Chapter 12 the Nature of Matter

8/12/2019 Chapter 12 the Nature of Matter

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Heinemann Physics 12(3e)Copyright Pearson Education Australia 2008 Teachers Resource and Assessment Disk(a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

1

Worked solutionsChapter 12 The nature of matter

12.1 Matter waves

1 B is the correct answer. The de Broglie wavelength of a particle is defined by = h/mv.

2 a = h/mv

= (6.63 ! 1034

J s)/(9.11 !1031

kg)(1.00 ! 107m s

1)

= 7.28 ! 1011

m

b = h/p

= (6.63 ! 1034

J s)/1.67 ! 1021

kg m s1

= 3.97 ! 1013

m

c p = (2 mEk)!

= [2(1.67 ! 1027

kg)(3.00 ! 1018

J)] != 1.00 ! 1022

kg m s1

and = h/p

= (6.63 ! 1034

J s)/1.00 ! 1022

kg m s1

= 6.62 ! 1012

m

d = (6.63 ! 1034

J s)/(0.400 kg)(10 m s1

)

= 1.66 ! 1034

m

3 a = (6.63 ! 1034

J s)/(40 ! 103

kg)(1.0 !103m s

1)

= 1.7 ! 1035

m

b There is no slit small enough for such a wave to pass through in order to be diffracted.

This wavelength is many times smaller than the radius of an atom.

c No. The objects that we encounter in our daily lives are simply too massive to have a

momentum small enough to produce a detectable matter wave.

4 a The wavelength of the gamma ray is given by:

= hc/E

= (6.63 ! 1034

J s)(3.00 ! 108m s

1)/(6.63 !10

14J)

= 3.00 ! 1012

mThen (1.67 !10

27kg) v= (6.63 ! 10

34J s)/3.00 ! 10

12m

and v = 1.32 ! 105 m s 1

b The wavelength of the yellow light is given by:

= hc/E

= (6.63 ! 1034

J s)(3.00 ! 108m s

1)/(2.15)(1.60 ! 10

19J)

= 5.78 ! 107

m

Then (1.67 !1027

kg) v= (6.63 ! 1034

J s)/5.78 ! 107

m

and v = 0.69 m s1

c The proton must have the same momentum as the electron.

Then (1.67 !1027

kg) v= 1.82 ! 1024

kg m s1

and v = 1.09!

10

3

m s

1


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Worked solutions Chapter 12 The nature of matter


2

5 a "EK = q"V

= (1.60 ! 1019

C)(50.0 V) = 8.0 !1018

m s1

b p = (2 m" Ek)!

= [2(9.11 ! 1031

kg)(8.0 ! 1018

m s1

)]!

= 3.82!

10

24

kg m s

1

c = h/p

= (6.63 ! 1034

J s)/3.82 ! 1024

kg m s1

= 1.74 ! 1010

m

6 "Ek= q"V

then p = (2 m" Ek)!= (2 mq"V)!

and = h/p = h/(2mq V)!

7 D. In order to undergo appreciable diffraction, the bombarding particle must have a wavelength

of around the same order of magnitude as the interatomic spacing. The other alternatives do not

satisfy these criteria.

The wavelength of the 1.0 keV electrons is:= (6.63 ! 10

34J s)/[(2(9.11 !10

31kg)(1.0 ! 10

3)(1.60 !10

19J)] !

= 3.9 ! 1011

m

8 a "V= h2/2mq(!)

= (6.63 ! 1034

J s)2/2(9.11 !10

31kg)(1.60 ! 10

19C)(2.0 !10

9m)

2

= 0.38 V

b "V= h2/2mq(!)

= (6.63 ! 1034

J s)2/2(1.67 !10

27kg)(1.60 ! 10

19C)(2.0 !10

9m)

2

= 2.0 ! 104

V

c "V= h2/2mq(!)

= (6.63!

10

34

J s)

2

/2(6.67!

10

27

kg)(3.20!

10

19

C)(2.0!

10

9

m)

2

= 2.57 ! 10

5V


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3

12.2 Photons shed light on atom structure1 a The term quantisation refers to the fact that the energy levels in an atom cannot assume a

continuous range of values but are restricted to certain discrete values, i.e. the levels are

quantised.

b The ground state is the lowest energy state that an atom can exist in, and represents thestable state of the atom before any external energy has been added.

c Excited states are the possible energy levels that an atom can move to after it has

absorbed energy from an external source.

d The ionisation energy is the least amount of energy required to eject an electron from an

atom, and represents the highest energy state of an atom.

2 a E3 E1 = h(6.00 ! 1014

Hz)

and E2 E1 = h(4.00 ! 1014

Hz)

then E3 E2= h (2.00 ! 1014 Hz) = h (3.00 ! 10

8 m s 1)/

and = (3.00 ! 108m s

1)/2.00 ! 10

14Hz

= 1.5 ! 106

m

b D is correct. This wavelength of light corresponds to the infrared part of the

electromagnetic spectrum.

3 a E2 E1 = (3.39 eV) (13.6 eV)

= 10.21 eV

= (10.21 eV)(1.60 ! 1019

C)

= 1.633 !1018

J = hf

Then f= 1.6336 ! 1018

J/(6.63 !1034

J s)

= 2.46 ! 1015

Hz

b E3 E1 = (1.51 eV) (13.6 eV)

= 12.09 eV

= (12.09 eV)(1.60 ! 10 19 C)= 1.934 !10

18J = hc/

Then = (6.63 ! 1034

Js)(3.00 ! 108m s

1)/1.934 !10

18J

= 1.03 ! 107

m

c The ionisation energy = 0 (13.6 eV) = 13.6 eV.

4 a The hydrogen atom can accept any fraction of the electrons energy that corresponds to

an exact difference in its energy levels. By accepting 12.09 eV of the electrons energy,

the atom can be excited to the n = 3 state.

b No. The difference between the ground state and n = 3 is 12.09 eV. A photon will not

give up a fraction of its energy to the atom.

cThe atom would be ionised and the ejected electron would be emitted with kinetic energy0.40 eV.

5 a E4 E1 = (0.88 eV) (13.6 eV) = 12.72 eV

Consequently the n= 4 state is the highest energy level that the atom can be excited to by

such a collision.

b There are six possible transitions as the hydrogen atom moves from the n = 4 state back

to the ground state. Each transition produces a photon energy defined by the difference in

energy levels. The possible photon energies are: 0.63 eV, 2.51 eV, 1.88 eV, 12.72 eV,

12.09 eV, 10.21 eV.

6 The excess energy is given to the electron in the form of kinetic energy. A freed electron may

have kinetic energy of any value, i.e. it is not quantised.


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4

7 Bohrs model states that electrons may only have specific energy values. The energy quanta

absorbed or emitted by an atom correspond to the differences between the allowed energy

states.

8 The shortest wavelength of light corresponds to the greatest frequency and therefore the greatest

energy change for the electron, E= 3.19 eV.

= hc/E=15 8

4.14 10 3.00 10

3.19

!

" " "

= 3.89 ! 107

m

9 Consider all of the possible jumps that can occur during de-excitation. These are n = 5 to 4, 5 to

3, 5 to 2, 5 to 1, 4 to 3, 4 to 2, 4 to 1, 3 to 2, 3 to 1, and 2 to 1. Therefore there are 10 different

photon energies emitted.

10 Like all atoms sodium atoms are able to emit more frequencies of light than they can absorb as

the electrons can fall down through orbits in stages or in a single fall. Atom excitation occurs in

single jumps from the ground state only.


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5

12.3 Bohr, de Broglie and standing waves1 The shortest wavelength corresponds to the highest frequency of light, which is produced by the

greatest energy level transition by an electron, i.e. from E= 0 eV to E= 3.4 eV.

2 Line 1: E= 3.4 eV to E= 13.6 eVSo " E= 10.2 eV

"E=!

hc

#1=E

hc

!=

10.2

1000.3104.14 815

!!! "

$ #1= 1.2 ! 107

m

Line 2: E= 1.5 eV to E= 13.6 eV

So " E= 12.1 eV

#2=E

hc

!

=12.1

1000.3104.14 815

!!! "

= 1.0 ! 107

m

Line 3: E= 0.85 eV to E= 13.6 eV

So " E= 12.75 eV

#3=E

hc

!

=12.75

1000.3104.14 815

!!! "

= 9.7 ! 108

m

3 The longest wavelength in the Balmer series occurs for "E= 3.4 1.5 = 1.9 eV.

#=E

hc

!

=1.9

1000.3104.14 815

!!! "

= 6.54 ! 107

m

= 654 nm

This is within the visible spectrum.

The shortest wavelength in the Balmer series occurs for "E= 3.4 0 = 3.4 eV.

#=E

hc

!

=3.4

1000.3104.14 815

!!! "

= 3.65 ! 107

m

= 365 nm

Therefore this highest frequency lies just above the highest visible frequency.

4 The longest wavelength corresponds to the smallest frequency and therefore the smallest

absorption energy.

From the ground state, the smallest "Eoccurs from E= 13.6 eV to E= 3.4 eV.

Hence " E= 3.4 13.6 = 10.2 eV.


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6

$ #1= 1.2 ! 107

m

5 de Broglie proposed a model of the atom in which electrons were viewed as matter waves with

resonant wavelengths.

6 The circumference of an electron orbit must correspond to an integer multiple of the de Broglie

wavelength of the electron, just as the violin string can only support waves of resonant

wavelengths that are integer multiples of the strings fundamental wavelength.

7 a E5 =2

5

6.13!

= 0.544 eV

E10 =2

10

6.13!

= 0.136 eV

E15 =2

15

6.13!

= 0.060 eV

E20 =2

20

6.13!

= 0.034 eV

b The differences between energy levels become smaller.


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7

Chapter review

1 The maximum amount of energy a mercury atom can accept from the beam = 8.8 eV. Then the

highest energy level that the mercury atom can be excited to is the n = 4 state.

There are six possible photon energies which can be produced as the mercury atom falls back tothe ground state. Each photon energy must correspond to an exact difference between levels.

Alternative C is the only photon energy that cannot be produced by such a transition.

2 There are three possible transitions as the mercury atom moves from the second excited state ( n

= 3) back to the ground state. They are:

E3 E1= (3.70 eV) (10.40 eV)

= 6.7 eV

= (6.7 eV)(1.60 ! 1019

C)

= 1.072 ! 1018

J

= (6.63 ! 1034

J s) f

and f= 1.62 ! 1015

Hz

E3 E2= (3.70 eV) (5.50 eV)

= 1.80 eV

= 2.88 ! 10 19 J

= (6.63 ! 1034

J s) f

and f= 4.34 ! 1014

Hz

E2 E1= (5.50 eV) (10.40 eV)

= 4.90 eV

= 7.84 ! 1019

J

= (6.63 ! 1034

J s) f

and f= 1.18 ! 1015

Hz

No other photon energies are possible.

3 a A high density of electrons

b A low density of electrons

c As de Broglie suggested, particles such as electrons can demonstrate a wavelength and

wave behaviour.

4 The pressure that a beam exerts on a surface is proportional to the change in momentum of each

particle in the beam as it strikes the surface. Assuming that the number of electrons striking the

target per second is the same, B is the correct answer.

5 a E =

hc

=9

834

10580

1000.31063.6

!

!

"

"""

= 3.429 !1019

J

p = (2 mEk)!

= (2 ! 9.11 ! 1031

) !(3.429 !1019

)!

= 7.90 ! 1025

kg m s1

b # =p

h

=25

34

1090.7

1063.6

!

!

"

"


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8

= 8.39 ! 1010

m

c Since F=area

pressureand F=

F

pn!

n =p

PA

!

=125

242

smkg1090.7

m104mN553.0!!

!!

"

""

= 2.80 ! 1020

6 The shortest wavelength of light corresponds to the highest possible frequency and therefore the

highest energy of light. Since all energies of light above the ionisation energy of hydrogen can

be absorbed, there is no limit.

7 Electrons in circular orbits around nucleus.

Positive attraction to negative provides centripetal force.

Quantised allowable orbit radii.Electron ordinarily occupies lower orbit.

Electron doesnt radiate energy while in stable orbit.

Only incident photons of energy equivalent to a difference between electron energy levels

can be absorbed.

As excited electrons drop from high to low energy levels, a photon of this energy

(difference) is emitted.

8 a Hydrogen can only absorb photons carrying precisely the right amount of energy to lift an

electron from one energy level to a higher level.

b If an electron is given 13.6 eV of energy by an incident photon, the electron will surpass

the highest possible orbital level.

9 The photon is absorbed, and the emitted photons go in all directions. Therefore in the direction

that the incident light shines, the intensity of this photon is largely reduced, producing the black

line.

10 Objects (for example, strings in musical instruments) have a set of possible modes or

frequencies of vibration. Standing waves are established with particular frequencies for each

object. The quantised states of the atom involve electrons only existing with particular energy

values also. The allowed energy values are particular to the type of atom. These allowed energy

values can be compared to the standing waves formed when physical objects vibrate. This is

considered evidence for the wavenature of matter (and therefore matter has a dual nature).


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9

Exam-style questions (Interactions of light andmatter)

1 a P= nE, where E= hc/

= (6.63 ! 1034

J s)(3.00 ! 108m s

1)/3.0 !10

6m

= 6.63 ! 1020

J

and n = P/ E= 60 W/6.63 ! 1020

J

= 9.0 ! 1020

b Infrared

2 a P= nE= n(hf)

then 100 W = n[(6.63 !1034

J s)(7.0 ! 1014

Hz)]

and n = 2.2 ! 1020

b p = E/c

= (6.63 ! 1034

J s)(7.0 ! 1014

Hz)/(3.00 !108

m s1

)= 1.5 ! 10

27kg m s

1

c F= n("p)

= 2.15 ! 1020

[2(1.547 !1027

kg m s1

)]

= 6.7 ! 107

N

3 a Energy of one photon, E= hf = 6.63 ! 1034

! 5.2 ! 1014

= 3.448 !1019

J

No. of photons per second = total energy delivered per second/energy per photon

=19

1000

3.448 10!

"

= 2.9 ! 1021

photons per second

b 1000 joules per second = 1000 watts

4 B. V0 is proportional to the energy of the incident photons. Since blue light has a higherfrequency than yellow light, its photons have more energy.

5 C. The retarding potential difference does work against the electrons as they try to reach the

collector.

A and B are incorrect because the potential difference between emitter and collector does not

affect these quantities.

6 A is the correct answer because the colour of the incident light is indicated by the value of V0,

while the intensity of the incident light is indicated by the size of the current.

7 Ek(max) = hc/ W

= (6.63 ! 10 34 J s)(3.00 ! 10 8 m s 1)/(3.00 ! 10 8 m s 1) W= (1.60 ! 10

19C)(36.43 eV)

then W= 8.012 !1019

J

= 5.01 eV

8 a Only certain frequencies of light will emit photoelectrons.

b There is no time difference between the emission of photoelectrons by light of different

intensities.

c The maximum kinetic energy of the ejected photoelectrons is the same for different light

intensities of the same frequency.

9 a = h/p, where p = mv


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10

= (9.11 ! 1031

kg)(3.0 ! 105m s

1)

= 2.733 !1025

kg m s1

then = (6.63 ! 1034

J s)/2.733 ! 1025

kg m s1

= 2.4 ! 109

m

b A series of bright and dark fringes

c The high-speed electrons are exhibiting wave-like behaviour.

10 2.42 ! 109

m = (6.63 ! 1034

J s)/(1.67 !1027

kg) v

and v= 1.64 !102m s

1

11 a E4 E1

= (1.60 eV) (10.40 eV)

= 8.80 eV

= 1.40 ! 1018

J

= (6.63 ! 1034

J s) f

then f = 2.11 ! 1015

Hz

and = (3.00 ! 108m s

1)/2.112 ! 10

15Hz

= 1.41 ! 10 7 m

b E2 E1

= (5.50 eV) (10.40 eV)

= 4.90 eV

= 7.84 ! 1019

J

= (6.63 ! 1034

J s) f

then f= 1.15 ! 1015

Hz

and = (3.00 ! 108m s

1)/1.15 ! 10

15Hz

= 2.60 ! 107

m

c E4 E3

= (1.60 eV) (3.70 eV)

= 2.10 eV= 3.36 ! 10

19J

= (6.63 ! 1034

J s) f

then f= 5.07 ! 1014

Hz

and = (3.00 ! 108m s

1)/5.07 ! 10

14Hz

= 5.92 ! 107

m

12 a W= QV

= 1.60 ! 1019

!65

= 1.04 ! 1017

= 1.0 ! 1017

J

b Ek = !mv2

1.04 !10

17= ! !9.11 !10

31 ! v

2

2.28 !1013

= v2

v = 4.8 !106m s

1

c # =mv

h

=631

34

108.4109.11

106.63

!!!

!

"

"

= 1.5 ! 1010

m


11/14



11

13 Neils Bohr would state that if incident light had an energy value less than the minimum energy

difference between the lowest and next orbital levels, the light would not result in any orbital

changes. Therefore the light would not be absorbed by the atom.

14 a Ionisation energy is 1810 eV.

b E= hf

1810 = 4.14 ! 10 15 ! f

f= 4.4 ! 1017

hz

c For excitation n = 1 to n = 2,

E= E2 E1

= 703 1810

= 1107 eV = hfand f = 2.67 ! 1017

Hz

d Yes, the atom would be ionised.

e v = 2% of 3.00 ! 108

= 6.00 ! 106m s

1

# =mv

h=

)1000.6()1011.9(207

106.63631

34

!!!!

!

"

"

= 5.9 ! 1013

m

15 de Broglie would say that the electrons (with their associated wavelengths) had diffracted as

they passed through the gaps between the atoms of the crystal, creating a diffraction pattern.

16 In addition to their particle properties, by existing in their quantised energy states they mimic

the standing waves (quantised energy) formed in many physical objects.

17 B, C, D, E. The photoelectric and Compton effects treat light as having particle-like properties

as well as wave properties.

18 An electronvolt is the energy that a single electron would gain after being moved through apotential of 1 V.

19 The electron absorbs energy required for release; excess energy results in extra kinetic energy of

the electron.

20 The surprise that light can display both particle and wave properties was repeated when

electrons were found to have particle properties and (when very fast moving) wave properties as

well.

21 a Fringe spacing doubles.

b Fringe spacing doubles.

c Fringe spacing doubles.

d White central band; fringe separates colours.

e Wider central bank

22 a W= QV

= 1.60 ! 1019

!10 ! 103

= 1.6 ! 1015

J

From W= Ek= !mv2

v =m

W2


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12

=31

15

109.11

101.62

!

!

"

""

= 5.9 ! 107m s

1

# =

mv

h

=731

34

109.5109.11

106.63

!!!

!

"

"

= 1.2 ! 1011

m

b Resolution is affected by diffraction. Significant diffraction occurs if object size %

wavelength. Electrons can have much smaller wavelengths than visible light.

23 For a given accelerating voltage, the proton would have slower speed, but its much greater mass

means # is shorter.

Velectron=m

QV2 Vproton =

m

QV2

= 5.9 ! 10 7 = 1.4 ! 10 6

#=mv

h # =

mv

h

=731

34

109.5109.11

106.63

!!!

!

"

"

=627

34

104.1101.67

106.63

!!!

!

"

"

= 1.2 ! 1011

m = 2.8 ! 1013

m

24 C. The de Broglie wavelength of a particle is given by = h/p and therefore depends only on

the momentum of the particle.

25 a The kinetic energy of the electron = 18 keV 16 keV = 2.0 keV.

b All photons travel at the speed of light c= 3.00 ! 108m s

1.

26 a Photon energy > ionisation energy, i.e. enough energy to free the electron.

b 14.0 13.6 = 0.4 eV = 6.4 !1020

J

c v = k2E

m=

31

20

109.11

106.42

!

!

"

""

= 3.7 ! 105m s

1

p = mv = 3.4 !1025

kg m s1

d # =p

h=

25

34

103.4

106.63

!

!

"

"

= 1.9 ! 109

m

16 keV

18 keV

2 keV


13/14



13

27 Since there is no energy level 10.0 eV above the ground state, the photon cannot be absorbed.

28 The correct answer is D. 12.0 eV is not sufficient energy to ionise the atom. As 12.0 eV does

not correspond to an exact difference between energy levels, the electron remains at the lowest

energy level.

29 a E= 10.40 1.60 = 8.8 eV

b # =E

hc=

8.8

1000.3104.14 815

!!! "

= 1.4 ! 107

m

c 10.40 eV

30 a 7.0 eV = 1.12 !1018

J

b n = 3

c E= 5.50 3.70 = 1.80 eV

E= 10.40 3.70 = 6.70 eV

E= 10.40 5.50 = 4.90 eV

d # =E

hc=

6.70

1000.3104.14 815

!!! "

= 1.9 ! 107

m

31 Incident energy = 30.4 eV + 10.40 eV = 40.8 eV

#=E

hc=

40.8

1000.3104.14 815

!!! "

= 3.0 ! 108

m

32 a Incident energy is insufficient for any excitation. At least 4.90 eV is required.

b

Electrons have escaped mercury atoms and conduct current across the tube.c First excitation level is reached. As electron falls, photon energy = 4.90 eV.

# =E

hc=

4.90

1000.3104.14 815

!!! "

= 2.5 ! 107

m

33 Since three full wavelengths fit into a circumference, n = 3.

34 A. Electron may only reach n = 3.

35 Extremely unlikely: electron only remains in excited state for less than a millionth of a second.

36 p=

h=

12

34

103.0

106.63

!

!

"

"

= 2.2 ! 1022

kg m s1

37 To produce diffraction patterns with the same fringe separation, they must have equivalent

wavelengths.

38 = c/f=8

18

3.00 10

8.3 10

!

!

= 3.6 ! 1011

m

393.6

!10

11

m, since they must have equivalent wavelengths.


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14

40 P= h/ =34

11

6.63 10

3.6 10

!

!

"

"

= 1.8 ! 1023

kg m s1

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Chapter 12 the Nature of Matter

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