle (Equation 29.10) in the form (Δpy)(Δy) = h/(4π). Since momentum is mass m times velocity v, the uncertainty in the velocity Δv is related to the uncertainty in the momentum by Δv = (Δpy)/m.

SOLUTION a. Using the uncertainty principle, we find the minimum uncertainty in the momentum as

follows:

( )( )

( )34

35

4

6.63 10 J s 2.1 10 kg m/s4 4 2.5 m

y

y

hp y

hpy

π

π π

−−

Δ Δ =

× ⋅Δ = = = × ⋅

Δ

b. For a golf ball this uncertainty in momentum corresponds to an uncertainty in velocity

that is given by 35

342.1 10 kg m/s 4.7 10 m/s0.045 kg

yy

pv

m

−−

Δ × ⋅Δ = = = ×


c. For an electron this uncertainty in momentum corresponds to an uncertainty in velocity that is given by

355

312.1 10 kg m/s 2.3 10 m/s

9.11 10 kgy

yp

vm

−−

−

Δ × ⋅Δ = = = ×

×

41. SOLUTION The minimum uncertainty Δy in the position of the particle is related to the

minimum uncertainty ypΔ in the momentum via the Heisenberg uncertainty principle. To

cast this relationship into a form that gives us the desired percentage for the minimum uncertainty in the speed, we note that the minimum uncertainty in the position is specified as the de Broglie wavelength λ. We can then express the de Broglie wavelength in terms of Planck’s constant h and the magnitude py of the particle’s momentum. The magnitude of the momentum is related to the mass m and the speed vy of the particle.

SOLUTION The percentage minimum uncertainty in the speed is

Percentage 100%y

y

v

v

Δ= × (1)

According to the Heisenberg uncertainty principle, the minimum uncertainty ypΔ in the

momentum and the minimum uncertainty Δy in the position of the particle are related according to

( )( )4yhp yπ

Δ Δ = (29.10)

We know that Δy is equal to the de Broglie wavelength / yh pλ = (Equation 29.8), where

the magnitude of the momentum is y yp mv= (Equation 7.2). Thus, we have

y y

h hyp mv

λΔ = = =

Substituting this result for Δy into Equation 29.10, we obtain

( )( ) ( ) 4y yy

h hp y pmv π

⎛ ⎞⎜ ⎟Δ Δ = Δ =⎜ ⎟⎝ ⎠

(2)

The last step in our transformation of the uncertainty principle is to realize that

( )y y yp mv m vΔ = Δ = Δ , since the mass is constant. Substituting this expression for ypΔ

into Equation (2) shows that


( ) ( ) 1 or 4 4

yy y

y y y

vh h hp m vmv mv vπ π

⎛ ⎞ ⎛ ⎞ Δ⎜ ⎟ ⎜ ⎟Δ = Δ = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Using this result in Equation (1), we find that

1Percentage 100% 100% 8.0%4

y

y

v

v π

Δ= × = × =

42. REASONING The mass m of the particle is related to its rest energy E0 by 2

0E mc= (Equation 28.5). Therefore, if there is a minimum uncertainty ΔE0 in measuring the rest energy of the particle, there will be a corresponding uncertainty Δm in measuring its mass:

( ) 2 00 2 or

EE m c m

cΔ

Δ = Δ Δ = (1)

The minimum uncertainty ΔE0 in the particle’s rest energy is related to the length of time Δt

the particle exists in a state by the Heisenberg uncertainty principle: ( )( )0 4hE tπ

Δ Δ =

(Equation 29.11), where h = 6.63×10−34 J·s is Planck’s constant. SOLUTION Solving Equation 29.11 for the minimum uncertainty ΔE0, we obtain

( )0 4hE

tπΔ =

Δ (2)

Substituting Equation (2) into Equation (1), we find that

( ) ( )( )34

3302 2 220 8

6.63 10 J s 7.9 10 kg4 4 7.4 10 s 3.00 10 m/s

E hmc t cπ π

−−

−

Δ × ⋅Δ = = = = ×

Δ × ×


43. SSM REASONING In order for the person to diffract to the same extent as the sound wave, the de Broglie wavelength of the person must be equal to the wavelength of the sound wave.

SOLUTION a. Since the wavelengths are equal, we have that

λ λsound person=

λ soundperson person

=h

m v

Solving for vperson , and using the relation λ sound sound sound= v f/ (Equation 16.1), we have

v hm v f

h fm vperson

person sound sound

sound

person sound

4

= =

=× ⋅

= ×

( / )

( . ) .6 63 1055.0 kg)(343 m

50 10–34

–36 J s)(128 Hz( / s)

m / s

b. At the speed calculated in part (a), the time required for the person to move a distance of

one meter is

t xv

= =×

FHG

IKJFHG

IKJFHG

IKJ = ×

1.0 m4.50

1.0 h3600 s

1 day24.0 h

1 year

365.25 days7.05

Factors to convertseconds to years

1010–36

27

m / s years

44. REASONING The energy of a photon is related to its frequency and Planck’s constant.

The frequency, in turn, is related to the speed and wavelength of the light. Thus, we can relate the energy to the wavelength. The given relationship between the wavelengths will then allow us to determine the unknown energy.

SOLUTION The energy E of a photon with frequency f is

E hf= (29.2) where h is Planck’s constant. The frequency is related to the speed c and wavelength λ of

the light according to cfλ

= (16.1)


Substituting this expression for f into Equation 29.2 gives

cE hf hλ

= =

Applying this result to both sources, we have

B AB A

and c cE h E hλ λ

= =

Dividing the two expressions gives

B B A

A B

A

chE

cE h

λ λλ

λ

= =

Using the given value for EA and the fact that B A3λ λ= in this result shows that

( )18 19A AB A

B A2.1 10 J 7.0 10 J

3E E

λ λλ λ

− −⎛ ⎞= = × = ×⎜ ⎟⎜ ⎟

⎝ ⎠

45. REASONING The de Broglie wavelength λ is related to Planck’s constant h and the

magnitude p of the particle’s momentum. The magnitude of the momentum is related to the mass m and the speed v at which the bacterium is moving. Since the mass and the speed are given, we can calculate the wavelength directly.

SOLUTION The de Broglie wavelength is

hp

λ = (29.8)

The magnitude of the momentum is p mv= (Equation 7.2), which we can substitute into

Equation 29.8 to show that the de Broglie wavelength of the bacterium is

( )( )

3418

156.63 10 J s 1 10 m

2 10 kg 0.33 m/sh hp mv

λ−

−−

× ⋅= = = = ×

×


46. REASONING AND SOLUTION a. We know E = hc/λ for a photon. The energy of the photon is

E =×F

HGIKJ = ×

−−5 0. eV

1.60 10 J1eV

8.0 10 J19

19

The wavelength is

λ = =× ⋅ ×

×= ×

−

−−hc

E

6 63 10 3 00 102 5 10

34 87

. ..

J s m / s

8.0 10 Jm19

c hc h

b. The speed of the 5.0-eV electron is

v Em

= =×

×= ×

−

−

2 2 8 0 101 3 10

196

..

J

9.11 10 kgm / s31

c h

The de Broglie wavelength is

λ = =× ⋅

× ×= ×

−

−−h

mv6 63 10

1 3 105 6 10

34

610.

..

J s9.11 10 kg m / s

m31c hc h

47. REASONING AND SOLUTION In the first case, the energy of the incident photon is given

by Equation 29.3 as

hf W= +KE max 0 = 0.68 eV + 2.75 eV = 3.43 eV In the second case, a rearrangement of Equation 29.3 yields

KE max –= hf W0 = 3.43 eV – 2.17 eV = 1.26 eV

48. REASONING The speed v of a particle is related to the magnitude p of its momentum by

v = p/m (Equation 7.2). The magnitude of the momentum is related to the particle’s de Broglie wavelength λ by p = h/λ (Equation 29.8), where h is Planck’s constant. Thus, the speed of a particle can be expressed as v = h/(mλ). We will use this relation to find the speed of the proton.

SOLUTION The speeds of the proton and electron are

proton electronproton proton electron electron

and h hv vm mλ λ

= =


Dividing the first equation by the second equation, and noting that λelectron= λproton, we

obtain proton electron electron electron

electron proton proton proton

v m mv m m

λλ

= =

Using values for melectron and mproton taken from the inside of the front cover, we find that

the speed of the proton is

( )31

6 3electronproton electron 27

proton

9.11 10 kg4.50 10 m/s 2.45 10 m/s1.67 10 kg

mv v

m

−

−

⎛ ⎞⎛ ⎞ ×= = × = ×⎜ ⎟⎜ ⎟⎜ ⎟ ×⎝ ⎠⎝ ⎠

49. REASONING The width of the central bright fringe in the diffraction patterns will be

identical when the electrons have the same de Broglie wavelength as the wavelength of the photons in the red light. The de Broglie wavelength of one electron in the beam is given by Equation 29.8, λ electron = h p/ , where p mv= .

SOLUTION Following the reasoning described above, we find

λ λred light electron=

λ red lightelectron electron

=h

m v

Solving for the speed of the electron, we have

v hmelectron

electron red light

1.10= =× ⋅

× ×= ×

λ6 63 10

10 1010. –34

–31 –93 J s

(9.11 kg)(661 m) m / s

50. REASONING We will first calculate the potential energy of the system at each of the two

separations, and then find the energy difference for the two configurations. Since the electric potential energy lost by the system is carried off by a photon that is emitted during the process, the energy difference must be equal to the energy of the photon. The wavelength of the photon can then by found using Equation 29.2 with Equation 16.1: E hc= / λ .


SOLUTION The initial potential energy of the system is (see Equations 19.3 and 19.6)

EPE

C) (8.99 10 N m / C )( C)0.420 m

J

1

9 2 2 –6–14

= =FHGIKJ

= ×× ⋅ ×L

NMOQP = ×−

eV ekqr1

1

191 6 10 8 30 10 2 84 10( . . .

The final potential energy is

EPE C) (8.99 10 N m / C )( C)1.58 m

J2

9 2 2 –6–15= = ×

× ⋅ ×LNM

OQP = ×−eV2

191 6 10 8 30 10 7 56 10( . . .

The energy difference, and therefore the energy of the emitted photon, is

ΔE = − = × × ×EPE EPE J – J = .08 J1 2–14 –15 –142 84 10 7 56 10 2 10. .

The wavelength of this photon is

λ = =× ⋅ ×

×= ×

hcEΔ

(6.63 J s)(3.00 m / s)2.08 J

9.56 m–34 8

–14–1210 10

1010

51. REASONING Since the net external force acting on the system (the photon and the

electron) is zero, the conservation of linear momentum applies. In addition, there are no nonconservative forces, so the conservation of total energy applies as well. Since the photon scatters at an angle of θ = 180.0° in Figure 29.10, the collision is "head-on." Thus, the motion takes place entirely along the horizontal direction, which we will take as the x axis, with the right as being the positive direction.

The conservation of linear momentum gives rise to Equation 29.7, which relates the

difference ′ −λ λ between the scattered and incident X-ray photon wavelengths to the scattering angle θ of the electron as

′ − = − = − ° =λ λ θhmc

hmc

hmc

1 1 180 0 2cos cos .b g b g (1)

The conservation of total energy is written as

hc hc mvλ λ

Energyof incident

photon

Initialkineticenergy

of electron

Energyof scattered

photon

Finalkineticenergy

of electron

4 4+ =

′+0 1

22 (2)


Equations (1) and (2) will permit us to find the wavelength λ of the incident X-ray photon. SOLUTION Solving Equation (1) for ′λ and substituting the result into Equation (2) gives

hc hch

mc

mvλ λ

=+

+2

12

2

Algebraically rearranging this result, we obtain a quadratic equation for λ:

λ λ22

12

2

2 2 0

4 85 10 9 70 1012

20

+ FHGIKJ − =

× ×−

−

hmc

hm mv

. .m m2

c h

where we have used h = 6.63 × 10–34 J⋅s, m = 9.11 × 10–31 kg, c = 3.00 × 108 m/s, and

v = 4.67 × 106 m/s. Solving this quadratic equation for λ, we obtain

λ = × −3 09 10 10. m

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Ch 29 ISM 8e Solutions