PROBLEMS

Note to Instructors: Most of the homework problems in this chapter are available for assignment via an online homework management program sllchas

WileyPLUS or WebAssign. and those marked with the icon @ are presented in a guided IlItorialformatthat provides enhanced interactivity. See Prefacefor additional details.

In working these problems, ignore relativistic effects.

ssm Solution is in the Student Solutions Manual.

www Solution is available on the World Wide Web at www.wiley.comlcollege/cutnell

936 I CHAPTER 29 I PARTICLES AND WAVES

11. A stone is dropped from the top of a building. As the stone falls,does its de Broglie wavelength increase, decrease, or remain thesame? Provide a reason for your answer.

12. An electron and a neutron have different masses. Is it possible,

according to Equation 29.8, that they can have the same de Brogliewavelength? Account foryour answer.

Section 29.3 Photons and the Photoelectric Effect

1. ssm Ultraviolet light with a frequency of 3.00 X 1015Hz strikesa metal surface and ejects electrons that have a maximum kinetic en­ergy of 6.1 eV. What is the work function (in eV) of the metal?

2. Two sources produce electromagnetic waves. Source B producesa wavelength that is three times the wavelength produced by sourceA. Each photon from source A has an energy of 2.1 X 10-181. Whatis the energy of a photon from source B?

3. An FM radio station broadcasts at a frequency of 98.1 MHz. The

power radiated from the antenna is 5.0 X 104 W. How many pho­tons per second does the antenna emit?

4. The maximum wavelength that an electromagnetic wave can haveand still eject electrons from a metal surface is 485 nm. What is thework function Wo of this metal? Express your answer in electron volts.

5. ssm Ultraviolet light is responsible for sun tanning. Find the wave­length (in nm) of an ultraviolet photon whose energy is 6.4 X 10-191.

6. Light is shining perpendicularly on the surface of the earth with

an intensity of 680 W/m2• Assuming all the photons in the light havethe same wavelength (in vacuum) of.730 nm, determine the number

of photons per second per square meter that reach the earth.

o Radiation of a certain wavelength causes electrons with a maxi­mum kinetic energy of 0.68 eV to be ejected from a metal whosework function is 2.75 eV. What will be the maximum kinetic energy(in eV) with which this same radiation ejects electrons from anothermetal whose work function is 2.17 eV?

~Multiple-Concept Example 3 provides pertinent information forproblems such as this. The maximum wavelength for which an elec­tromagnetic wave can eject electrons from a platinum surface is196 nm. When radiation with a wavelength of 141 nm shines on thesurface, what is the maximum speed of the ejected electrons?

* 9. ssm Consult Interactive LearningWare 29.1 at www.wiley.com/college/cutnell for background material relating to this problem. Anowl has good night vision because its eyes can detect a light inten­sity as small as 5.0 X 10-13 W/m2• What is the minimum number of

photons per second that an owl eye can detect if its pupil has a diam­eter of 8.5 mm and the light has a wavelength of 510 nm?

* 10. A proton is located at a distance of 0.420 m from a point chargeof +8.30 /Le. The repulsive electric force moves the proton until it

is at a distance of 1.58 m from the charge. Suppose that the electricpotential energy lost by the system is carried off by a photon that isemitted during the process. What is its wavelength?

** 11. ssm www A laser emits 1.30 X 1018 photons per second in abeam of light that has a diameter of 2.00 mm and a wavelength of

13. ssm In Figure 29.1 replace the electrons with protons that have

the same speed. With the aid of Equation 27.1 for the bright fringesin Young's double-slit experiment and Equation 29.8, decide

whether the angular separation between the fringes would increase,decrease, or remain the same when compared to that produced bythe electrons.

TThiS icon represents a biomedical application.

514.5 nm. Determine (a) the average electric field strength and(b) the average magnetic field strength for the electromagnetic wavethat constitutes the beam.

** 12. (a) How many photons (wavelength = 620 nm) must be ab­sorbed to melt a 2.0-kg block of ice at 0 °C into water at 0 °C?(b) On the average, how many H20 molecules does one photon con­vert from the ice phase to the water phase?

Section 29.4 The Momentum of a Photon

and the Compton Effect

13. A light source emits a beam of photons, each of which has amomentum of 2.3 X 10-z9 kg' m/so (a) What is the frequency of thephotons? (b) To what region of the electromagnetic spectrum do thephotons belong? Consult Figure 24.10 if necessary.

14. A photon of red light (wavelength = 720 nm) and a Ping-Pongball (mass = 2.2 X 10-3 kg) have the same momentum. At whatspeed is the ball moving?

15. ssm In a Compton scattering experiment, the incident X-rayshave a wavelength of 0.2685 nm, and the scattered X-rays have awavelength of 0.2703 nm. Through what angle e in Figure 29.10 arethe X-rays scattered?

ti(;) Incident X-rays have a wavelength of 0.3120 nm and are scat­~d by the "free" electrons in graphite. The scattering angle in Fig­ure 29.10 is e = 135.0°. What is the magnitude of the momentum of(a) the incident photon and (b) the scattered photon? (For accuracy,use h = 6.626 X 10-34 J . sand c = 2.998 X 108 m/s.)

17. Refer to Interactive Solution 29.17 at www.wiley.com/college/

cutnell for help in solving this problem. An incident X-ray photon ofwavelength 0.2750 nm is scattered from an electron that is initially atrest. The photon is scattered at an angle of e = 180.0° in Figure29.10 and has a wavelength of 0.2825 nm. Use the conservation of

linear momentum to find the momentum gained by the electron.

* 18. The X-rays detected at a scattering angle of e = 163° in Figure29.10 have a wavelength of 0.1867 nm. Find (a) the wavelength of anincident photon, (b) the energy of an incident photon, (c) the energyof a scattered photon, and (d) the kinetic energy of the recoil electron.(For accuracy, use h = 6.626 X 10-34 J. sand c = 2.998 X 108 m/s.)

* 19. ssm www What is the maximum amount by which the wave­

length of an incident photon could change when it undergoes Comp­ton scattering from a nitrogen molecule (Nz)?

** 20. An X-ray photon is scattered at an angle of e = 180.0° from an

electron that is initially at rest. After scattering, the electron has a speedof 4.67 X 106 m/so Find the wavelength of the incident X-ray photon.

Section 29.5 The De Broglie Wavelength'and the Wave Nature of Matter

21. The interatomic spacing in a crystal of table salt is 0.282 nm.This crystal is being studied in a neutron diffraction experiment,similar to the one that produced the photograph in Figure 29.13a.

How fast must a neutron (mass = 1.67 X 10-27 kg) be moving to

have a de Broglie wavelength of 0.282 nm?

22. A bacterium (mass = 2 X 10-15 kg) in the blood is moving at0,33 mfs. What is the de Broglie wavelength of this bacterium?

23. ssm As Section 27.5 discusses, sound waves diffract, or bend,around the edges of a doorway. Larger wavelengths diffract morethan smaller wavelengths. (a) The speed of sound is 343 m/so Withwhat speed would a 55.0-kg person have to move through a doorwayto diffract to the same extent as a 128-Hz bass tone? (b) At the speedcalculated in part (a), how long (in years) would it take the person tomove a distance of one meter?

@ How fast does a proton have to be moving in order to have thesame de Broglie wavelength as an electron that is moving with a

~eed of 4.50 X 106 m/s?25 A photon has the same momentum as an electron moving with a

peed of 2.0 X 105 m Is. What is the wavelength of the photon?

* 26. A particle has a de Broglie wavelength of 2.7 X 10-10 m. Thenits kinetic energy doubles. What is the particle's new de Brogliewavelength, assuming that relativistic effects can be ignored?

* 27. ssm From a cliff that is 9.5 m above a lake, a young woman(mass = 41 kg) jumps from rest, straight down into the water. Atthe instant she strikes the water, what is her de Broglie wavelength?

@ The width of the central bright fringe in a diffraction pattern ona screen is identical when either electrons or red light (vacuum

wavelength = 661 nm) pass through a single slit. The distancebetween the screen and the slit is the same in each case and is largecompared to the slit width. How fast are the electrons moving?

* 29. Consult Interactive Solution 29.29 at www.wiley.comlcollege/cutnell to explore a model for solving this problem. In a televisionpicture tube the electrons are accelerated from rest through a potentialdifference V. Just before an electron strikes the screen, its de Brogliewavelength is 0.900 X 10-11 m. What is the potential difference?

** 30. The kinetic energy of a particle is equal to the energy of a photon.The particle moves at 5.0% of the speed of light. Find the ratio of thephoton wavelength to the de Broglie wavelength of the particle.

IADDITIONAL PROBLEMS

37. A particle has a speed of 1.2 X 106 m/so Its de Broglie wave­

length is 8.4 X 10-14 m. What is the mass of the particle?

38. The dissociation energy of a molecule is the energy required to

break apart the molecule into its separate atoms. The dissociationenergy for the cyanogen molecule is 1.22 X 10-18 J. Suppose thatthis energy is provided by a single photon. Determine the (a) wave­

length and (b) frequency of the photon. (c) In what region of theelectromagnetic spectrum (see Figure 24.10) does this photon lie?

39. ssm The de Broglie wavelength of a proton in a particle accel­erator is 1.30 X 10-14 m. Determine the kinetic energy (in joules) ofthe proton.

40. What is (a) the wavelength of a 5.0-eV photon and (b) thede Broglie wavelength of a 5.0-eV electron?

ADDITIONAL PROBLEMS I 937

Section 29.6 The Heisenberg Uncertainty Principle

/31) Consider a line that is 2.5 m long. A moving object is some­~ere along this line, but its position is not known. (a) Find the min­imum uncertainty in the momentum of the object. Find the mini­

mum uncertainty in the object's velocity, assuming that the object is(b) a golf ball (mass = 0.045 kg) and (c) an electron.

d32) An electron is trapped wit~in a sphere that has a diameterL;,(6.0 X 10-15 m (about the size of the nucleus of an oxygenatom). What is the minimum uncertainty in the electron'smomentum?

33. ssm www In the lungs there are tiny sacs of air, which are

called alveoli. The average diameter of one of these sacs is 0.25 mm.Consider an oxygen molecule (mass = 5.3 X 10-26 kg) trappedwithin a sac. What is the minimum uncertainty in the velocity of thisoxygen molecule?

34. Review Conceptual Example 7 as background for this problem.When electrons'pass through a single slit, as in Figure 29.15, theyform a diffraction pattern. As Section 29.6 discusses, the centralbright fringe extends to either side of the midpoint, according to anangle ()given by sin () = MW, where A is the de Broglie wavelengthof the electron and W is the width of the slit. When A is the same

size as W, A = 90°, and the central fringe fills the entire observationscreen. In this case, an electron passing through the slit has roughlythe same probability of hitting the screen either straight ahead oranywhere off to one side or the other. Now, imagine yourself in aworld where Planck's constant is large enough so you exhibit similareffects when you walk through a 0.90-m-wide doorway. Your massis 82 kg and you walk at a speed of 0.50 m/so How large wouldPlanck's constant have to be in this hypothetical world?

35. ssm www Particles pass through a single slit of width

0.200 mm (see Figure 29.15). The de Broglie wavelength of eachparticle is 633 nm. After the particles pass through the slit, theyspread out over a range of angles. Use the Heisenberg uncertaintyprinciple to determine the minimum range of angles.

* 36. The minimum uncertainty /:::"y in the position y of a particle isequal to its de Broglie wavelength. Determine the minimumuncertainty in the speed of the particle, where this minimum uncer­

tainty /:::"vy is expressed as a percentage of the particle's speed vy

(Percentage = /:::"vy X 100%). Assume that relativistic effects canvy

be ignored.

41. Recall from Section 14.3 that the average kinetic energy of an atom

in a monatomic ideal gas is given by KE = ~kT, where k = 1.38 X10-23 J/K and T is the Kelvin temperanlre of the gas. Determine the deBroglie wavelength of a helium atom (mass = 6.65 X 10-27 kg) thathas the average kinetic energy at room temperature (293 K).

42. A magnesium surface has a work function of 3.68 eV. Electro­

magnetic waves with a wavelength of 215 nm strike the surface andeject electrons. Find the maximum kinetic energy of the ejectedelectrons. Express your answer in electron volts.

* 43. ssm An electron, starting from rest, accelerates through a po­tential difference of 418 V. What is the final de Broglie wavelengthof the electron, assuming that its final speed is much less than thespeed of light?

e e/L=-=-f (E I h)

Chapter 29 Problems 1483

he (6.63x10-34 J·s)(3.0x108 m/s) = 3.1 x 10-7 m = j310nm I= E = 6.4x 10-19 J

6. REASONING Light intensity I is the power that passes perpendicularly through a surfacedivided by the area of that surface (Equation 16.8). The number N of photons per second persquare meter that reaches the earth is the light intensity divided by the energy E of a singlephoton; N = liE. The energy of a single photon is E = hf (Equation 29.2), where h isPlanck's constant and f is the frequency of the photon. The frequency of the photon isrelated to its wavelength /L by Equation 16.1 as f = e/ /L, where e is the speed of light in avacuum. Therefore, we have

SOLUTION The number of photons per second per square meter that reach the earth is

N=I/L= (680W/m2)(730x10-9m) =!2.5X1021 photons/(s.m2)1he (6.63x10-34 J ·s)(3.00x10s m/s) --------

7. REASONING AND SOLUTION In the first case, the energy of the incident photon is givenby Equation 29.3 as

hf = KE + Wo = 0.68 eV + 2.75 eV = 3.43 eVmax

In the second case, a rearrangement of Equation 29.3 yields

KEmax=hf-Wo=3.43eV-2.17eV= 11.26eVI

8. REASONING AND SOLUTION The work function of the material (using /L = 196 nm) isfound from

The maximum kinetic energy of the ejected electron is (using /L= 141 nm)

KE = hf- W = he//L- W = 3.96 X 10-19Jmu 0 0

The speed of the electron is then

2(3.96X103~191) = !9.32X105 m/s [9.11 x 10 kg

Chapter 29 Problems 1487

15. I SSM I REASONING The angle e through which the X-rays are scattered is related to thedifference between the wavelength A' of the scattered X-rays and the wavelength /L of theincident X-rays by Equation 29.7 as .

A' - /L= ~ (1- cos e)me

where h is Planck's constant, m is the mass of the electron, and e is the speed of light in avacuum. Wecan use this relation directly to find the angle, since all the other variables areknown.

SOLUTION Solving Equation 29.7 for the angle e, we obtain

cose=l- me (A' -/L)h

(9.11x10-31 kg) (3.00x108 m1s) (0.2703x10-9 m _ 0.2685x10-9 m) = 0.26=1- 6.63x10-34 J.s

16. REASONING AND SOLUTION

a. According to Equation 29.6, the magnitude of the momentum of the incident photon is

=12.= 6.626x10-34 J,s=12.124XlO-24 k ·m/s IP /L 0.3120x10-9 m g

b. The wavelength of the scattered photon is, from Equation 29.7,

/LI = /L+ ~ (1- cos e)me

where e is the scattering angle. Combining this expression with Equation 29.6, we find thatthe magnitude of the momentum of the scattered photon is

I h h

p =;.;= (h J/L+ me (I-cos e)

=6.626x10-34 J·s

I 6.626xI0-3' J·s }I-COS 135.0°)0.3120x10-9 m+ (9.109x10-31kg)(2.998x108m/s)

= !2.096X10-24 kg·m/s [

--~~-------------~-------------------~--------------------~-- ------ -----

j

Chapter 29 Problems 1491

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

A =Asound person

h

A sound = m v personperson

Solving for v , and using the relation A d = V d / f d (Equation 16.1), we haveperson soun soun soun

Vperson

h

m person (v sound / fsound )

h fsound

m person V sound

s

s

(6.63 X10-34 J. s)(128 Hz)

(55.0 kg)(343 m/ s)j4.50xlO-36 m/s I

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

tx

v

1.0 m

4.50x10-36 m/s ( 10 h ) C day) ( 1 year J3600 s 24.0 h 365.25 daysv

Factors to convertseconds to years

j7.05X1027 years I

)

)

24. REASONING The speed v of a particle is related to the magnitude p of its momentum byv =p/m (Equation 7.2). The magnitude of the momentum is related to the particle'sde Broglie wavelength A by p = h/A (Equation 29.8), where h is Planck's constant. Thus,the speed of a particle can be expressed as v = h/(mA). We will use this relation to find thespeed of the proton.

SOLUTION The speeds of the proton and electron are

h

v proton = mproton Aproton

andh

velectron = melectron Aelectron

ii

Dividing the first equation by the second equation, and noting that A I t = At, wee ec ron pro on

obtain

vproton _ melectronAelectron = melectron

velectron - mproton Aproton mproton

(29.8)

(7.2)p=mv

or

(6.2)1 2KE = - mv2

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

Substituting this result for p into Equation 29.8 gives

h h

A = p = ~2m(KE)

v - [melectron: (450 106 )(9.11X10-31 kg] I 3 Iproton - velectron m =. x m/s -27 = 2.45xlO m/S

proton 1.67 X 10 kg

The kinetic energy and the magnitude of the momentum are

SOLUTION The de Broglie wavelength is

Using values for melectron and mproton taken from the inside of the front cover, we find thatthe speed of the proton is

1492 PARTICLES AND WAVES

25. REASONING AND SOLUTION The momentum of the photon is p = h/A and that of theelectron is p = mv. Equating and solving for the wavelength of the photon,

26. REASONING The de Broglie wavelength A is related to Planck's constant h and themagnitude p of the particle's momentum. The magnitude of the momentum can be relatedto the particle's kinetic energy. Thus, using the given wavelength and the fact that thekinetic energy doubles, we will be able to obtain the new wavelength.

or

and

h

A, = ~2m(KE)f = /KE),~ h (KE)f

~2m(KE\

Using the given value for Aiand the fact that KEf = 2 (KEi ) , we find

Dividing the two equations and rearranging reveals that

Chapter 29 Problems 1493

Applying this expression for the final and initial wavelengths Af and Ai' we obtain

27. I SSMI REASONING AND SOLUTION The de Broglie wavelength A of the woman is

given by Equation 29.8 as A = h / p, where p is the magnitude of her momentum. The

magnitude of the momentum is p = mv , where m is the woman's mass and v is her speed.

According to Equation 3.6b of the equations of kinematics, the speed v is given by

v = ~, since the woman jumps from rest. In this expression, a = -9.80 m/ S2 and~~UyY y

Y = -9.5 m. With these considerations we find that

= 6.63 X 10-34 J ·s = 11.2 X 10-36 m I(41 kg)~2( -9.80 m/ S2)( -9.5 m) ----

28. REASONING The width of the central bright fringe in the diffraction patterns will beidentical 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 byEquation 29.8, A I t \ = h / p, where p = mv.e ec ron

SOLUTION Following the reasoning described above, we find

A =Ared light electron

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

1- mv2. The conservation of energy states that the final total energy of the electron equals

the initial total energy:

!I.86XI04 vi

h

A red light = m n V electronelectro

= 6.63xlO-34 J·s(9.l1xlO 31 kg)(661xIO-9 m) =11.10XI03 m/s I

(6.63x10-34 J .s)2

2(9.11xlO-31 kg)(1.60xI0-19 C)(0.900xlO-11 m)2

velectron = n1 A red lightelectron

h

h2V-- - ..

2meA 2

SOLUTION The potential difference that accelerates the electron is

Solving this equation for the potential difference gives V = mv2 I (2e) .

Solving for the speed of the electron, we have

The speed of the electron can be expressed in terms of the magnitude p of its momentum byv = plm (Equation 7.2). The magnitude of the electron's momentum is related to itsde Broglie wavelength A by p = hi A (Equation 29.8), where his Planck's constant. Thus, the

speed can be written as v = hl(mA). Substituting this expression for v into V = mv2 I (2e)

gives V = h2 I (2meA2 ).

REASONING AND SOLUTION The energy of the photon is E = hi = he! A h ' while thep oton

kinetic energy of the particle is KE = (l12)mv2 = h2/(2mA2). Equating the two energies and

rearranging the result gives AphotO/ A = (2me!h)A. Now the speed of the particle is v ==

0.050e, so A = hl(0.050 me), and

Ah IA=2/0.050= [40XI01p oton .

I 2"2 mv = e V

'------y------J '--v-----'Final total Initial total

energy energy

30.

1494 PARTICLES AND WAVES

~~"

.;-.,~~

";',~nl~,

Chapter 29 Problems 1495

31. REASONING We know that the object is somewhere on the line. Therefore, theuncertainty in the object's position is ~y = 2.5 m. The minimum uncertainty in the object's

momentum is ~Py and is specified by the Heisenberg uncertainty principle (Equation 29.10)

in the form (~p )(~y) = hl(4Jr). Since momentum is mass m times velocity v, the uncertaintyy

in the velocity ~v is related to the uncertainty in the momentum by ~v = (~p)lm.

SOLUTION

a. Using the uncertainty principle, we find the minimum uncertainty in the momentum asfollows:

h

(~py )(~y) = 4Jr

~p= _h__ 6.63X10-34].sy 4ff~- 4ff(2.5m) -12.1X10-35kg.m/sl

b. For a golf ball this uncertainty in momentum corresponds to an uncertainty in velocitythat is given by

~py _ 2.lx10-35 kg·m/s =14.7X10-34 m/s I~vy = ---;;: - 0.045 kg

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

_ ~py = 2.1x10-35 kg·m/s -12.3X10-5 m/s I~vy - m 9.1lxlO-31 kg

32. REASONING We assume that the electron is moving along the y direction, and that it canbe anywhere within the sphere. Therefore, the uncertainty in the electron's position is equalto the diameter d of the sphere, so ~y = d. The minimum uncertainty ~p in the y componenty

of the electron's momentum is given by the Heisenberg uncertainty principle as

~py = hi (4Jr~y) (Equation 29.10).

SOLUTION Setting ~y = d in the relation ~Py = hi (4Jr~Y) gives

~p = _h __ h 6.63x10-34 ].y 4ff~y - 4ffd = 4ff(6.0x10-15 ~) =18.8X10-21 kg.m/sl

33. I SSMII wwwl REASONING AND SOLUTION According to the uncertainty principle,

the minimum uncertainty in the momentum can be determined from ~py~y = hi ( 4ff) .