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2

Relativity II

2-1

p=

mv éë1 - ( v 2 c 2 ) ùû

12

( 1.67 ´ 10-27

kg ) ( 0.01c )

= 5.01 ´ 10 -21 kg × m s

(a)

p=

(b)

1.67 ´ 10-27 ( p=

kg ) ( 0.5c )

(c)

p=

( 1.67 ´ 10-27

kg ) ( 0.9c )

(d)

1.00 MeV 1.602 ´ 10 -13 J = = 5.34 ´ 10 -22 kg × m s so for (a) 8 c 2.998 ´ 10 m s

2 12

é1 - ( 0.01c c ) ù ë û

2 12

é1 - ( 0.5c c ) ù ë û

2 12

é1 - ( 0.9c c ) ù ë û

= 2.89 ´ 10 -19 kg × m s

= 1.03 ´ 10 -18 kg × m s

5.01 ´ 10 -21 ( p=

kg × m s ) ( 100 MeV c )

5.34 ´ 10 -22 kg × m s

= 9.38 MeV c

Similarly, for (b) p = 540 MeV c and for (c) p = 1 930 MeV c . 2-2

(a)

Scalar equations can be considered in this case because relativistic and classical velocities are in the same direction. p = g mv = 1.90 mv =

mv 2 12

é1 - ( v c ) ù ë û

Þ

1 2 12

é1 - ( v c ) ù ë û

= 0.85c (b)

No change, because the masses cancel each other.

( ) ùúû

é 1 = 1.90 Þ v = ê1 1.90 ë

2 12

c

18

CHAPTER 2

RELATIVITY II

2-3

As F is parallel to v, scalar equations are used. Relativistic momentum is given by mv p = g mv = 1 2 , and relativistic force is given by é1 - ( v c ) 2 ù ë û ì ü dp d ï mv ï = í ý dt dt ï é1 - ( v c ) 2 ù1 2 ï îë û þ dp m æ dv ö F= = ç ÷ dt é1 - ( v 2 c 2 ) ù 3 2 è dt ø ë û F=

2-4

(a)

æ v2 ö Using the results of Problem 2-3, qE = m ç 1 - 2 ÷ c ø è a=

(b)

(c)

dv æ qE ö æ v2 ö =ç ÷ ç1- 2 ÷ dt è m ø è c ø

-3 2

æ dv ö , or ç ÷ è dt ø

32

. Here v is a function of t and q, E, M, and c are parameters.

As expected; as v ® c , a ® 0 because in general no speed can exceed c, the speed of light. Separating variables,

dv

( 1- v

2

c

)

2 32

qE = æç ö÷ dt , or èmø v

v

( 1- v

2

c

)

=

2 12 0

v2

( 1 - v2

c2 )

12

v

ò

0

t

dv

( 1- v

2

c

)

2 32

0

qE m

dt ,

qEt m

qEt ö 2 qEt = æç ÷ = m è m ø

qEt ö 2 æ v 2 v2 = æç ÷ -ç è m ø è c2 é qEt ö2 ù æ qEt ö 2 v 2 ê1 + æç ÷ ú=ç ÷ ë è m ø û è m ø

ö æ qEt ö 2 ÷ ÷ç øè m ø

( qEt mc ) 2 2 1 + ( qEt mc ) ( qEct ) 2 2 v = ( mc ) 2 + ( qEt ) 2 v2 =

Note that the limiting behavior of v as t ® 0 and t ® ¥ is reasonable. As qEct dx v= = , dt é( mc ) 2 + ( qEt ) 2 ù 1 2 ë û

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2

x = qEc éë( mc ) + ( qEt ) ùû

é 1 ê 2 êë ( qE )

t

{

}

ù 12 c é ( mc ) 2 + ( qEt ) 2 ù - mc . ú = ë û úû 0 qE

As t ® 0 , x ® 0 , and t ® ¥ , x ® ct ; reasonable results. 2-5

This is the case where we use the relativistic form of Newton’s second law, but unlike dp Problem 2-3 in which F is parallel to v, here F is perpendicular to v and F = so that dt

F = qv ´ B =

dp d ìï mv = í dt dt ï 1 - ( v c ) 2 î

üï ý. ïþ

Assuming that B is perpendicular to the plane of the orbit of q, the force is radially inward, and we find F = qvB radial =

d ìï mv í dt ï 1 - ( v c ) 2 î

üï ý. ïþ

As the force is perpendicular to v, it does no work on the charge and the magnitude (but not the direction) of v remains constant in time. Thus, d dt

ìï mv í 2 îï 1 - ( v c )

üï m dv . ý= 2 1 - ( v c ) dt þï

dv æ v 2 ö æ dv ö =ç ÷ Identifying ç ÷ as the centripetal acceleration where the scalar equation dt è r øradial è dt ø 12

12

2 æ v2 ö æ qBr ö æ 1 - v ö or v = ç ÷ç ç ÷ ÷ . Finally, the period T is è m øè c2 ø è r ø radial 12 2p m 2p r 1 æ qB ö æ v2 ö T = = f = 12 1 2 . As and , f =ç ÷ ç1 - 2 ÷ . ( qB ) ( 1 - v 2 c 2 ) v ( qBr m ) ( 1 - v 2 c 2 ) T c ø è 2p m ø è

2-6

é ù m =ê 2 2 ú ë1- v c û 2p r

-19 C ) BR kg × m C × s = 1.60 ´ 10 -19 BR kg × m s . To Using Equation 2.4 p = e BR = ( 1.60 ´ 10

convert kg × m s to MeV c , use 1 MeV c = p=

( 106 ) ( 1.60 ´ 10-19 C ) ( 1 3.00 ´ 10

( 1.60 ´ 10-19 BR

8

J C)

ms

kg × m s ) ( 1 MeV c )

5.34 ´ 10 -22 kg × m s

= 5.34 ´ 10 -22 kg × m s , so that = 300 BR MeV c .

20

CHAPTER 2

RELATIVITY II

2-7

2 E = g mc 2 , p = g mu ; E2 = ( g mc 2 ) ; p 2 = ( g mu ) 2 ;

{ ( mc

E2 - p 2 c 2 = ( g mc 2 ) - ( g mu ) c 2 = g 2 2

2

2 æ u2 ö æ u2 ö = ( mc 2 ) ç 1 - 2 ÷ ç 1 - 2 ÷ c øè c ø è

E2 = p 2 c 2 + ( mc 2 ) 2-8

(a)

-1

)

2 2

= ( mc 2 )

- ( mc ) 2 u2 2

}

Q.E.D.

2

ER = mc 2 = ( 1.67 ´ 10-27 kg ) ( 3 ´ 10 8 m s ) = 1.503 ´ 10 -10 J = 939.4 MeV (Numerical 2

round off gives a slightly larger value for the proton mass)

2-9

1.503 ´ 10-10 J

(b)

E = g mc 2 =

(c)

K = E - mc 2 = 4.813 ´ 10-10 J - 1.503 ´ 10 -10 J = 3.31 ´ 10-10 J = 2.07 ´ 10 3 MeV

(a)

When K = ( g - 1) mc 2 = 5mc 2 , g = 6 and E = g mc 2 = 6 ( 0.511 0 MeV ) = 3.07 MeV .

(b)

1 é v = 1g êë c

( )

( 1 - ( 0.95c c ) )

2 12

2 12

ù ú û

= 4.813 ´ 10 -10 J » 3.01 ´ 10 3 MeV

é æ 1 ö2 ù and v = c ê1 - ç ÷ ú ëê è g ø ûú

2-10

E = g mc 2 ; 1.5mc 2 = g mc 2 ; g = 1.5 Þ 1.5 =

2-11

(a)

12

( ) ùúû

1 é = c ê1 6 ë 1

éë1 - ( v 2 c 2 ) ùû

12

2 12

= 0.986 c

( ) ùúû

1 é ; v = c ê1 1.5 ë

2 12

= 0.75c

K = 50 ´ 109 eV ; mc 2 = 938.27 MeV ; E = K + mc 2 = ( 50 ´ 109 eV ) + ( 938.27 ´ 106 eV ) = 50 938.3 MeV é E2 - m 2 c 4 ù E2 = p 2 c 2 + m2 c 4 Þ p = ê ú c2 ë û

12

12

é( 50 938.3 MeV ) 2 - ( 938.27 MeV ) 2 ù ë û p= = 5.09 ´ 1010 eV c c 5.09 ´ 1010 eV = 1.6 ´ 10 -19 J eV ) = 2.71 ´ 10 -17 kg × m s ( 8 3 ´ 10 m s

2

E = g mc = (b)

mc 2 é1 - ( v c ) 2 ù ë û

12

é æ mc 2 ö2 ù Þ v = c ê1 - ç ÷ ú êë è E ø úû

é æ 938.27 MeV ö2 ù = ( 3 ´ 10 8 m s ) ê1 - ç ÷ ú ëê è 50 938.3 MeV ø ûú

12

12

= 2.999 5 ´ 10 8 m s

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2-12

(a)

2 2 When K e = K p , me c ( g e - 1) = mp c ( g p - 1) . In this case me c 2 = 0.511 0 MeV and

mp c 2 = 938 MeV , g e = éë1 - ( 0.75 ) 2 ùû first equation, we find g p = 1 +

gp =

(b)

1 é1 - ( up c ) 2 ù ë û

12

12

= 1.511 9 . Substituting these values into the

2

me c ( g - 1) mp c

2

= 1+

; therefore up = c ( 1 - g p-2 )

(a)

12

939

= 1.000 279 . But

= 0.023 6c .

æ g e ö æ me When pe = pp , g p mp up = g e me ue or up = çç ÷÷ çç è g p ø è mp 2 æ 1.511 9 ö é 0.511 0 c up = ç ÷ê 2 è 1.000 279 ø ë 939 c

2-13

( 0.511 0 ) ( 1.511 9 - 1)

ö ÷÷ ue , ø

ù -4 ú ( 0.75c ) = 6.17 ´ 10 c . û

E = 400mc 2 = g mc 2 æ v2 g = ç1 - 2 c è 2 æ v ö ç1- 2 ÷ = c ø è

ö ÷ ø

-1 2

= 400

( 4001 )

2

v é 1 ù1 2 = ê1 c ë 400 2 úû v = 0.999 997 c

2-14

(b)

K = E - mc 2 = ( 400 - 1) mc 2 = 399mc 2 = ( 399 ) ( 938.3 MeV ) = 3.744 ´ 105 MeV

(a)

E = mc 2 E 4 ´ 10 26 J m= 2 = = 4.4 ´ 109 kg 2 8 c ( 3.0 ´ 10 m s )

(b)

t=

(a)

Vq ö2 Vq ö-2 ü v ì æ K = g mc 2 - mc 2 = Vq and so, g 2 = æç 1 + and = í1 - ç 1 + ÷ ý 2 ÷ c î è è mc ø mc 2 ø þ

( 2.0 ´ 1030 ) 4.4 ´ 10

9

kg

kg s

= 4.5 ´ 10 20 s = 1.4 ´ 1013 years 12

2-15

v ìï 1 = í1 c ï 1 + ( 5.0 ´ 104 eV 0.511 MeV ) 2 î or v = 0.413c .

12

ü ï ý ïþ

= 0.412 7

22

CHAPTER 2

(b)

K=

RELATIVITY II

1 mv 2 = Vq 2 12

2Vq ö v = æç ÷ è m ø

(c)

4 ïì 2 ( 5.0 ´ 10 eV ) ïü =í 2 ý ïî 0.511 MeV c ïþ

12

= 0.442c

The error in using the classical expression is approximately

3 ´ 100% or about 7.5% 40

in speed. 2-16

(a)

In S the speed of the particle is u and p =

mu

( 1 - v2

c2 )

12

, E=

mc 2

( 1 - u2

c2 )

12

, and

u-v

( E2 - p2 c 2 ) = m 2 c 4 . In S¢ , u¢ = 1 - uv c2 , and mu¢

p¢ =

1 - ( u¢ c ) mc 2

E¢ =

(b)

2

1 - ( u¢ c )

2

=

=

m éë( u - v ) ( 1 - uv c 2 ) ùû 1 - éë( u - v ) ( 1 - uv c 2 ) ùû

2

( 1 c2 )

mc 2 1 - éë( u - v ) ( 1 - uv c 2 ) ùû

2

( 1 c2 )

=

=

m éë( u - v ) ( 1 - uv c 2 ) ùû 1 - ( u - v) 2

( 1 - uv c 2 )

2

mc 2 1 - ( u - v) 2

( 1 - uv c 2 )

2

2 2 2 2 4 Using these expressions for E ¢ and p ¢ , one obtains ( E ¢ - p ¢ c ) = m c , and since

E2 - p 2 c 2 = m2 c 4 , it follows that, E ¢2 - p ¢2 c 2 = E2 - p2 c 2 . 2-17

Dm = mRa - mRn - mHe (an atomic unit of mass, the u, is one-twelfth the mass of the or 1.660 54 ´ 10

-27

kg )

Dm = ( 226.025 4 - 22.017 5 - 4.002 6 ) u = 0.005 3 u E = ( Dm ) ( 931 MeV u ) = ( 0.005 3 u ) ( 931 MeV u ) = 4.9 MeV 2-18

(a)

The mass difference of the two nuclei is Dm = 54.927 9 u - 54.924 4 u = 0.003 5 u DE = ( 931 MeV u ) ( 0.003 5 u ) = 3.26 MeV.

(b)

The rest energy for an electron is 0.511 MeV. Therefore, K = 3.26 MeV - 0.511 MeV = 2.75 MeV .

12

C atom

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2-19

Dm = 6mp + 6mn - mC = [ 6 ( 1.007 276 ) + 6 ( 1.008 665 ) - 12 ] u = 0.095 646 u , DE = ( 931.49 MeV u ) ( 0.095 646 u ) = 89.09 MeV . Therefore the energy per nucleon =

2-20

89.09 MeV = 7.42 MeV . 12

Dm = m - mp - me = 1.008 665 u - 1.007 276 u - 0.000 548 5 u = 8.404 ´ 10 -4 u E = c 2 ( 8.404 ´ 10-4 u ) = ( 8.404 ´ 10 -4 u ) ( 931.5 MeV u ) = 0.783 MeV . E, p

2-21 e(–)

g

e(+)

 

K, p(e–) positron at rest

g

E, p

Conservation of mass-energy requires K + 2 mc 2 = 2 E where K is the electron’s kinetic energy, m is the electron’s mass, and E is the gamma ray’s energy. E=

K + mc 2 = ( 0.500 + 0.511) MeV = 1.011 MeV . 2

Conservation of momentum requires that pe - = 2 p cos  where pe - is the initial momentum E = 1.011 MeV c . Using of the electron and p is the gamma ray’s momentum, c Ee2- = pe2- c 2 + ( mc 2 ) pe - =

Finally, cos  = 2-22

(a)

2

2 where Ee - is the electron’s total energy, Ee - = K + mc , yields

( 1.00 ) 2 + 2 ( 1.00 ) ( 0.511) MeV 1 K 2 + 2Kmc 2 = = 1.422 MeV c . c c

pe -

2p

= 0.703 ;  = 45.3° .

Using the results of Problem 2-6 and substituting numerical values p ( in MeV c ) = 300 BR = ( 300 ) ( 2.00 T ) ( 0.343 m ) = 206 MeV c .

24

CHAPTER 2

RELATIVITY II

Since the momentum of the K 0 is zero before the decay, conservation of momentum requires the pion momenta to be equal in magnitude and opposite in direction. The pion’s speed u may be found by noting that

p E

=

mu mc

2

1 - u2 c 2 2

1-u c

2

=

u c

2

or

u pc = c E

where p is the pion momentum and E is the pion’s total energy. Thus for either pion, pc u pc = = 2 1 2 where m is the pion’s mass. Finally, c E é 2 2 p c + ( mc 2 ) ù ë û u 206 MeV = = 0.827 . c ( 206 MeV ) 2 + ( 104 MeV ) 2 (b)

Conservation of mass-energy requires that EK 0 = 2 E where EK 0 is the total energy of a pion. As the K 0 pion decays at rest, EK0 = mK0 c 2 = 2 p2 c 2 + ( mc 2 )

2

= 2 ( 206 ) 2 + ( 140 ) 2 MeV = 498 MeV ,

2 or mK 0 = 498 MeV c .

2-23

In this problem, M is the mass of the initial particle, ml is the mass of the lighter fragment, vl is the speed of the lighter fragment, mh is the mass of the heavier fragment, and vh is the speed of the heavier fragment. Conservation of mass-energy leads to Mc 2 =

ml c 2 1 - vl2 c 2

+

mh c 2 1 - vh2 c 2

From the conservation of momentum one obtains

( ml ) ( 0.987 c ) ( 6.22 ) = ( mh ) ( 0.868c ) ( 2.01) ( mh ) ( 0.868c ) ( 2.01) ml =

( 0.987 ) ( 6.22 )

= 0.284 mh

Substituting in this value and numerical quantities in the mass-energy conservation equation, one obtains 3.34 ´ 10-27 kg = 6.22 ml + 2.01mh which in turn gives 3.34 ´ 10-27 kg = ( 6.22 ) ( 0.284 ) ml + 2.01mh or mh =

3.34 ´ 10 -27 kg 3.78

= 8.84 ´ 10 -28 kg and

ml = ( 0.284 ) mh = 2.51 ´ 10 -28 kg . 2-24

The moving observer sees the charge as stationary, so she says it feels no magnetic force. q ( E + v ´ B ) = q ( E ¢ + zero ) , E ¢ = E + v ´ B .

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2-25

(a)

The x component of the gravitational force between a light particle of mass m and the GMS m GMS mb 3 2 . The change in momentum in the x Sun is given by Fx = r 2 sin f = 2 ( b + y2 ) direction is given by Dpx =

¥

¥

ò Fx dt = ò

GMS mb

( b2 + y 2 )

32

dt . To convert dt to dy, assume the

deflection is very small and that the position of the light particle is given by y = -ct dy for x = 0 . Thus dt = and we get c dy dy y GMS mb -¥ 2GMS mb +¥ 2GMS mb Dpx = = = ò ò 32 32 12 2 2 2 2 2 2 c c c +¥ ( b + y ) 0 (b +y ) b ( y + b2 ) 2GMS mb æ 1 = ç 2 c èb

ö = 2GM S m ÷ cb ø

From Figure P2.25(b),  @ (b)

2GMS m 2GMS = . cb ( mc ) bc 2

2 ( 6.67 ´ 10 -11 N × m 2 kg 2 ) ( 1.99 ´ 10 30 kg )

( 6.96 ´ 10

Energy conservation:

8

1 1- 0

2

Momentum conservation: 0 +

2-27

mc

so we find  @

For b = RS = 6.96 ´ 108 m and MS = 1.99 ´ 10 30 kg

=

2-26

Dpx

m ) ( 3.00 ´ 10

1 400 kgc 2 +

8

900 kgc 2

900 kg ( 0.85c ) 1 - 0.85

2

v 1 452 = = 0.467 c 3 108

m s)

2

1 - 0.85 =

Mc 2

=

2

2

1- v c

Mv 2

1- v c

= 4.24 ´ 10 -6 rad = 2.43 ´ 10-4 deg

2

2

; 3 108 kg 1 -

; 1 452 kg 1 -

v2 c

2

c2

=M.

Mv . c

v = 0.467 c .

(a)

Dividing gives

(b)

Now by substitution 3 108 kg 1 - 0.467 2 = M = 2.75 ´ 10 3 kg .

If the energy required to remove a mass m from the surface is equal to its rest energy mc 2 , -11 N × m 2 kg 2 ) ( 1.99 ´ 10 30 kg ) GMS m GMS ( 6.67 ´ 10 2 = mc R = = then R and g , 2 c2 g ( 3.00 ´ 108 m s )

Rg = 1.47 ´ 10 3 m = 1.47 km . 2-28

=

v2

(a)

The charged battery stores energy E = P t = ( 1.20 J s ) ( 50 min ) ( 60 s min ) = 3 600 J

¥

0

26

CHAPTER 2

RELATIVITY II

3 600 J E = 4.00 ´ 10 -14 kg . 2 so its mass excess is Dm = c 2 = 8 ( 3 ´ 10 m s )

(b)

2-29

-14 kg Dm 4.00 ´ 10 = = 1.60 ´ 10 -12 too small to measure. -3 m 25 ´ 10 kg

The energy of the first fragment is given by E12 = p12 c 2 + ( m1 c 2 ) = ( 1.75 MeV ) 2 + ( 1.00 MeV ) 2 ; 2

E1 = 2.02 MeV . For the second, E22 = ( 2.00 MeV ) 2 + ( 1.50 MeV ) 2 ; E2 = 2.50 MeV . (a)

Energy is conserved, so the unstable object had E = 4.52 MeV . Each component of momentum is conserved, so the original object moved with p2 = px2 + p y2 =

(

1.75 MeV c

) ( 2

+

2.00 MeV c

). 2

Then for ( 4.52 MeV ) 2 = ( 1.75 MeV ) 2 + ( 2.00 MeV ) 2 + ( mc 2 ) ; m = 3.65 MeV c 2 . 2

(b)

Now E = g mc 2 gives 4.52 MeV = v = 0.589c .

2-30

1

2

3.65 MeV ; 1 - v = 0.654 and 2 2 1- v c c2

Take the two colliding protons as the system p 1

E1 = K + mc 2 E2 = mc 2

in itia l p f

E12 = p12 c 2 + m2 c 4 p2 = 0.

fi n a l

In the final state, Eff = K + Mc 2 : Eff2 = p 2 c 2 + M 2 c 4 . By energy conservation, E1 + E2 = Ef , so E12 + 2 E1 E2 + E22 = Ef2

p

p

Initial (beams)

p12 c 2 + m2 c 4 + 2 ( K + mc 2 ) mc 2 + m 2 c 4 = pf2 c 2 + M 2 c 4 Final (beams)

By conservation of momentum, p1 = pf . Then M 2 c 4 = 2Kmc 2 + 4m 2 c 4 =

4 Km2 c 4 2mc

2

+ 4m 2 c 4

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Mc 2 = 2 mc 2 1 +

K 2 mc 2

.

By contrast, for colliding beams in the original state, we have E1 = K + mc 2 and E2 = K + mc 2 . In the final state, Ef = Mc 2 E1 + E2 = Ef : K + mc 2 + K + mc 2 = Mc 2 K ö Mc 2 + 2mc 2 æç 1 + ÷. è 2mc 2 ø 2-31

Conservation of momentum g mu : mu 2

+

2

1-u c

m ( -u ) 2

3 1-u c

2

=

Mvf 1 - vf2

c

2

=

2mu 3 1 - u2 c 2

.

Conservation of energy g mc 2 : mc 2

+

1 - u2 c 2

mc 2 3 1 - u2 c 2

To start solving we can divide: vf = M 2

1 - u 4c

2

=

2J

1 - vf2 c 2

4m 2

3 1- u c

=

4mc 2 3 1 - u2 c 2

2

=

M

( 1 2 ) 4 - u2 c 2

3 1 - u2 c 2 4m , in agreement with the classical result. 3

= 2.00 ´ 1013 W

(a)

r=

(b)

The kinetic energy of one electron with v = 0.999 9c is

Dt

=

100 ´ 10 -15 s

.

2 m 4 - u2 c 2

Note that when v << c , this reduces to M = energy

Mc 2

2u u = . Then 4 2

M=

2-32

=

æ 1 ( g - 1) mc 2 = ç ç 1 - 0.999 92 è = 5.72 ´ 10-12 J

ö 2 ÷ ( 9.11 ´ 10 -31 kg ) ( 3 ´ 108 m s ) = 69.7 ( 8.20 ´ 10 -14 J ) ÷ ø

28

CHAPTER 2

RELATIVITY II

Then we require

0.01 ( 2 J ) = N ( 5.72 ´ 10 -12 J ) 100 N=

2-33

2 ´ 10 -4 J 5.72 ´ 10 -12 J

= 3.50 ´ 107 .

The energy that arrives in one year is E = P Dt = ( 1.79 ´ 1017 J s ) ( 3.16 ´ 107 s ) = 5.66 ´ 10 24 J . E 5.66 ´ 1024 J m = = = 6.28 ´ 107 kg . Thus, c 2 ( 3.00 ´ 108 m s ) 2

Solution manual modern physics 3rd edition serway

solution manual modern physics 3rd edition serway. Full file at http://testbank360.eu/solution-manual-modern-physics-3rd-edition-serway