Physics-12-Ch 11 & 12 & 13-Solns

Copyright 2003 Nelson Chapter 11 Einsteins Special Theory of Relativity 639

Unit 5 Matter-Energy Interface

ARE YOU READY? (Pages 558559)

Knowledge and Understanding 1.

2.

Frequency (Hz) Wavelength (m) Product f (m/s) Spectrum Location

5.0 1010

6.0 103

3.0 108 radio, TV

3.8 1014

8.0 107

3.0 108 infrared

1.2 1015

2.5 107

3.0 108 ultraviolet

1.0 1018

3.0 1010

3.0 108 X ray (a) The product is the same for each example of electromagnetic radiation, 3.0 108 m/s.

(b) The results conclude that all electromagnetic waves travel at exactly the same speed. (c) See table for spectrum location.

3. (a) The source of all electromagnetic waves is an oscillating electric or magnetic field. (b) Electromagnetic waves obey all of the laws of optics and travel through a vacuum. (c) 3.0 108 m/s 4. 1 positive; 2 negative; 3 no charge 5. An electron accelerated through an electrical potential of 258 V would gain 258 eV or (257 eV)(1.6 1019 J/eV).

me = 9.11 1031 kg v = ?

2K

K

19

31

6

12

2

2(258 eV)(1.6 10 J/eV) = 9.11 10 kg

= 9.5 10 m/s

E mv

Evm

v

=

=

The speed of the electron just before it hits the second plate is 9.5 106 m/s. 6. Law of conservation of energy: For an isolated system, energy can be converted into different forms, but it cannot be

created or destroyed. 7. (a) The two quantities conserved in elastic collisions are the total kinetic energy of the system and momentum. (b) For an elastic collision, we know that

1 1 2 2 1 1 2 2

p pm v m v m v m v

=

+ = +G G G G

Since m1 = m2 and v2 = 0, we can simplify to 1 1 2v v v = +

G G G

640 Unit 5 MatterEnergy Interface Copyright 2003 Nelson

We can apply the conservation of momentum to the x-components:

( )1 1 2

1

1

10.0 cm/s = (cos 30 ) 5.0 cm/s cos 609.05 cm/s

x x x

x

x

v v vv

v

= +

+

=

The speed of ball 1 after the collision is 9.05 cm/s. A scale vector diagram would yield the same result. 8. (a) The speed of light is a specific magnitude of speed (c = 3.00 108 m/s). A light-year is the distance light would travel

in one year. (b) v = (0.5)3.00 108 m/s = 1.5 108 m/s d = 4.3 ly = ( )8 16365 d 24 h 3600 s4.3 y 3.00 10 m/s 4.068 10 my d h = t = ?

16

8

8

4.068 10 m1.50 10 m/s2.712 10 s

dvtdtv

t

=

=

=

=

To convert to days:

8 3 31 h 1 d2.712 10 s 3.139 10 d, or 3.14 10 d3600 s 24 h

=

It would take you 3.14 103 days to travel to Alpha Centauri. (c) t = 8.3 min = 8.3 min 60 s

min = 498 s

8

11 8

= (3.00 10 m/s)(498 s)

1.49 10 m, or 1.49 10 km

d v t

d

=

=

The Sun is 1.49 108 km from Earth. 9. (a) 226 4 22288 2 86Ra He + Rn

(b) 214 0 21482 1 83Pb e + Bi

Math Skills 10. (a) y = mx + b (b) f-intercept = 6.1 1014 Hz E-intercept = 2.5 eV

(c) riseslope = run

1415

5.5 eV= 13.5 10 Hz

slope = 4.1 10 eV/Hz

The slope of the line on the graph is 4.1 1015 eV/Hz (d) y = 4.1 10-15x 2.5


11. (a) x = 0.5y t1 = 1.0 h

12 2

2

2

2

2

22

1

3.6 10 m/s

(0.5 )1

4.8 10 s

ttxy

yy

t

=

=

=

The value of t2 is 4.8 102 s.

(b) 4200125 220

2

t =

4

1log 25 log 220 log4200 2

log 25log 2204200

1log2

13 177 a, or 1.3 10 a

t

t

t

= +

= =

The value for t is 1.3 104 a.

Technical Skills and Safety 12. Some precautions when working with a source of ultraviolet light:

Do not stare directly at the light or light source. Wear protective eye wear, such as dark goggles or glasses with side shields that are specifically coated to block out

UV light. 13. Some precautions when using a high-voltage power supply:

Before connecting or disconnecting a high-voltage power supply, always make sure the power supply is in the OFF mode.

When plugging or unplugging cords, always grasp the base of the cord closest to the outlet, never yank on the cord. Keep all cords and plugs off the floor to avoid stepping on them.

CHAPTER 11 EINSTEINS SPECIAL THEORY OF RELATIVITY

Reflect on Your Learning (Page 560) 1. All motion is relative. To say that the ball is moving at 16 m/s, you would have to also state what it is relative to.

For example, you could state that the ball is moving at 16 m/s relative to the boat, the water, or Earth. 2. We must have some reference point (position, size, personal experience, etc.) in order to relate most things and opinions

we encounter in life. Thus, everything is relative to some frame of reference, whether physical or nonphysical. 3. Student answers will vary. What the student should understand from this question is that the perception of a time interval

is always relative to the observer; each observer can have a different view of the rate of the flow of time. 4. It is expected, under Newtonian physics, that the velocity vectors would add together and the speed of the laser light beam

relative to Earth would be 0.9c + c = 1.9c. For the speed of the laser light beam relative to the ship, it is expected that it would be c 0.9c = 0.1c. This is actually not the case.

5. On cursory analysis, the answer would be no. But, if it is known that the mass of the sun is slowly diminishing, the answer is yes, since mass is not conserved.


6. E = energy m = mass c = speed of light (3.00 108m/s)

The equation implies that as mass decreases or increases, energy inversely increases or decreases. Since energy cannot be created or destroyed, we conclude that mass is another form of energy.

Try This Activity: A Thought Experiment (Page 561) (a) Intuitive reasoning based on Newtons laws of motion tells us that we should catch up with the light waves. (b) The light waves should appear stationary, that is, light will stand still relative to the observer. (c) Classically, the beam should be some form of a wave front of energy. (d) According to Maxwells theory, and all reliable observations, there is no such thing as stationary light; no one has ever

held a stationary clump of light in the palm of his or her hand. Light always travels are 3.00 108 m/s in a vacuum no slower, no faster. (And as we will see in this chapter, in all frames of reference.)

(e) If the light is stationary relative to the observer, it would not flow to the observer nor be reflected from a mirror. 11.1 FRAMES OF REFERENCE AND RELATIVITY

Section 11.1 Questions (Page 568)

Understanding Concepts 1. There are no physics experiments or measurements that will determine whether you are moving. This is because no

physics experiment can determine whether one inertial frame is moving with respect to another inertial frame. 2. An inertial frame of reference is one at rest, or one moving with a constant velocity, while a noninertial frame of reference

is one that is accelerating. The Newtonian laws of physics are valid in all inertial frames of references, but break down in noninertial frames. Earth is strictly a noninertial frame of reference, but its acceleration (rotating on its axis while moving along its Sun orbit) is so small that it may be considered very nearly inertial. A rapidly accelerating car, an accelerating elevator, or a merry-go-round are all noninertial frames of reference where the normal laws of physics are found not to hold. We often account for unexpected motions that occur in noninertial frames by introducing fictitious inertial forces into the picture. The centrifugal force in a rotating frame of reference is an example of such a fictitious force. Such forces are not necessary if we pick an inertial frame.

3. The ball will fall in front of you. The ball will retain the horizontal component of velocity, v1, which the train had when the ball was thrown, and will travel a horizontal distance d1 = v1t while in the air. You and the train are slowing down while the ball is in the air. The ball will travel a horizontal distance d2 = vavgt, where vavg < v1. Therefore, the ball will land in front of you, since d1 > d2.

4. The Michelson-Morley experiment was designed to detect the so-called ether or absolute frame of reference against which the speed of light could be measured. Its great significance lies in the fact that it failed to detect such an effect, allowing for the conclusion, by Einstein, that the speed of light was absolute, regardless of the frame of reference from which it is measured.

5. (i) The laws of physics are valid in all inertial frames of reference. (ii) Light travels through empty space with a speed of 3.00 108 m/s, relative to all frames of reference. 6. Yes, in certain circumstances. Consider a stationary

observer, Os, standing midway between two lampposts in a straight line, as shown. If the lamps come on at the same time, he will observe them as simultaneous events since he is equidistant from them. If observer Om is moving towards Os as shown, he may be at Om1 when the lights come on, and at Om2 when he observes them. But since Om2L1 = Om2L2 , he will see them as simultaneous events. This will only be true if Om is moving toward, or away from Os. If he were moving in any other direction, he would not see the two events as simultaneous.


11.2 RELATIVITY OF TIME, LENGTH, AND MOMENTUM

PRACTICE (Page 573)

Understanding Concepts 1. Airline pilots watches are running slow relative to Earth. However, the change is so small that the watches do not record

any change at all when the pilots return to Earth. 2. v = 2.0 108 m/s

tm = 1.6 108 s ts = ?

sm 2

2

2

s m 2

89

8

8s

1

1

2.0 10 m/s(1.6 10 s) 13.0 10 m/s

1.2 10 s

ttvc

vt tc

t

=

=

=

=

The particles average lifetime when at rest is 1.2 108 s. 3. v = 0.60c

tm = 32.0 h v = 0.94c tm = ?

First we must determine the time interval relative to Earth:

( )

sm 2

2

s

2

2

s

1

32.0 h0.60

1

25.6 h

tt

vc

t

cc

t

=

=

=

To calculate the time interval relative to the Vulcans for their new speed:

( )

sm 2

2

2

2

m

1

25.6 h

0.941

75.0 h

ttvc

cc

t

=

=

=

The new time interval would be 75.0 h.


4. v = 2.6 108 m/s

( )

sm 2

2

m

2s

2

28

2

m

s

1

1

1

1

2.6 10 m/s1

2

ttvc

tt v

c

ctt

=

=

=

=

Its lifetime has increased by a factor of 2.

PRACTICE (Page 576)

Understanding Concepts 5. v = 0.90c

Lm = 50.0 m Ls = ?

( )

sm 2

2

ms 2

2

2

2

s

1

1

50.0 m

0.901

114.7 m, or 115 m

LL

vc

LLvc

cc

L

=

=

=

=

The spaceships length at rest is 115 m. 6. v1 = 0.60c

Lm = 8.0 ly v2 = 0.80c Lm = ?

21

m s 2

ms 2

12

1

1

vL Lc

LLvc

=

=


Since Ls is the same in both cases:

m m

2 21 22 2

m

m

1 1

8.0 ly0.80 0.60

6.0 ly

L L

v vc c

L

L

=

=

=

The friend measures the length of the trajectory to be 6.0 ly. 7. v = 0.65c

Lm =3.00 102 m Ls = ?

( )

2

m s 2

sm 2

2

2

2

2

2s

1

1

3.00 10 m

0.651

3.95 10 m

vL Lc

LLvc

cc

L

=

=

=

=

8. L0 = 40.0 ly (a) v = 1.00 108 m/s

L = ?

( )

2

0 2

28

2

1

1.00 10 m/s40.0 ly 1

37.7 ly

vL Lc

cL

=

=

=

You would determine the distance to be 37.7 ly. (b) t = ?

For you as the observer, v = 1.00 108 m/s, which is 3c .

37.7 ly

3113 a

dtv

c

t

=

= =

You would determine the journey to take 113 a.


9. v1 = 0.40c v2 = ?

2

m s 21vL Lc

=

Since Lm = Lm, and Ls = 2Ls:

2 22 1

s s2 2

2 222 2

222

222

2 2 2

2 2

2 1 1

0.402 1 1

44 0.84

0.8440.84

40.841

40.89

v vL Lc c

v cc cvcvc

c v c

v c

v c

=

=

=

=

=

= =

The speed of the faster spaceship is 0.89c relative to Earth.

PRACTICE (Page 578)

Understanding Concepts 10. v = 0.999c

me = 9.11 1031 kg p = ?

( )( )

( )

2

2

31

2

2

21

1

9.11 10 kg 0.999

0.9991

6.11 10 kg m/s

mvpvc

c

cc

p

=

=

=

The relativistic momentum of the electron is 6.11 1021 kgm/s. 11. m = 2.00 kg

v = 0.40c p = ?

( )

( )

2

2

2

2

8

1

2.00 kg 0.40

0.401

2.62 10 kg m/s

mvpvc

c

cc

p

=

=

=

The relativistic momentum in the Earth frame is 2.62 108 kgm/s.


12. v = 0.60c mp = 1.67 1027 kg p = ?

( )( )

( )

2

2

27

2

2

19

1

1.67 10 kg 0.60

0.601

3.76 10 kg m/s

mvpvc

c

cc

p

=

=

=

The relativistic momentum in that system is 3.76 1019 kgm/s.


Understanding Concepts 1. The clocks run at the same time if they are in the same inertial frame. Moving clocks affect the time of an event, as

measured by an observer in another inertial frame, where the object (with the clock) is moving relative to the observer. 2. To an observer moving with the process, the observer is stationary relative to the clock measuring the time. Proper time

for an observer moving relative to the process, the time interval (tm) will be longer.

3. Given that sm 221

tt

vc

=

and 2

m s 21vL Lc

= , if c approaches infinity, the expression 2

21vc

approaches 1. In this

case, there would be no relativistic effects for both time and length. 4. If the speed of light were 100 km/h, relativistic effects would become noticeable at much smaller speeds, and we would

be confined to a much smaller space in which we would move more slowly and time would pass more slowly. Travelling at 60 km/h (0.6c) would introduce time and length changes that play havoc with daily life. For example, people in a car would age slower than those at rest, and the car could not exceed 100 km/h. That is, 110 km/h would be impossible.

5. As discussed in Section 11.2, exceeding the speed of light is impossible. Matter can exceed the speed of sound, but this is not true for light.

6. ts = 2.2 106 s (a) v = 0.99c tm = ?

( )

sm 2

2

6

2

2

5m

1

2.2 10 s

0.991

1.6 10 s

tt

vc

cc

t

=

=

=

The average lifetime in the laboratory inertial frame of muons is 1.6 105 s. (b) d = ?

( )( )62

0.99 2.2 10 s

6.5 10 m

d v t

c

d

=

=

=

The muons could travel an average distance of 6.5 102 m in the laboratory inertial frame. (c) Lm = ?


2

m s 2

22

2

m

1

0.996.5 10 m 1

92 m

vL Lc

cc

L

=

=

=

The muons travel at an average distance of 92 m in the laboratory frame. 7. tm = 40 a

v = 2.5 108 m/s

( )

sm 2

2

2

s m 2

28

2

s

1

1

2.5 10 m/s40 a 1

22 a

ttvc

vt tc

ct

=

=

=

=

Therefore, her age upon return is 20 a + 22 a = 42 a. 8. v = 0.90c

Lm = 50.0 m Ls = ?

Since a stationary observer measures the relativistic length of the spaceship:

( )

2

m s 2

ms 2

2

2

2

s

1

1

50.0 m

0.901

115 m

vL Lc

LLvc

cc

L

=

=

=

=

The tape measure shows a length of 115 m. 9. v = 0.80c

Lm = 40.0 m L0 = 2.0 106 m

(a) Ls = ?

Since a stationary observer measures the relativistic length of the spaceship:

( )

2

m s 2

ms 2

2

2

2

s

1

1

40.0 m

0.801

66.7 m

vL Lc

LLvc

cc

L

=

=

=

=

The tape measure gives a value of 66.7 m.


(b) L = ? Since the observer in the spaceship observes the relativistic diameter of the planet:

( )

2

0 2

26

2

6

1

0.802.0 10 m 1

1.2 10 m

vL Lc

cc

L

=

=

=

The resulting value of the diameter of the planet is 1.2 106 m. (c) t = 8.0 s

t0 = ?

( )

02

2

2

0 2

2

2

0

1

1

0.808.0 s 1

4.8 s

tt

vc

vt tc

cc

t

=

=

=

=

The astronaut considers the journey to take 4.8 s. 10. bs = 1.00 m

v = 0.90c D = 2.70 103 kg/m3

(a) Only dimensions in the direction of motion are affected, therefore dimension b is affected. (b) Vm = ? First we must calculate the relativistic length of dimension b, since it is the only one affected by the motion:

( )

2

m s 2

2

2

m

b b 1

0.901.00 m 1

b 0.436 m

vc

cc

=

=

=

To calculate volume:

( )( )( )0 0

3

a bc1.00 m 0.436 m 1.00 m

0.436 m

V

V

=

=

=

The relativistic volume of the cube is 0.436 m3. (c) p = ? First we must calculate the rest mass of the cube:

( )( )3 3 33

2.70 10 kg/m 1.00 m

2.70 10 kg

m DV

m

=

=

=


To calculate the relativistic momentum of the cube:

( )( )

( )

2

2

3

2

2

12

1

2.70 10 kg 0.90

0.901

1.67 10 kg m/s

mvpvc

c

cc

p

=

=

=

The relativistic momentum of the cube is 1.67 1012 kgm/s. 11. m = 6.68 1017 kg

v = 0.400c p = ?

( )( )

( )

2

2

17

2

2

9

1

6.65 10 kg 0.400

0.4001

8.71 10 kg m/s

mvpvc

c

cc

p

=

=

=

Applying Inquiry Skills 12. Table 1

vc

221

vc

ts tm Ls Lm

0.050 0.9987 100 s 100.13 s 100 m 99.87 m 0.100 0.9950 100 s 100.50 s 100 m 99.50 m 0.150 0.9887 100 s 101.14 s 100 m 98.87 m 0.200 0.9798 100 s 102.06 s 100 m 97.98 m 0.250 0.9682 100 s 103.28 s 100 m 96.82 m 0.300 0.9539 100 s 104.83 s 100 m 95.39 m 0.350 0.9367 100 s 106.75 s 100 m 93.67 m 0.400 0.9165 100 s 109.11 s 100 m 91.65 m 0.450 0.8930 100 s 111.98 s 100 m 89.30 m 0.500 0.8660 100 s 115.47 s 100 m 86.60 m 0.550 0.8352 100 s 119.74 s 100 m 83.52 m 0.600 0.8000 100 s 125.00 s 100 m 80.00 m 0.650 0.7599 100 s 131.59 s 100 m 75.99 m 0.700 0.7141 100 s 140.03 s 100 m 71.41 m 0.750 0.6614 100 s 151.19 s 100 m 66.14 m 0.800 0.6000 100 s 166.67 s 100 m 60.00 m 0.850 0.5268 100 s 189.83 s 100 m 52.68 m 0.900 0.4359 100 s 229.42 s 100 m 43.59 m 0.950 0.3122 100 s 320.26 s 100 m 31.22 m

(a) 0.89c (b) 0.40c (c) 0.10c to 0.15c; Objects that travel at this speed are typically particles.


(d) Graph of tm versus v:

Graph of Lm versus v:

11.3 MASS AND ENERGY: E = mc2

PRACTICE (Page 583)

Understanding Concepts 1. mp = 1.67 1027 kg

E = ?

( )( )2

227 8

10

1.67 10 kg 3.00 10 m/s

1.50 10 J

E mc

E

=

=

=

To convert to electron volts:

10

191.50 10 J

1.60 10 J/eV939 MeV

E

E

=

=

The rest energy of the proton is 1.50 1010 J, or 939 MeV.


2. E = 2.00 102 MJ m = ?

( )

2

2

8

28

9

2.00 10 J

3.00 10 m/s

2.22 10 kg

E mcEmc

m

=

=

=

=

The rest mass of the small object is 2.22 109 kg. 3. v = 0.950c (a) Etotal = ?

( )( )

( )

2

total 2

2

227 8

2

2

10total

1

1.67 10 kg 3.00 10 m/s

0.9501

4.813 10 J

mcEvc

cc

E

=

=

=

To convert to MeV:

10

total 19

9 3total

4.813 10 J1.60 10 J/eV3.00 10 eV, or 3.00 10 MeV

E

E

=

=

The total energy is 3.00 103 MeV. (b) EK = ?

Since EK = Etotal Erest, we must first calculate Erest:

( )( )2

rest227 8

10rest

1.67 10 kg 3.00 10 m/s

1.503 10 J

E mc

E

=

=

=

After converting to MeV, the rest energy is 939 MeV. To calculate kinetic energy:

K total rest

3

3K

3.00 10 MeV 939 MeV

2.07 10 MeV

E E E

E

=

=

=

The kinetic energy is 2.07 103 MeV.


4. vi = 0c vf = 0.990c E = ?

( )( ) ( )

total rest2

2

2

2

2

2

2

231 8

2

2

13

1

1 1

1

19.11 10 kg 3.00 10 m/s 10.990

1

4.99 10 J

E E E

mc mcvc

mcvc

cc

E

=

=

=

=

=

The work required is 4.99 1013 J. 5. E = 9.80 1018 J

m = ?

( )

2

2

18

28

2

9.8 10 J

3.00 10 m/s

1.09 10 kg

E mcEm

c

m

= =

=

=

The amount of mass that would have to be converted to energy is 1.09 102 kg. 6. Ep = 938.3 MeV

En = 939.6 MeV m = ?

( )( )

( )( )( )

( )

2

2

28

28

6 19

28

30

939.6 MeV 938.3 MeV

3.00 10 m/s

1.3 MeV

3.00 10 m/s

1.3 10 eV 1.60 10 J/eV

3.00 10 m/s

2.31 10 kg

E mcEm

c

m

= =

=

=

=

=

The difference in their rest mass energy is 2.31 1030 kg



Understanding Concepts 1. The equation E = mc2 is consistent with the law of conservation of energy in that it predicts that whenever an amount of

mass m disappears, an amount of energy mc2 appears in some form or another. Mass is simply another form of energy to be considered when applying the law of conservation of energy.

2. m = 1.0 kg E = ?

( )( )2

28

16

1.0 kg 3.00 10 m/s

9.0 10 J

E mc

E

=

=

=

The amount of energy that can be produced is 9.0 1016 J. 3. mE = 5.98 1024 kg

v = 2.96 104 m/s m = ? First we must calculate the energy of Earth revolving around the Sun:

( )( )2

224 4

33

121 5.98 10 kg 2.96 1022.62 10 J

E mc

E

=

=

=

To calculate the amount of mass required:

( )

2

33

28

16

2.62 10 J

3.00 10 m/s

2.91 10 kg

Emc

m

=

=

=

The amount of mass required to accelerate Earth from rest to its orbital speed is 2.91 1016 kg. 4. E = 5.0 GW = 5.0 109 W

m = ? First we must convert the amount of energy produced per second in one year:

( )917

365 d 24 h 60 min 60 s5.0 10 J/sa d h min

1.577 10 J

E

E

= =

To calculate the amount of fuel converted:

( )

2

17

28

1.577 10 J

3.00 10 m/s

1.75 kg, or 1.8 kg

Emc

m

=

=

=

The amount of fuel converted to energy in one year is 1.8 kg. 5. v = 0.90c

m = 1.673 1027 kg EK = ?


( )( ) ( )

K total rest2

22

2

22

2

227 8

2

2

30K

1

1 1

1

11.673 10 kg 3.00 10 m/s 10.90

1

1.95 10 kg

E E E

mc mcvc

mcvc

cc

E

=

=

=

=

=

The energy required to accelerate the particle is 1.95 1010 J. 6. E = 1.60 MeV = ( )( )6 191.60 10 eV 1.60 10 J/eV = 2.56 1013 J ET = 3.20 MeV (a) m = ?

( )

2

2

13

28

30

2.56 10 J

3.00 10 m/s

2.84 10 kg

E mcEmc

m

=

=

=

=

The rest mass of the particle is 2.84 1030 kg. (b) EK = ?

K total rest

K

3.20 MeV 1.60 MeV1.60 MeV

E E E

E

=

=

=

The kinetic energy of the particle is 1.60 MeV.

Making Connections 7. E = 9.0 1016 J (from question 2)

t = 1 h = 3600 s cost = ? Recall that a 1-W power source delivers 1 J per second.

169.0 10 W

P E

P = =

Therefore,

( )( )16

16

9.0 10 W 1 s

9.0 10 W s

EPt

E P t

E

=

=

=

=


To convert to kWh: 16 103

1 h 1 kW9.0 10 W s 2.5 10 kW h3600 s 1.0 10 W

=

To calculate the cost:

10

10

$0.15cost 2.5 10 kW hkW h

cost $0.38 10

=

=

The cost of the rest energy would be $3.8 109. 8. Let the subscript g represent gas and the subscript a represent the Aspirin tablet.

m = 325 mg = 0.000325 kg, or 3.25 104 kg Eg = 1.05 108 J dg = 30.0 km da = ? First calculate the energy provided by the Aspirin tablet:

( )( )2

a a24 8

13a

3.25 10 kg 3.00 10 m/s

2.925 10 J

E m c

E

=

=

=

To determine how far the car could travel:

( )( )

aa g

g

13

8

6a

2.925 10 J 30.0 km1.05 10 J

8.36 10 km

Ed d

E

d

=

=

=

The distance the car could travel on a single Aspirin tablet is 8.36 106 km. 9. Erest for deuteron = 1876 MeV

Total rest mass of proton and neutron = 911 MeV + 941 MeV E = 1876 MeV 1842 MeV E = 24 MeV

The amount of energy liberated is 24 MeV. 11.4 THE LIFE AND TIMES OF ALBERT EINSTEIN

PRACTICE (Page 586)

Making Connections (a) A description of Einsteins early education, his lack of success in school, why he was not accepted at universities, why he

ended up as a clerk in the patent office, etc. (b) This topic emphasizes his humanitarian interests and outlines his pacifism, with some emphasis on his intervention during

the war years. (c) Although he was not involved in the project, he was instrumental in encouraging the projects beginning and asking that

its results (the atomic bomb) be demonstrated, not used on Japan. The role of others, such as Leo Szilard, who cajoled and convinced Einstein to write letters to President Roosevelt, would be part of the report.

(d) In the post war period, with nuclear proliferation collimating in the testing of the hydrogen in 1951, many concerned scientists opposed these developments, often being accused as being communist sympathizers. This campaign and Einsteins role should be the emphasis in this report.

(e) An overview of what Einstein was attempting would begin this report, followed by why he was unsuccessful. An overview of developments since his death could complete the report.


CHAPTER 11 SUMMARY

Make a Summary (Page 587) (a) Although time will move at the same rate, as seen by the occupants of the spacecraft, the time on the space craft, relative

to the people left behind, will be slower. As a result, the occupants of the spacecraft will age at a slower rate than their friends and relatives in Earth.

(b) Barnards star is far enough away that the round trip would be over 12 ly. Assuming the spacecraft does not stop on the way, and it is travelling at close to the speed of light, the twin paradox will prevail. All your friends and relatives will be much older that you.

(c) Although your spacecraft may end up travelling at close to the speed of light, it must accelerate to that speed, requiring more and more energy the faster it travels. To achieve the speed of light, an infinite amount of energy would be required, which is impossible. Nevertheless, the maximum speed will be limited by the large amount of energy required, and the inability to carry enough fuel to reach these speeds.

(d) Since it is impossible to carry enough fuel or provide enough propulsion even from ion engines, somehow mass must be converted to energy, using the mass-energy conversion relationship. The fuel would have to be swept up from the free hydrogen in space and converted into energy in some type of thermal nuclear system. (Many physicists are skeptical that these types of space travel will possible at all, given the present state of knowledge.)

CHAPTER 11 SELF QUIZ (Pages 588589)

True/False 1. F The speed of light in water is less than the speed of light in a vacuum. This fact does not violate the speed-of-light

postulate of the special theory of relativity because it states ... light travels through empty space with a speed of c 2. T 3. T 4. (a) T (b) F Since both observers are moving relative to each other, and simultaneously moving relative to a clock, they are

measuring two-position time, and will probably never agree on the rate at which the clock ticks. (c) T 5. F If you are in an inertial frame at a point a few thousand kilometres above the North Pole, the clock at the pole is

stationary relative your inertial frame, while the equator is moving relative to your frame, As a result, there will be time dilation at the equator and the clock will run slower at the equator than at the North Pole, relative to your frame of reference.

6. F She could be addressing her son (see twin paradox, Section 11.2). 7. (a) T (b) T 8. T 9. T 10. F The rest mass energy is contained in the molecules of the spring. When the spring is compressed the mass does not

change, only the potential energy in the spring changes (increases). 11. F The classical laws of conservation of energy and conservation of mass need to be modified mass-energy conversion, but

this is usually only a factor at speeds greater than 0.1c

Multiple Choice 12. (d) In Newtonian mechanics, no physics experiment can identify which inertial frame is truly at rest and which is moving.

Since the train is windowless, there is no external frame of reference and thus there cannot be any relative velocity. 13. (c) Second postulate: Light travels through empty space with a speed of c = 3.00 108 m/s in all inertial frames of

reference no more no less. Thus, (c) is the only correct option. 14. (d) Simultaneity is relative to the inertial frame of reference. Simultaneity is a relative concept (see Section 11.1). 15. (a) The MichelsonMorley experiment established that there is no observable ether wind. All of the other options are

incorrect. 16. (b) Speed of light is constant in all inertial frames of reference (see question 13). Thus, (b) is the only correct answer.


17. (e) None of the options will inform you that you are moving, since none of the changes indicated will be observable to the astronaut.

18. (a) Since you are observing the duration of two events (two position time), there will be time dilation and the clock will tick slower, relative to the observer

19. (a) Proper time is the duration of a process measured by the observer who sees the entire process as occurring in one position. Thus, the correct option is (a).

20. (b) On the spaceship, time would run at the same rates as those seen by the occupants. Thus, the value does not change relative to those on the spaceship. In other words, the hearts beat at the same rate. Of course, to an observer in another inertial frame of reference, the rate could be slower.

21. (d) sm 221

ttvc

=

s2

2

s

s

m s

(0.5 )1

1 0.25

0.8771.2 or 1.2

t

cc

t

t

t t T

=

=

=

=

The period is 1.2T, therefore (d) is correct. 22. (b) Length contraction only occurs along the length of the moving object as seen by an external observer. Thus, (b) is the

only correct answer. 23. (a) 2E mc =

2

6

2

9

3.7 10 J

4.1 10 kg

Emc

cm

=

=

=

Matter is converted to energy at the rate of 4.1 109 kg/s, therefore (a) is the correct answer. CHAPTER 11 REVIEW (Pages 590591)

Understanding Concepts 1. (a) The passenger measures the proper time since the passenger is stationary relative to the clock. (b) Again, the passenger is stationary relative to the train, and would measure the proper length. (c) You would measure the proper length between the railway ties, since you are stationary relative to the railway ties. 2. Only observers stationary relative to the event would record the proper time interval. (a) The spectator is stationary relative to the event and is in the same inertial frame. The spectator would record the proper

time. (b) The fan sitting on the couch is stationary relative to the event, because the fan is watching the game through the TV

cameras eyes, which is also stationary relative to the event. The fan would record the proper time. (c) The short stop is moving relative to the event since he is running in to cover the play. He would not record the proper

time. 3. Yes, all motion is relative to the observer. Therefore, stationary and moving can vary between an observer on the plane

and an observer on Earth. 4. It means it does move slower. This is the actual behaviour, not just the perception of the observer. 5. As discussed in the text, the clocks in orbit will run slower by a factor predicted by the time dilation relationship.

Although the difference will be small, the two clocks will no longer by synchronizedthe clock on the airplane will be slower.


6. To measure proper time, since the observer is at rest relative to the event, only one clock is needed. To measure time intervals, where the object or event is moving relative to the observer, the beginning and ending of the event are in two positions. Thus, two synchronized clocks are required.

7. One instance is the experiment where synchronized clocks were sent around the world on passenger jets and then compared with another at rest on Earth. Another instance is the lifetime of high speed particles (muons) increased when accelerated to velocities approaching c.

8. v = 0.92c Lm = 48.2 m Ls = ?

( )

2

m s 2

ms 2

2

2

2

s

1

1

48.2 m

0.921

123 m

vL Lc

LLvc

cc

L

=

=

=

=

The ship is 123 m long at rest. 9. Ls = 3.20 103 km

v = 0.70c Lm = ?

( )

2

m s 2

23

2

3m

1

0.703.20 10 km 1

2.28 10 km

vL Lc

cc

L

=

=

=

The voyagers aboard the UFO measure the distance as 2.28 103 km. 10. v = 0.90c

Lm = 228 m Ls = ?

( )

ms 2

2

2

2

s

1

228 m

0.901

523 m

LLvc

cc

L

=

=

=

The length you now obtain is 523 m. 11. vX = 0.70c

vY = 0.86c X

Y?

LL

=


( )

( )

2X2

X m2

Y m Y2

2

2

2

2

X

Y

1

1

0.701

0.861

0.7140.510

1.4

vLsL L cL L vLs

c

cc

cc

LL

= =

=

=

=

The ratio of the values for the planets diameter is 1.4. 12. v = 0.866c p = ?

( )( )

( )

2

2

31

2

2

30

1

9.11 10 kg 0.866

0.8661

1.58 10 kg m/s

mvpvc

c

cc

p

=

=

=

The relativistic momentum of the electron is 1.58 1030 kgm/s. 13. E = 4.9 MeV = 4.9 106 eV = 7.84 1013 J

m = ?

( )( )

2

13

28

30

7.84 10 J

3.00 10 m/s

8.7 10 kg

Emc

m

=

=

=

The rest-mass equivalent of the energy is 8.7 1030 kg. 14. P = 3.00 109 W = 3.00 109 Js t = 1 a = 7365 d 24 h 3600 s 3.15 10 s

a d h

= m = ?

9 93.00 10 W 3.00 10 J/s

EPt

=

=

To calculate Etotal:

( )9 7total

16total

3.00 10 J 3.15 10 ss

9.45 10 J

E

E

= =


To calculate the change in mass:

( )

2

16

28

9.45 10 J

3.00 10 m/s

1.05 kg

Emc

m

=

=

=

The change in mass is 1.05 kg. 15. m = 9.11 1031 kg

v = 0.20c Etotal = ?

Total energy before the collision = EK electron + EK positron + Erest electron + Erest positron, but rest masses are equal and thus rest energies of the two particles are also equal.

( )

( )( )( )

2 2total

2 2

2 2

231 2

14total

12 22

9.11 10 kg 0.20

9.84 10 J

E mv mc

mv mc

m v c

c c

E

= + = +

= +

= +

=

Converting to MeV: 14

519

9.84 10 J 6.15 10 eV, or 0.615 MeV1.6 10 J/eV

=

The energy of the electromagnetic radiation is 0.615 MeV. 16. Erest = 105.7 MeV

Etotal = 106.7 MeV EK = ?

K total rest

K

106.7 MeV 105.7 MeV1.0 MeV

E E E

E

=

=

=

The kinetic energy of the muon is 1.0 MeV. 17. (a) m = 1.20 103 kg

v = 28.0 m/s c = 47.0 m/s EK = ?

( )( ) ( )( )

K total rest

22

2

2

2

2

2

23

2

2

5K

1

1 11

11.20 10 kg 47.0 m/s 128.0 m/s

147.0 m/s

6.49 10 J

E E E

mc mcvc

mcvc

E

=

=

=

= =

The relativistic kinetic energy of the car would be 6.49 105 J.


(b) K relK nonrel

?E

E=

First we must calculate the nonrelativistic kinetic energy:

( )( )2

K nonrel

23

5K nonrel

121 1.2 10 kg 28.0 m/s24.70 10 J

E mv

E

=

=

=

To calculate the ratio:

5K rel

5K nonrel

K rel

K nonrel

6.49 10 J4.70 10 J

1.4

EEE

E

=

=

The ratio is 1.4:1. 18. Etotal = 1.00 1068 J 1

2Etotal = 5.00 1067 J

m = 4.00 1030 kg number of stars = ? First we must calculate the change in mass:

( )

2

67

28

50

5.00 10 J

3.00 10 m/s

5.56 10 kg

Emc

m

=

=

=

To calculate the number of stars:

50

30

20

5.56 10 kgnumber of stars4.00 10 kg/star

number of stars 1.39 10 stars

=

=

The number of stars half of the energy could create is 1.39 1020 stars. 19. m = 1.97 1031 kg

E = 1.02 1044 J (a) m = ?

( )

2

44

28

27

1.02 10 J

3.00 10 m/s

1.13 10 kg

Emc

m

=

=

=

1.13 1027 kg of mass are converted to energy in the explosion. (b)

star

?mm

=

27

31star

5

star

1.13 10 kg1.97 10 kg

5.75 10

mm

mm

=

=

The ratio of the mass destroyed to the mass of the star is 5.75 105:1.


20. E = 1.00 1034 J (a) E = 1.00 1033 J

m = ?

( )

2

33

28

16

1.00 10 J

3.00 10 m/s

1.11 10 kg

Emc

m

=

=

=

The decrease in the mass of the oceans would be 1.11 1016 kg. (b) Dw = 1.00 103 kg/m3

V = ?

16

3 3

13 3

1.11 10 kg1.00 10 kg/m1.11 10 m

mVD

V

=

=

=

The volume of water this corresponds to is 1.11 1013 m3. 21. (a) energy consumption = 485 kWh m = 1.00 g efficiency = 40% t = ?

5

94.85 10 J h 3600 s485 kW h 1.746 10 J/months h

= =

For 12 months total energy: 9 101.746 10 J 12 2.095 10 J =

To calculate the change in energy:

( )( )2

23 8

13

1.00 10 kg 3.00 10 m/s

9.00 10 J

E mc

E

=

=

=

To calculate time:

( )13

9

4

0.40 9.00 10 J

1.746 10 J2.06 10 months

t

t

=

=

A mass of 1.0 g would last 2.06 104 months. (b) number of homes?

13

310

9.00 10 Jnumber of homes 4.30 10 homes2.095 10 J/homes

= =

The number of homes that could be supplied with this energy is 4.30 103 homes.

Making Connections 22. Science fiction authors utilize some of the concepts of special relativity in their writing to suggest time as a fourth

dimension, time standing still, space warp, and vapourization of mass into energy (ray guns).

23. Since v = cn

, and n =1.33 for water:

299 792 500 m/s of light in water 230 609 315 m/s1.33

v = =

Pavel Cerenkov had succeeded in speeding electrons up to 257 500 000 m/s by 1934, thus sending particles at speeds greater than light. particles with kinetic energies of 0.26 MeV will travel at speeds in excess of 2.30 106 m/s in a dense medium such as water. Frank and Tamm also conducted this work in 1937 and the three shared the 1958 Nobel prize for their work in this area.


While no particle can exceed the speed of light in a vacuum (3.00 108 m/s), it is possible for a particle to travel faster than the speed of light in certain media, such as water. This is not a violation of the Einsteins prediction (Section 11.2) that no object can equal or exceed the speed of light since it refers to the situation in a vacuum or air, that is, 3.00 108 m/s. When the charged particle moves through the water it tends to polarize the water molecules in a direction.

Extension 24. p = ?

First we must calculate the value for v:

total rest2

total2

2

2

2

2

2

2

2

2

2

2

2

2

5

5

5

1

1151125

1125

0.96

0.960.98

E E

E mc

mc mcvc

vcvcvcvc

v cv c

=

=

=

=

=

=

=

=

=

To calculate momentum:

( )( )

2

2

31

21

1

9.11 10 kg 0.9815

1.34 10 kg m/s

mvpvc

c

p

=

=

=

The momentum of the electron in that frame is 1.34 1021 kgm/s. 25. s m

12

t t=

m2

s2

2

2

2

2

2

2

12

1

11434

0.75

0.750.87

tt v

cvcvcvc

v cv c

= =

=

=

=

=

=

The value of v, expressed as a fraction of c is 0.87c.


26. pR = 3p

R 2

2

2

2

2

2

2

2

2 2 2

1

3

1

113119

9 9

89

0.943

mvpvc

ppvc

vcvc

c v c

v c

v c

=

=

=

=

=

=

=

The speed of the particle is 0.943c, or 2.83 108 m/s. 27. m = 1.40 105 kg

p = 3.15 1013 kgm/s v = ?

( )

( )

( ) ( )( ) ( )

( )

2

2

513

2

2

13

52

2

8

2

2

2 282

2

22 22 8 82

2 22 8 2 8 2

2 17 2 2

1

1.40 10 kg3.15 10 kg m/s

1

3.15 10 kg m/s1.40 10 kg

1

2.25 10 m/s

1

2.25 10 m/s1

2.25 10 m/s 2.25 10 m/s

2.25 10 m/s 2.25 10 m/s

1.406 10 m /s 4

mvpvc

v

vc

v

vc

v

vc

vvc

vvc

cv c v

v

=

=

=

=

=

=

=

= 33 2 2

8

.556 10 m /s

1.80 10 m/sv

=

The rocket is travelling at a speed of 1.80 108 m/s. 28. (a) In the special theory of relativity, Einstein concluded that there is no way for an observer to determine whether a given

frame of reference is at rest or is moving at constant velocity in a straight line. Therefore, Einstein said, the laws of physics must be the same in different inertial reference frames. To consider only uniformly moving reference frames is somewhat restricting. What about the general case of motion, where reference frames can be accelerating?

It is in the general theory of relativity that Einstein tackled the problem of accelerating reference frames and developed a theory of gravity. You sense that you are accelerating when in a car moving around a sharp curve. You


could not prove by any experiment that in fact you werent simply experiencing the pull of a gravitational field. Conversely, we might think we are being pulled by gravity when in fact we are undergoing an inertial acceleration having nothing to do with gravity. For example, pilots making a steep bank often have this experience, and cannot tell in which direction Earth lies without their instruments.

As a thought experiment, consider a person in a freely falling elevator near Earths surface. If an observer held out a ball and let go of it, gravity would pull it downward toward Earth. But, it would fall at the same rate (g = 9.8 m/s2) at which the person and elevator were falling. So the ball would float right next to the persons hand. The effect is exactly the same as if the elevator was at rest and no forces were acting on it. On the other hand, suppose the elevator were far out in space where there is no gravitational field. If the person released the ball, it would float. If instead, the elevator in space were accelerating upward at an acceleration of 9.8 m/s2, the ball as seen by an observer would fall to the floor with an acceleration of 9.8 m/s2, just as if it were falling because of gravity on Earth. According to the principle of equivalence, the observer could not do an experiment to determine whether the ball fell because the elevator was accelerating upward at 9.8 m/s2 in the absence of gravity, or because a gravitational field with g = 9.8 m/s2 was acting downward and the observer was at rest on Earth. The two descriptions are equivalent.

The principle of equivalence is related to the concept of mass and to the idea that there are two types of mass. For any force, Newtons second law says that F = ma, where m is the inertial mass. The more inertial mass a body has, the less it is affected by a given force and the less acceleration it undergoes. You might say that inertial mass represents resistance to any type of force. The second type of mass is gravitational mass, present when one body attracts another by the gravitational force. Up to this point, inertial and gravitational masses are the same because no experiment, not even high-precision experiments, has been able to discern any measurable difference between inertial and gravitational mass. This, then, is another way to state the equivalence principle: gravitational mass is equivalent to inertial mass.

(b) The principle of equivalence can be used to show that light ought to be deflected due to the gravitational force of a massive body. Let us consider a thought experiment to get the idea. Consider an elevator in free space where no gravity acts. If there is a hole in the side of the elevator and a beam of light enters from outside, the beam travels straight across the elevator and makes a spot on the opposite side if the elevator is at rest. If the elevator is accelerating upward, the light beam still travels straight across in a reference frame at rest. In the upward accelerating elevator, however, the beam is observed to curve downward. Why? Because during the time the light travels from one side of the elevator to the other, the elevator is moving upward at with an increasing speed. Now, according to the equivalence principle, an upwardly accelerating reference frame is equivalent to a downward gravitational field. Hence, we can picture the curved light path in as being the effect of a gravitational field. Thus we expect gravity to exert a force on a beam of light and to bend it out of a straight-line path.

(c) That light is affected by gravity is an important prediction of Einsteins general theory of relativity, and it can be tested. The amount a light beam would be deflected from a straight-line path must be small even when passing a massive body. For example, light near Earths surface after travelling 1.0 km is predicted to drop only about 1010 m, which is not detectable. The most massive body near us is the Sun, and it was calculated that light from a distant star would be deflected by 1.75 of arc as it passed near the Sun. However, such a measurement could be made only during a total eclipse of the Sun, so the Suns intensity would not overwhelm the starlight passing near its edge. An eclipse occurred in 1919 and scientists journeyed to the South Atlantic to observe it. Their photos of stars around the Sun revealed shifts in accordance with Einsteins prediction, producing headlines in the popular press at the time and making Einstein famous beyond the scientific community.

Copyright 2003 Nelson Chapter 12 Waves, Photons, and Matter 667

CHAPTER 12 WAVES, PHOTONS, AND MATTER

Reflect on Your Learning (Page 592) 1. In both cases, if white light is emitted, most of the wavelengths of the visible spectrum originate in the source of light.

However, the sources of energy are different. In the incandescent lamp, thermal energy agitates the atoms and molecules to the point that their changing electric and magnetic fields radiate electromagnetic energy. In the fluorescent lamp, changes in the energy levels within the atom produces radiation, not heat. It is this radiation that causes fluorescence in the phosphors on the inside of the tube, therefore producing white light.

2. Light energy causes changes and eventually a breakdown in the colour pigments in the printing ink. A red colour printing appears red because most of the other wavelengths in the white light spectrum are absorbed, and the red wavelength is reflected. For a blue surface, most of the wavelengths of the white line spectrum are absorbed, and the blue wavelength is reflected. Since the emissions of red and green wavelengths from the surface are of a lower energy than that of blue, the source of energy reflected by these pigments or dyes require less energy to be damaged by the incident radiation, and thus fade sooner than that for blue and violet.

3. Quantum is defined as an abrupt change, sudden increase, or dramatic advance. A quantum leap is an abrupt change or advance, particularly in method, information, or knowledge.

4. An electron microscope is a microscope that uses electron rather than visible light to produce images. 5. (a) The force holding the satellite in orbit is the centrally acting gravitational force between the satellite and Earth. (b) Friction on the satellite and irregularities in the orbit eventually cause its speed to decrease and it enters successively

low orbits. Eventually, it enters Earths atmosphere and either burns up or crashes on Earths surface. (c) The Coulomb force of attraction between the positively charged nucleus and negatively charged electron holds the

electron in orbit around the nucleus of the atom. The electron stays in a stable orbit and does not 'crash' into the nucleus.

Try This Activity: Discharging with Light (Page 593) The discharge times will vary, depending on the level of humidity in the air. The negatively charged zinc plate will typically discharge within 2 min., as water molecules adjacent to the plate absorb

electrons. With the glass filter in place, the ultraviolet light incident on the zinc plate should have little affect and the discharge rate

will be similar to that without the ultraviolet light. Without the glass filter in place, the ultraviolet light causes the zinc plate to discharge quite quickly, typically in one-half to

one-quarter of the time without the ultraviolet light. (a) For the negatively charged electroscope to discharge, the electrons on the zinc plate must have escaped into the

atmosphere. (b) The ultraviolet light has provided energy to the electrons making it easier for them to escape. 12.1 FOUNDATIONS OF QUANTUM THEORY

PRACTICE (Page 597598)

Understanding Concepts 1. Discrete quantities include money, matter (atoms), and scores because all are exact. Nondiscrete quantities include time, energy, and length because all are measured and therefore error is introduced. 2. h = 6.63 1034 Js

c = 3.00 108 m/s E = ?


(a) = 941 nm = 9.41 107 m

( )( )34 8

7

19

6.63 10 J s 3.00 10 m/s

9.41 10 m2.1 10 J

hcE

E

=

=

=

The energy is 2.1 1019 J, or 1.32 eV. (b) f = 4.4 1014 Hz

( )( )34 1419

6.63 10 J s 4.4 10 Hz

2.9 10 J

E hf

E

=

=

=

The energy is 2.9 1019 J, or 1.8 eV (c) = 435 nm = 4.35 107 m

( )( )34 8

7

19

6.63 10 J s 3.00 10 m/s

4.35 10 m4.57 10 J

hcE

E

=

=

=

The energy is 4.57 1019 J, or 2.86 eV. (d) f = 1.2 1018 Hz

( )( )34 1816

6.63 10 J s 1.2 10 Hz

7.96 10 J

E hf

E

=

=

=

The energy is 7.96 1016 J, or 5.0 103 eV. 3. E = 3.20 1019 J = ?

( )( )34 8

19

7

6.63 10 J s 3.00 10 m/s

3.20 10 J6.22 10 m

hcE

=

=

=

The wavelength is 6.22 107 m, or 622 nm. The colour is red (from Section 9.6). 4. E = 2.25 eV

To convert energy into joules: E = (2.25 eV)(1.60 1019 J/eV) E = 3.60 1019 J

To calculate frequency:

1934

14

3.60 10 J6.63 10 J s5.43 10 Hz

E hfEfh

f

=

=

=

=

The frequency is 5.43 1014 Hz. 5. 1 = 3.80 107 m 2 = 1.14 107 m

21

= ?


Since photon 11

hcE

= and photon 22

hcE

= ,

photon 1 1

photon 2

2

2

17

7

photon 1

photon 2

3.80 10 m1.14 10 m3.33

1

hcEE hc

EE

= =

=

=

The ratio of a quantum of soft ultraviolet radiation to a quantum of hard ultraviolet radiation is 3.33:1.

Making Connections 6. Let the subscript R stand for radio waves, and C stand for cosmic rays.

R = 1.0 102 m C = 1.0 1016 m

Since, RR

hcE

= and Cc

hcE

= ,

RR

C

C

C

R16

2

18R

C

1.0 10 m1.0 10 m

1.0 10

hcEE hc

EE

= =

=

=

It would take 1018 quanta of radio waves. Radio waves have a long wavelength (low frequency), thus they have low penetration and low energy. Cosmic radiation has a high frequency, therefore, they have high energy and high penetrating power, which would be a biological hazard.

PRACTICE (Pages 604605)

Understanding Concepts 7.

8. Intensity is the quantity or number of protons. Frequency determines the energy (E = hf). For photoemission to occur, the

photoelectron needs sufficient energy to be released from the surface. If the photons, no matter how many, do not have sufficient energy, no photoemission occurs.


9. The energy of the emitted photons is directly related to the energy of the incident photon (E = hf) by the constant h. Thus, all the graphs have the same slope (h).

10. According to classical wave theory, the energy would have to build up to the point where the photoelectron is released from the surface, which isnt the case. The photoelectron is released immediately if there is sufficient energy.

11. W = 2.4 eV f = ?

( )( )

photon

19

34

14

2.4 eV 1.6 10 J/eV

6.63 10 J s5.8 10 Hz

E W

hf WWfh

f

=

=

=

=

=

The minimum frequency of the photon is 5.8 1014 Hz. 12. W = 3.33 eV f = ?

( )( )19

34

14

3.33 eV 1.6 10 J/eV

6.63 10 J s5.02 10 Hz

Wfh

f

=

=

=

The threshold frequency is 5.02 1014 Hz. 13. W = 2.48 eV

= 450 nm = 4.50 10-7 m EK = ? First we must calculate the energy of the incident photons:

( )( )34 8

7

19

6.63 10 J s 3.00 10 m/s

4.50 10 m4.42 10 J

hcE

E

=

=

=

Converted to electron volts, Ephoton = 2.76 eV. We can now calculate the kinetic energy:

K,max photon

K,max

2.76 eV 2.48 eV0.28 eV

E E W

E

=

=

=

The maximum kinetic energy of the ejected electrons is 0.28 eV. 14. = 3.50 102 nm = 3.50 107 m

EK = 1.20 eV W = ?

( )( )( )( )

K photon

photon K

34 8

7 19

1.20 eV

6.63 10 J s 3.00 10 m/s1.20 eV

3.50 10 m 1.6 10 J/eV

2.35 eV

E E WW E E

hc

W

=

=

=

=

=

The work function of the metal is 2.35 eV.


15. f = 8.0 1014 Hz W = 1.2 eV v = ?

First we must calculate the kinetic energy:

( )( )( )

K photon

34 14

19

K

6.63 10 J s 8.0 10 Hz1.2 eV

1.6 10 J/eV

2.115 eV

E E W

hf W

E

=

=

=

=

For the maximum energy of photoelectrons that reach the plate:

( )( )( )K

19

31

5

2.115 eV 1.0 eV 1.115 eV

2

2 2.115 eV 1.6 10 J/eV

9.11 10 kg

6.3 10 m/s

EVm

V

=

=

=

=

The maximum speed with which an electron reaches the plate is 6.3 105 m/s.

Making Connections 16. Students answers will vary depending on their choice of device. Some choices may include smoke detectors, fibreoptic

detectors, security systems, garage door openers, or automatic light switches.

PRACTICE (Page 607)

Understanding Concepts 17. The units of momentum are kgm/s. Since, ( )2 2 2J = N m = kg m/s m = kg m /s , therefore:

2 2kg m /s sm

kg m/s

h

h

=

=

Therefore, the units of h

are the units of momentum.

18. = 5.00 102 nm = 5.00 107 m p = ?

34

7

27

6.63 10 J s5.00 10 m

1.33 10 kg m/s

hp

p

=

=

=

The momentum of the photon is 1.33 1027 kgm/s.


19. f = 4.5 1015 Hz p = ?

( )( )34 15

8

27

6.63 10 J s 4.5 10 Hz

3.00 10 m/s9.9 10 kg m/s

hp

hfc

p

=

=

=

=

The momentum of the photon is 9.9 1027 kgm/s. 20. E = 1.50 102 eV p = ?

( )( )2 19

8

26

1.50 10 eV 1.6 10 J/eV

3.00 10 m/s8.00 10 kg m/s

Epc

p

=

=

=

The momentum of the photon is 8.00 1026 kgm/s. 21. v = 1.0 106 m/s = ?

First we must calculate momentum:

( )( )31 625

9.11 10 kg 1.0 10 m/s

9.11 10 kg m/s

p mv

p

=

=

=

To calculate wavelength:

34

25

10

6.63 10 J s9.11 10

7.3 10 m

hp

=

=

=

The wavelength of the photon is 7.3 1010 m, or 0.73 nm.

Section 12.1 Questions (Pages 608609)

Understanding Concepts 1. The historical discoveries and interpretations that led to the confirmation of the photon theory of light are

blackbody radiation ultraviolet catastrophe quanta Planck photoelectric effect Einstein momentum of a photon Compton

2. To escape, the photoelectron requires sufficient energy to overcome being bonded to the surface. Since Ep f, the photoelectron is only emitted when the frequency is high enough, that is when Ep > W.

3. Students may include some of the following: security system detectors TV, audiosystem remote controls outdoor security lights automatic night lights garage door detectors


cameras light control digital still and movie cameras scanners for computers

4. EK = the maximum energy of the photoelectron hf = the energy of the incident photon W = the work function, the energy binding the electron to the photoelectric surface

5.

Photoelectric Effect Compton Effect photon disappears photon re-emerges with lower frequency low energy photon high energy photon result electron with kinetic energy result lower energy photon and electron with

kinetic energy energy conserved energy conserved

6. Five interactions are reflection, the photoelectric effect, the Compton effect, an increase in the internal energy of an atom, and pair production.

7. (a) The surface temperature of the emitting source determines the relative magnitude of the colours of the spectrum present in the radiation. (The surface temperature is often called the colour temperature for this reason.) The chemicals in daylight colour film are chosen to match the relative spectral density of the light radiated from the Sun (6000 K), therefore only artificial light of similar composition will work. Since incandescent light and fluorescent light have lower colour temperatures, the radiation they produce is improperly balanced. For example, pictures taken with daylight film under incandescent light tend to have unnatural colours (skin tones are orange instead of white).

(b) A common flash in cameras today, the xenon flash, provides radiation similar to the radiation produced by the Sun. The resulting colours on the film are correct.

8. = 122 nm E = ?

( )( )34 8

7

18

6.63 10 J s 3.00 10 m/s

1.22 10 m1.63 10 J

hcE

E

=

=

=

The energy of an ultraviolet photon is 1.63 1018 J, or 10.2 eV.

9. hcE

=

( )( )( ) ( )

( ) ( )

34 8

9 19

3

6.63 10 J s 3.00 10 m/s

nm 10 m/nm 1.6 10 J/eV

1.24 10eVnm

E

=

=

10. All three curves intersect on the x-axis and at this point the photocurrent is zero. The maximum retarding potential (V0) represents the maximum energy, and, as a result, the maximum velocities of the photoelectrons. Since all three intensity curves intersect at V0, the maximum velocities of the photoelectrons are independent of the intensity of light directed at the photoelectric surface.

11. From Table 1 we obtained the following data: sodium = 2.26 eV copper = 4.46 eV potassium = 1.60 eV barium = 2.48 eV

The work function represents the energy that binds the electron to the photoelectric surface. The minimum energy required for an electron to be released from the surface to become a photoelectron, is the value given to the work function for a particular metal. From the values above it is apparent that it takes 2.20 eV more energy to release a photoelectron from copper than it does for one from sodium.

The larger the value of the work function, the larger the threshold frequency. We can therefore conclude that since barium has a higher work function than potassium, barium will have a higher threshold frequency.


12. W = 7.2 1019 J f = ?

Since the work function is the energy required to release an electron from an illuminated metal, we can use the following relationship:

1934

15

7.2 10 J6.63 10 J s1.1 10 Hz

E hfEfh

f

=

=

=

=

The minimum frequency is 1.1 1015 Hz. 13. W = 4.52 eV

EK = 1.68 eV = ?

First we must calculate energy:

K

1.68 eV 4.52 eV6.20 eV

E E W

E

= +

= +

=

To calculate wavelength (using the relationship from question 9):

3

3

3

2

1.24 10 (nm)

1.24 10

1.24 106.20 eV

2.00 10 nm

E

E

=

=

=

=

The wavelength of light required is 2.00 102 nm. 14. = 482 nm W = ?

K photon

photon K

3

7

1.24 10 1.2 eV4.82 10 m1.38 eV

E E W

W E E

W

=

=

=

=

The work function of the metal is 1.38 eV. 15. (a) = 2.0 107 m f = ?

87

15

3.00 10 m/s2.0 10 m

1.5 10 Hz

c fcf

f

=

=

=

=

The frequency of the photon is 1.5 1015 Hz.


(b) E = ?

To calculate energy in electron volts:

( )3

3

1.24 10nm

1.24 10200 nm

6.2 eV

E

E

=

=

=

To calculate energy in joules:

( )( )1919

6.2 eV 1.6 10 J/eV

9.9 10 J

E

E

=

=

The energy of the same photon is 6.2 eV, or 9.9 1019 J. (c) p = ?

( )( )34 15

8

27

6.63 10 J s 1.5 10 Hz

3.00 10 m/s3.3 10 kg m/s

Epchfc

p

=

=

=

=

The momentum of the same photon is 3.3 1027 kgm/s. 16. (a) = 2.50 10-9 m p = ?

34

9

25

6.63 10 J s2.50 10 m

2.65 10 kg m/s

hphp

p

=

=

=

=

The momentum of the photon is 2.65 1025 kgm/s. (b) me = 9.11 10-31 kg

p = 2.65 10-25 kgm/s v = ?

2531

5

2.65 10 kg m/s9.11 10 kg

2.91 10 m/s

p mvpvm

v

=

=

=

=

The speed of the electron is 2.91 105 m/s. (c) EK = ?

Ephoton = ? To calculate the kinetic energy of the electron:

( )( )2

K electron

231 5

20K electron

121 9.11 10 kg 2.91 10 m/s23.85 10 J

E mv

E

=

=

=


To calculate the energy of the photon:

( )( )

photon

34 8

9

17photon

6.63 10 J s 3.00 10 m/s

2.50 10 m7.96 10 J

hcE

E

=

=

=

The kinetic energy of the electron is 3.86 1020 J, which is substantially smaller than the energy of the photon, which is 7.96 1017 J.

17. Let the subscript 1 represent the first radio station and 2 represent the second radio station. f1 = 5.70 102 kHz = 5.70 105 Hz f2 = 102 MHz = 102 106 Hz E1 = ? E2 = ?

Using the relationship photon :E hf=

( )( )34 51

281

6.63 10 J s 5.70 10 Hz

3.78 10 J

E

E

=

=

( )( )34 62

262

6.63 10 J s 102 10 Hz

6.76 10 J

E

E

=

=

The energy of the photons emitted from radio station 1 is 3.78 1018 J, or 2.36 109 eV. The energy of the photons emitted from radio station 2 is 6.76 1026 J, or 4.23 107 eV.

Applying Inquiry Skills 18. (a) Leaves containing chlorophyll appear green in white light because the red and violet photons are absorbed and not

reflected, whereas the green photons are reflected and not absorbed. (b) As seen from the graph, green light has a much smaller effect on the rate of photosynthesis than does red and violet

light. (c) As seen from Figure 2 on page 595, a source with a surface temperature of 2500 K is very weak in the blue region of

the spectrum and does not produce any radiation in the violet region. The graph in this question shows the photosynthesis reaction is stimulated by light from the blue-violet region. Therefore, the rate of photosynthesis produced by an incandescent lamp will be too low for most green plants and will be unable to maintain the health of the plant.

Making Connections 19. The higher the value of Plancks constant, the steeper the slope and the lower the cutoff frequency. Thus, photoemission

would occur for a broad range of materials, creating a state of change in many of them. On the other hand, the smaller the constant, the lower the slope and the higher the cutoff frequency. It is possible that the cutoff frequency is high enough that the photoelectric effect does not occur. Therefore, solar cells, cameras (picture and digital), and detectors would not work.

20. = 552 nm = 5.52 107 m P = 1.0 mW = 1.0 103 W = 1.0 103 J/s number of light quanta = ?

To calculate the energy of each photon:

( )( )34 8

7

19

6.63 10 J s 3.00 10 m/s

5.52 10 m3.60 10 J

hcE

E

=

=

=

Since 15% of the solar energy is converted, 15%(3.60 1019 J) = 5.4 1020 J.


To calculate the number of photons:

3

20

16

photonsnumber of light quanta

1.0 10 J5.4 10 J

number of light quanta 1.85 10

E

=

=

=

Therefore, 1.85 1016 light quanta would be needed by the calculator each second. 21. (a) = 1.0 cm = 1.0 10-2 m E = ?

( )( )34 8

2

23

6.63 10 J s 3.00 10 m/s

1.0 10 m2.0 10 J

hcE

E

=

=

=

The energy of a single microwave quantum is 2.0 1023 J. (b) = 1.0 cm = 1.0 102 m

mw = 0.250 kg T = 100C 20C = 80C c = 4.2 103 J/kgC number of quanta = ?

First we must calculate the change in energy:

( )( )( )34

0.250 kg 4.2 10 J/kg C 80 C

8.4 10 J

E mc T

E

= =

=

To calculate the number of quanta:

4

23

27

8.4 10 Jnumber of quanta2.0 10 J

number of quanta 4.2 10

=

=

Therefore, 4.2 1027 quanta are required to raise the temperature of the water. 22. All the detectors on the Hubble telescope are photodetectors since the information must be digitized for transmission to

Earth. Some areas for the students to research are construction and type of detector, placement of detectors, and the sensitivity/range of detectors (infrared, ultraviolet, X ray).

23. Ultraviolet photons have shorter wavelengths and a higher energy than infrared photons. This energy is sufficient enough to kill the skin cells, which is the result we see when the skin is sunburned. On the other hand, the infrared photons have sufficient energy to warm the cells but not damage them. Students should be careful not to confuse individual photon energy with total energy. A few joules of ultraviolet radiation can do more damage than a few hundred joules of infrared radiation.

12.2 WAVEPARTICLE DUALITY

PRACTICE (Page 614)

Understanding Concepts 1. (a) m = 2.0 kg

v = 15 m/s = ?


( )( )34

35

6.63 10 J s2.0 kg 15 m/s

2.2 10 m

hmv

=

=

=

The de Broglie wavelength is 2.2 1035 m. (b) v = 1.3 105 m/s = ?

( )( )34

27 5

12

6.63 10 J s1.7 10 kg 1.3 10 m/s

3.0 10 m

hmv

=

=

=

The de Broglie wavelength is 3.0 1012 m. (c) v = 5.0 104 m/s = ?

( )( )34

31 4

8

6.63 10 J s9.11 10 kg 5.0 10 m/s

1.5 10 m

hmv

=

=

=

The de Broglie wavelength is 1.5 108 m. 2. Ephoton = 3.0 eV

Eelectron = 5.0 eV photon = ? electron = ?

To calculate the wavelength of the photon:

( )( )

( )( )

photonphoton

34 8

19

7photon

6.63 10 J s 3.00 10 m/s

3.0 eV 1.6 10 J/eV

4.1 10 m

hcE

=

=

=

To calculate the wavelength of the electron, we must first calculate the speed of the electron:

( )( )( )

electron

19

31

6

2

2 5.0 eV 1.6 10 J/eV

9.11 10 kg

1.325 10 m/s

Evm

v

=

=

=

To calculate the wavelength of the electron:

( )( )

electron

34

31 6

10electron

6.63 10 J s9.11 10 kg 1.325 10 m/s

5.5 10 m

hmv

=

=

=

The de Broglie wavelength of the photon is 4.1 107 m, and of the electron is 5.5 1010 m.


3. m = 0.50 kg v = 5.00 102 m/s = ?

( )( )34

2

36

6.63 10 J s0.50 kg 5.00 10 m/s

2.7 10 m

hmv

=

=

=

The de Broglie wavelength is 2.7 1036 m. 4. = 0.15 nm = 0.15 109 m EK = ?

Since EK = 212

mv , K2Evm

= . Plugging this value for v into the relationship p = mv:

( )

K

2K

2

K

2

234

9

31

17K

2

2

2

2

6.63 10 J s0.15 10 m

2 9.11 10 kg

1.07 10 J

Ep mm

p E m

pEmh

m

E

=

=

=

=

=

=

The energy required is 1.07 1017 J, or 67 eV. 5. EK = 1.00 102 V = 100 eV

= ? To convert EK to joules:

( )( )19K

17K

100 eV 1.6 10 J/eV

1.6 10 J

E

E

=

=

To calculate wavelength (using the relationship found in question 4):

( )( )( )

2

K

K

K

34

17 31

10

22

2

6.63 10 J s

2 1.6 10 J 9.11 10 kg

1.23 10 m

pEm

p E mhE m

=

=

=

=

=

The associated de Broglie wavelength is 1.23 1010 m.


6. (a) = 1.0 1010 m p = ?

34

10

24

6.63 10 J s1.0 10 m

6.6 10 kg m/s

hp

p

=

=

=

The momentum of the electron is 6.6 1024 kgm/s. (b) v = ?

( )( )34

31 10

6

6.63 10 J s9.11 10 kg 1.0 10 m

7.3 10 m/s

hmvhv

m

v

=

=

=

=

The speed of the electron is 7.3 106 m/s. (c) EK = ?

( )( )2

K

231 6

17K

121 9.11 10 kg 7.3 10 m/s22.4 10 J

E mv

E

=

=

=

The kinetic energy of the electron is 2.4 1017 J, or 1.5 102 eV.


Understanding Concepts 1. (a). Diffraction of electrons, because the behaviour is the same as for electromagnetic waves (X rays). (b) The momentum of a photon.

2. The equation sin nnw = for a single slit predicts the most probable location on the screen. For example, for

11.5sin

w = , the photon is most likely to land on the screen in a bright area, whereas 1sin w

= will most likely not

land on the screen at that point (a dark area). 3. Students may find it easier to summarize their points in a table:

Property Photon Electron energy 2 eV

(3.2 1019 J) 2 eV (3.2 1019 J)

mass m = 0

m = 9.11 1031 kg momentum

19

8

27

3.2 10 J3.00 10 m/s1.1 10 kg m/s

Ep

c

p

=

=

=

( )( )( )31 31

25

2

2 9.11 10 kg 3.2 10 kg

7.6 10 kg m/s

p mE

p

=

=

=


Property Photon Electron speed

83.00 10 m/sv cv=

=

( )( )1931

5

K2

2 3.2 10 J

9.11 10 kg

8.4 10 m/s

Ev

m

v

=

=

=

wavelength

( )( )34 819

7

6.63 10 J s 3.00 10 m/s

3.2 10 J6.2 10 m

hcE

=

=

=

( )( )( )34

31 19

10

2

6.63 10 J s2 9.11 10 kg 3.2 10 J

8.7 10 m

hp

h

mE

=

=

=

=

4. (a) m = 1.67 1027 kg

v = 1.5 104 m/s = ?

( )( )34

27 4

11

6.63 10 J s1.67 10 kg 1.5 10 m/s

2.6 10 m

hmv

=

=

=

The de Broglie wavelength of the neutron is 2.6 1011 m. (b) m = 9.11 1031 kg

v = 1.2 106 m/s = ?

( )( )34

31 6

10

6.63 10 J s9.11 10 kg 1.2 10 m/s

6.1 10 m

hmv

=

=

=

The de Broglie wavelength of the electron is 6.1 1010 m. (c) EK = 1.0 109 eV = ?

( )( )( )( )K

34

27 9 19

16

2

6.63 10 J s

2 1.67 10 kg 1.0 10 eV 1.6 10 J/eV

9.1 10 m

hp

hmE

=

=

=

=

The de Broglie wavelength of the proton is 9.1 1016 m.


5. EK = 5.00 104 eV = ?

Since 2K12

E mv= , then K2Evm

= .

Therefore,

hmv

=

( )( )( )

K

34

4 1931

31

12

2

6.63 10 J s

2 5.0 10 eV 1.60 10 J/eV9.11 10 kg

9.11 10 kg

5.49 10 m

hEmm

=

=

=

The de Broglie wavelength of the electrons is 5.49 1012 m. 6. = 0.20 nm = 0.20 109 m

p = ? m = ?

To calculate momentum:

34

9

24

6.63 10 J s0.20 10 m

3.3 10 kg m/s

hp

p

=

=

=

To calculate mass:

24

8

32

(for a photon)

3.3 10 kg m/s3.00 10 m/s

1.1 10 kg

p mvp mc

pmc

m

=

=

=

=

=

The momentum of the photon is 3.3 1024 kgm/s. The mass of the photon is 1.1 1032 kg. 7. v = 1.2 105 m/s

= 8.4 1014 m m = ?

( )( )34

5 14

26 26

6.63 10 J s1.2 10 m/s 8.4 10 m

6.57 10 kg, or 6.6 10 kg

hmvhm

v

m

=

=

=

=

The mass of the microscopic object is 6.6 1026 kg.


8. EK = 5.0 eV w = ?

( )( )( )( )

K

34

31 19

10

2

6.63 10 J s

2 9.11 10 kg 5.0 eV 1.60 10 J/eV

5.5 10 m

hmE

=

=

=

For a single slit, sin nnw = . For significant diffraction, 1

w

, or w . Therefore, slit width must be less than 5.5 1010 m.

9. v = 0.25c = 7.50 107 m/s (a) = ?

( )( )34

27 7

15

6.63 10 J s1.67 10 kg 7.50 10 m/s

5.29 10 m

hmv

=

=

=

The associated de Broglie wavelength is 5.29 1015 m. (b) EK = ?

( )( )2

K

227 7

12K

121 1.67 10 kg 7.5 10 m/s24.70 10 J

E mv

E

=

=

=

The kinetic energy is 4.70 1012 J. (c) E = ?

12

12

19

7

4.70 10 J4.70 10 J

1.60 10 J/eV2.94 10 eV

E

E

=

=

=

Therefore, the proton was accelerated through a potential difference of 2.94 107 eV (since electron and proton have the same charge).

10. = 1.0 1011 m

( )( )( )

K

22

K2

K 2

34

11 31

15K

2

2

26.63 10 J s

2 1.0 10 m 9.11 10 kg

2.41 10 J

hmE

hmE

hEm

E

=

=

=

=

=


To calculate the potential difference:

15

15

19

4

2.41 10 J2.41 10 J

1.60 10 J/eV1.5 10 eV

E

E

=

=

=

The potential difference is 1.5 104 eV.

Making Connections 11. The scanning tunnelling microscope (STM) allows scientists the ability of atomic vision where they can actually see

atoms. An STM works by passing an electric current between an experimentally manipulated scanning probe the size of an atom and the surface of the material under study. The STM is able to feel the topography of the object by detecting small variations in the current.

The STM can map the surface of atoms by probing a semi-conductive surface with a very small needle. When the electron shells overlap and a current is applied, the electrons participate in quantum leaps. If the surface of the atom has a depression, the electron shells are farther apart and the electrons do not leap. If the surface is raised, the electron shells are closer together and more electrons jump to the probe. By measuring the amount of electron jumps, the STM can pass the info to a computer which makes a false colour map of the surface

The word tunnelling, as it relates to the STM, refers to a small current being applied to a separate materials, and artificially joining them together so that an electron is able to move from one atomic landscape to another. The image on the IBM website http://www.almaden.ibm.com/vis/stm/images/stm.gif shows the standing-wave patterns in the local density of states of a copper surface.

12. In a transmission electron microscope, exposure time must be limited. The high-energy electrons bombarding the specimen can cause physical changes that will affect the resulting image. In scanning electron microscopes it is desirable that the specimen be rendered electrically conducting, otherwise a sharp picture will not be obtained. Conductivity is usually achieved by evaporating a film of metal 50 to 100 angstroms thick onto the specimen in a vacuum (such a thickness does not materially affect the resolution of the surface details).

13. Biographical information for Willard Boyle: physicist, Retired former Executive Director of Research, Communications Sciences Division, Bell Labs in

New Jersey. born: August 19, 1924 in Amherst Nova Scotia resides in Wallace, Nova Scotia degrees: BSc, McGill, 1947; MSc, McGill, 1948; PhD (Physics), McGill, 1950 co-inventor of the Charge Coupled Device, first continually operating ruby laser Awards: The Ballantyne Medal of the Franklin Institute, 1973; Morris Lieberman Award of the IEEE, 1974; Progress

Medal of The Photographic Society of America; Breakthrough Award by the Device Research Conference of the IEEE; Co-winner, C&C prize of the NEC Foundation, Tokyo, 1999; Edwin H. Land Medal, Optical Society of America, 2001

Biographical information for James Hillier:

physicist, Retired Executive Vice President and Sr. Scientist RCA Labs, Princeton, NJ born: August 22, 1915 in Brantford, Ontario degrees: BA, U of Toronto, 1937, MA, U of T, 1938, PhD, U of T, 1941 co-designer of the first commercially available electron microscope in North America Awards: Inductee, National Inventors Hall of Fame, 1980; Albert Lasker Award, 1960; Commonwealth Award, 1980

12.3 RUTHERFORDS MODEL OF THE ATOM


Understanding Concepts 1. The atomic model must have a very small, but very massive, positively charged nucleus, and the remainder of the atom

must be empty space. 2. The electrons are kept in orbital paths by the Coulomb force of electric attraction between the positive nucleus and

themselves.


3. ge

FF

= ?

( )( )( )( )( )

1 2

g 2

1 2e

2

1 2

1 2

11 2 2 31 27

29 2 2 19

g 40

e

6.67 10 N m /kg 9.11 10 kg 1.67 10 kg

9.00 10 N m /C 1.6 10 C

4.4 10

Gm mF d

kq qFd

Gm mkq q

FF

=

=

=

=

Therefore, Fg 1040 Fe, and we can ignore Fg due to its negligible effect. 4. EK = 4.5 MeV = 4.5 106 eV

m = 6.6 1027 kg q2 = +79e r0 = ?

At the point of closest approach, the initial kinetic energy has been totally converted into electric potential energy:

1 2K p0

kq qE Er

= =

To convert EK into joules:

( )( )K

6 19

13K

Physics-12-Ch 11 & 12 & 13-Solns

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Transcript of Physics-12-Ch 11 & 12 & 13-Solns