CHAPTER 29 INTRODUCTION TO QUANTUM PHYSICS

Find the longest-wavelength photon that can eject an electron from potassium, given that the binding energy is 2.24 eV.. Calculate the binding energy in eV of electrons in aluminum, if

CHAPTER 29: INTRODUCTION TO

QUANTUM PHYSICS

29.1 QUANTIZATION OF ENERGY

1 A LiBr molecule oscillates with a frequency of 1.7×1013Hz (a) What is the difference

in energy in eV between allowed oscillator states? (b) What is the approximate value

of n for a state having an energy of 1.0 eV?

Solution (a) ∆E =hf =(6.63×10− 34 J⋅s)(1.7×1013s− 1)=1.127×10− 20J

J1060.1

eV1J

10127

2

1s107.1sJ1063.6

J/eV10

60.1eV0.12

1

1 13 34

2 The difference in energy between allowed oscillator states in HBr molecules is 0.330

eV What is the oscillation frequency of this molecule?

sJ1063.6

J/eV10

60.1eV330

3 A physicist is watching a 15-kg orangutan at a zoo swing lazily in a tire at the end of a

rope He (the physicist) notices that each oscillation takes 3.00 s and hypothesizes that the energy is quantized (a) What is the difference in energy in joules between allowed oscillator states? (b) What is the value of n for a state where the energy is 5.00 J? (c) Can the quantization be observed?

Solution

(a)

1

sec333.0s00.3

1J10210.2

J00.52

12

E n

(c) No, ∆E is much too small and n is much too large.

29.2 THE PHOTOELECTRIC EFFECT

4 What is the longest-wavelength EM radiation that can eject a photoelectron from

silver, given that the binding energy is 4.73 eV? Is this in the visible range?

Solution

J/eV10

60.1eV73.4

m/s1000.3sJ1063.6BE

BE0

KE

;BEKE

7 19

8 34

λ

λ

No, this is UV

5 Find the longest-wavelength photon that can eject an electron from potassium, given

that the binding energy is 2.24 eV Is this visible EM radiation?

Solution

J/eV10

60.1eV24.2

m/s1000.3sJ1063.6BE

BE0

KE

;BEKE

7 19

8 34

λ

λ

Yes, it is visible as green light

6 What is the binding energy in eV of electrons in magnesium, if the longest-wavelength

photon that can eject electrons is 337 nm?

eV1J

10902.5

m1037.3

m/s1000.3sJ1063.6BE

BE0KE

;BEKE

19 19

9

8 34

7 Calculate the binding energy in eV of electrons in aluminum, if the longest-wavelength

photon that can eject them is 304 nm.

Solution The longest wavelength corresponds to the shortest frequency, or the smallest energy

Therefore, the smallest energy is when the kinetic energy is zero From the equation

0BE

KE=hf − = , we can calculate the binding energy (writing the frequency in

terms of the wavelength):

eV09.4J101.60

eV1.000J

106.543

m103.04

m/s1000.3sJ106.63BE

BE

19 19

7

8 34

8 What is the maximum kinetic energy in eV of electrons ejected from sodium metal by

450-nm EM radiation, given that the binding energy is 2.28 eV?

eV483.0eV28.2eV7625.2BEKE

eV7625.2J1060.1

eV000.1J1042.4

m1050.4

m/s1000.3sJ1063.6

19 19

7

8 34

λ

9 UV radiation having a wavelength of 120 nm falls on gold metal, to which electrons

are bound by 4.82 eV What is the maximum kinetic energy of the ejected

photoelectrons?

eV5.54eV

4.82eV10.359BE

KE

eV10.359J

101.602

eV1.000J

101.6575

m101.20

m/s103.00sJ106.63

19 18

7

8 34

λ

10 Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of

0.860 eV from sodium metal What is the binding energy of electrons to sodium metal?

eV2.25eV0.86eV3.1078KE

BEBEKE

eV1078.3J101.60

eV1.000J

104.9725

m104.00

m/s103.00sJ106.63

19 19

7

8 34

hc hf

λ

11 UV radiation having a 300-nm wavelength falls on uranium metal, ejecting 0.500-eV

electrons What is the binding energy of electrons to uranium metal?

eV3.64eV0.500eV

4.14KE

BEBEKE

eV4.14J

101.602

eV1.000J

106.63

m103.00

m/s103.00sJ106.63

19 19

7

8 34

hc hf

λ

12 What is the wavelength of EM radiation that ejects 2.00-eV electrons from calcium

metal, given that the binding energy is 2.71 eV? What type of EM radiation is this?

J107.536

m/s103.00sJ106.63

J107.536eV

1

J101.60eV

4.71eV2.00eV2.71KEBE

BEKE

7 19

8 34

19 19

=+

hf

hf

hf

λλ

This is ultraviolet radiation

13 Find the wavelength of photons that eject 0.100-eV electrons from potassium, given

that the binding energy is 2.24 eV Are these photons visible?

Solution KE =hf −BEandc=λf so

.nm531m105.313

J101.60

eV1.000eV

0.100eV

2.24

m/s103.00sJ106.63KE

BE

7

19

8 34

×

⋅

×

=+

Yes, these photons are visible

14 What is the maximum velocity of electrons ejected from a material by 80-nm photons,

if they are bound to the material by 4.73 eV?

J101.72922KE2

1KE

J101.7295

eV1

J101.604.73eVm

108.00

m/s103.00sJ106.63

BEBE

KE

6 2

1

31

18 2

18

19 8

8 34

mv

λ

hc hf

15 Photoelectrons from a material with a binding energy of 2.71 eV are ejected by

420-nm photons Once ejected, how long does it take these electrons to travel 2.50 cm to a detection device?

m102.50

m/sec10

2.962kg

109.11

J103.9972KE22

1KE

J103.997

eV1

J101.602.71eVm

104.20

m/s103.00sJ106.63

BEBE

KE

8 5

2

5 2

1 31

20 2

20

19 7

8 34

m v

mv

λ

hc hf

16 A laser with a power output of 2.00 mW at a wavelength of 400 nm is projected onto

calcium metal (a) How many electrons per second are ejected? (b) What power is carried away by the electrons, given that the binding energy is 2.31 eV?

m/s103.00sJ106.63λ

19 7

8 34

104.9725

J/s10

1

J101.60eV2.28J

104.9725

BEBE

KE

4 19

15

19

19 19

17 (a) Calculate the number of photoelectrons per second ejected from a 1.00-mm 2 area

of sodium metal by 500-nm EM radiation having an intensity of 1.30kW/m2 (the intensity of sunlight above the Earth’s atmosphere) (b) Given that the binding energy

is 2.28 eV, what power is carried away by the electrons? (c) The electrons carry away less power than brought in by the photons Where does the other power go? How can

it be recovered?

Solution (a) First calculate n, the number of photons per second hitting a square-meter area

Each photon has energy

2 21 19

2 3

19 7

8 34

ms/10272.3J103.973

mJ/s101.30

J103.973m

105.00

m/s102.998sJ106.626

( )( ) 3.27 10 /s

nm1000

m1nm1.00m /s10

2 2

N

J103.204eV

1

J101.60eV2.28J

103.973

BEλBEKE

4 20

15

20

19 19

(c) The other energy goes to the sodium metal to free the electrons (binding energy) This lost power can be recovered by the spontaneous absorption of electrons by sodium metal For each electron absorbed an energy of 2.28eV will be released

18 Unreasonable Results Red light having a wavelength of 700 nm is projected onto

magnesium metal to which electrons are bound by 3.68 eV (a) Use KEe=hf – BE to

calculate the kinetic energy of the ejected electrons (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

eV1J

10841.2

m1000.7

m/s1000.3sJ1063.6

19 19

7

8 34

and since KE =hf −BE=1.776eV−3.68eV=−1.90eV

(b) Negative kinetic energy is impossible

(c) The assumption that the photon can knock the electron free is unreasonable

19 Unreasonable Results (a) What is the binding energy of electrons to a material from

which 4.00-eV electrons are ejected by 400-nm EM radiation? (b) What is

unreasonable about this result? (c) Which assumptions are unreasonable or

inconsistent?

Solution (a) We want to use the equation KE =hf −BE to determine the binding energy, so

we first need to determine an expression of hf Using E =hf , we know:

J10602.1

eV1J

10966.4

m1000.4

m/s10998.2sJ10626.6

19 19

7

8 34

and since KE =hf −BE: BE=hf −KE =3.100eV−4.00eV=−0.90eV

(b) The binding energy is too large for the given photon energy

(c) The electron’s kinetic energy is too large for the given photon energy; it cannot be

greater than the photon energy.

29.3 PHOTON ENERGIES AND THE ELECTROMAGNETIC SPECTRUM

20 What is the energy in joules and eV of a photon in a radio wave from an AM station

that has a 1530-kHz broadcast frequency?

eV106.34eV106.340J

101.60

eV1J101.0144

J101.01J101.0144s

101.53J/s106.63

9 9

19 27

27 27

1 6 34

21 (a) Find the energy in joules and eV of photons in radio waves from an FM station that

has a 90.0-MHz broadcast frequency (b) What does this imply about the number of photons per second that the radio station must broadcast?

Solution

(a)

eV103.73J

101.60

eV1J105.97

J105.97s

109.00J/s106.63

7 19

26

26 1

8 34

(b) This implies that a tremendous number of photons must be broadcast per second

In order to have a broadcast power of, say 50.0 kW, it would take

photon/sec10

8.38J/photon

105.97

J/s10

eV104.14

eV10

23 (a) What is the wavelength of a 1.00-eV photon? (b) Find its frequency in hertz (c)

Identify the type of EM radiation.

Solution

nmeV

hc E

(b) 1.24 10 m 2.42 10 Hz

m/s10

(c) The radiation is in the infrared part of the spectrum

24 Do the unit conversions necessary to show that hc=1240eV⋅nm, as stated in the

text.

Solution ( )( ) 1240eV nm

J101.60

eV1.00m

1

nm10m/s103.00sJ10

9 8

25 Confirm the statement in the text that the range of photon energies for visible light is

1.63 to 3.26 eV, given that the range of visible wavelengths is 380 to 760 nm.

Solution

eV3.26nm

380

nm1240eVλ

eV1.63nm

760

nmeV1240λ

min max

max min

hc E

26 (a) Calculate the energy in eV of an IR photon of frequency 2.00×1013Hz (b) How

many of these photons would need to be absorbed simultaneously by a tightly bound molecule to break it apart? (c) What is the energy in eV of a γ ray of frequency

Hz1000

3 × 20 ? (d) How many tightly bound molecules could a single such γ ray break apart?

Solution

J101.60

eV1s

102.00sJ10

10.0

eV10

27 Prove that, to three-digit accuracy, h=4.14×10− 15eV⋅s, as stated in the text.

J101.60

eV1.00s

J10

28 (a) What is the maximum energy in eV of photons produced in a CRT using a 25.0-kV

accelerating potential, such as a color TV? (b) What is their frequency?

Solution

J101.60

eV1.00V

102.50C10

19 4

29 What is the accelerating voltage of an x-ray tube that produces x rays with a shortest

wavelength of 0.0103 nm?

Solution ( )( )

m101.03C101.60

m/s103.00sJ10

11 19

8 34

hc E qV

30 (a) What is the ratio of power outputs by two microwave ovens having frequencies of

950 and 2560 MHz, if they emit the same number of photons per second? (b) What is the ratio of photons per second if they have the same power output?

Solution (a) Let N be the number of photons per second

2.69MHz

950

MHz2560

2 2

1 1 2 2

1 1 2

f N hf N

hf N P

P Nhf NE

P γ

since N1 = N2(b)

0.371MHz

2560

MHz950

1

2 2

1 2 2

1 1 2

f

f N

N hf N

hf N P

P

since P1 =P2

31 How many photons per second are emitted by the antenna of a microwave oven, if its

power output is 1.00 kW at a frequency of 2560 MHz?

condphotons/se10

5.89J/photon

101.697

J/s101.00

J101.697s

102.5610

6.63

26 24

3

24 1

9 34

32 Some satellites use nuclear power (a) If such a satellite emits a 1.00-W flux of γ rays

having an average energy of 0.500 MeV, how many are emitted per second? (b) These

γ rays affect other satellites How far away must another satellite be to only receive one γ ray per second per square meter?

Solution

(a)

photons/s10

1.25J/photon10

8.00

J/s101.00

J1000.8J/eV10

1.60eV105.00

13 14

3

14 19

4π

photon/s10

1.254π

4πphotons/mΦ

5 2

2

13 2

2 2

r N

33 (a) If the power output of a 650-kHz radio station is 50.0 kW, how many photons per

second are produced? (b) If the radio waves are broadcast uniformly in all directions, find the number of photons per second per square meter at a distance of 100 km Assume no reflection from the ground or absorption by the air.

Solution

(a) E γ =hf =(6.63×10− 34 Js)(6.50×105 s− 1)=4.31×10− 28J

Then, 4.31 10 J/photon 1.16 10 photon/s 1.16 10 photon/s

J/s10

photons/s10

1.164

N N

34 How many x-ray photons per second are created by an x-ray tube that produces a flux

of x rays having a power of 1.00 W? Assume the average energy per photon is 75.0 keV.

photon/s10

8.33J/photon10

1.20

J/s1.00

J1020.1J/eV10

1.60eV107.50

13 14

14 19

35 (a) How far away must you be from a 650-kHz radio station with power 50.0 kW for

there to be only one photon per second per square meter? Assume no reflections or absorption, as if you were in deep outer space (b) Discuss the implications for

detecting intelligent life in other solar systems by detecting their radio broadcasts.

Solution

mphoton/s1

4π

photon/s10

1.161Φ

4π4

2

32 2

N r

r π N

(b) The distance calculated in part (a) is approximately 1/3 ly Therefore, if radio stations from intelligent life in other solar systems are to be detected, their broadcasts would have to have substantial power outputs Also, since there are stray radio waves in outer space, their signals would have to be large compared to the background radio waves This means it is rather unlikely for us to detect intelligent life by detecting their radio broadcasts

36 Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range

(typical for incandescent bulbs) with an average wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away would you be if 500 photons per second enter the 3.00-mm diameter pupil of your eye? (This number easily stimulates the retina.)

km181m101.81

mphoton/s10

7.0734π

photon/s10

2.924π

4Φ

mphotons/s10

7.073m

101.50π

500Φ

5

2 2 7

19 2

2

2 7

2 3

N

N r

r π N

29.4 PHOTON MOMENTUM

38 (a) Find the momentum of a 4.00-cm-wavelength microwave photon (b) Discuss why

you expect the answer to (a) to be very small.

Solution

sJ106.63λ

32 2

(b) The wavelength of microwave photons is large, so the momentum they carry is very small

39 (a) What is the momentum of a 0.0100-nm-wavelength photon that could detect

details of an atom? (b) What is its energy in MeV?

Solution

sJ106.63λ

23 11

MeV1m/s103.00m/skg106.63

Solution

(a) Using

m13.3m101.326m/s

kg105.00

J.s106.63

h p

(b) Using

eV109.38J

101.60

eV1J

101.50

m/s1000.3m/skg105.00

2 - 19

20

8 29

41 (a) A γ-ray photon has a momentum of 8.00×10−21kg⋅m/s What is its wavelength?

(b) Calculate its energy in MeV.

Solution

sJ10

(b)

MeV15.0J

101.602

MeV1J

102.40

m/s103.00kgm/s10

8.00

13 12

8 21

42 (a) Calculate the momentum of a photon having a wavelength of 2.50μm (b) Find

the velocity of an electron having the same momentum (c) What is the kinetic energy

of the electron, and how does it compare with that of the photon?

Solution

Js10

109.11

kgm/s10

m

p v v m p

28

26 2

2 31

2

102.06KE

J;

107.96m/s

1000.3kgm/s10

2.652

J103.86m/s

102.911kg109.110.52

1KE

γ γ

e e

E

c p E mv

43 Repeat the previous problem for a 10.0-nm-wavelength photon.

Solution

sJ106.63λ

26 8

109.11

m/skg10

26

21 2

4 31

2

108.25KE

J;

101.99m/s

103.00m/skg106.63

J102.41m/s

107.277kg109.110.52

1KE

γ γ

e e

E

c p E mv

44 (a) Calculate the wavelength of a photon that has the same momentum as a proton

moving at 1.00% of the speed of light (b) What is the energy of the photon in MeV? (c) What is the kinetic energy of the proton in MeV?

Solution

(a)

m101.32m/skg105.01

sJ106.63

m/skg105.01m/s

103.0010

1.00kg101.67

13 21

34

21 8

2 27

p p

p

h λ λ

h p mv p

(b)

MeV9.39J

101.60

MeV1J

101.503

m/s103.00m/skg105.01

13 12

8 21

E γ γ

(c)

MeV10

4.70J107.515

m/s103.0010

1.00kg101.670.52

1KE

2 15

2 8 2

27 2

45 (a) Find the momentum of a 100-keV x-ray photon (b) Find the equivalent velocity of

a neutron with the same momentum (c) What is the neutron’s kinetic energy in keV?

Solution

(a)

m/s103.00

J/eV10

1.60eV10

8

19 5

(b)

m/s103.19kg

101.67

m/skg10

(c)

keV105.32J

101.60

keV1J

108.516

m/s103.194kg101.670.52

1KE

3 16

19

2 4 27

E =γ 2and =γ

c mu

mc p

=

=

γγ

As the mass of particle approaches zero, its velocity u will approach c so that the

ratio of energy to momentum approaches

c c

c p

E

m→0 = 2 =

lim

, which is consistent with the equation c

E

p= for photons

48 Unreasonable Results A car feels a small force due to the light it sends out from its

headlights, equal to the momentum of the light divided by the time in which it is emitted (a) Calculate the power of each headlight, if they exert a total force of

N1000

2 × − 2 backward on the car (b) What is unreasonable about this result? (c)

Which assumptions are unreasonable or inconsistent?