Tài liệu Physics exercises_solution: Chapter 38 doc

The frequency depends on the energy of each photon and the number of photons per second depends on the power output of the source... c There is no energy level 8 eV above the ground stat

38.1: 5.77 10 Hz

m105.20

sm103.00λ

14 7

103.84)

sm10(3.00s)mkg10(1.28

smkg101.28m

105.20

sJ106.63λ

19 8

27

27 7

eV)J10(1.60eV)10

34

19 6

Hz105.91

sm103.00

P hf

P dN dE

dt dE dt

)(

eVJ10(1.60eV)(5.1m

102.35s

J10(6.63

20

19 7

34

) )

J)10

101.45

15 0

c h

hc hf hf

2.3010 19 J1.44 eV

38.7: a) f  c λ5.01014 Hz

b) Each photon has energy E  hf 3.3110 19 J

Source emits 75J sso(75J s) (3.3110 19 photons/s)2.31020photons s

c) No, they are different The frequency depends on the energy of each photon and the number of photons per second depends on the power output of the source

38.8: For red light λ700nm

eV1.77

J101.6

1eVJ

102.84

m)10(700

)sm10(3.00s)J10(6.626λ

19 19

9

8 34



38.9: a) For a particle with mass, K  p2 2m.p2 2p1 meansK2 4K1

b) For a photon, E pc p2 2p1 meansE2 2E1

38.10: Kmax  hf 

Use the information given for λ400nmtofind:

J10

3.204

eV)J10(1.602eV)

(1.10m

10400

)sm10(2.998s)

J10(6.626

19

19 9

8 34

103.418

J103.204m

10300

)sm10(2.998s)J10(6.626

19

19 9

8 34

38.11: a) The work function 0 eV0

V)(0.181C)

10(1.60m

102.54

s)m10(3.00s)J10(6.63

19

19 7

8 34

107.53

)sm10(3.00s)J10(6.63

19

8 34

10(2.50

s)m10(3.00s)eV10(4.136λ

1

7

8 15

V)(2.7C)102(1.602

J102.47J

s)J10(6.63λ

λ

7 28

p

This is infrared radiation

38.14: a) The threshold frequency is found by setting V = 0 in Eq (40.4), f0  h

m1072.3λ

19 7

hf



38.15: a) 2.31 10 J 1.44eV.

m108.60

)sm10(3.00s)J10(6.63λ

19 7

8 34

s)m10(3.00s)J10(6.63λ

19 7

8 34

2.96eV





E

38.16: a)  E1 20eV b) The system starts in the n = 4 state If we look at all paths

to n = 1 we find the 4-3, 4-2, 4-1, 3-2, 3-1, and 2-1 transitions are possible (the last three

are possible in combination with the others), with energies 3 eV, 8 eV, 18 eV, 5 eV, 15

eV, and 10 eV, respectively c) There is no energy level 8 eV above the ground state energy, so the photon will not be absorbed d) The work function must be more than 3

eV, but not larger than

1

n

R (Balmer series implies final state is n = 2)

nm433m104.33m

)1021(1.10

10021

100λ

100

2125

14

1λ

1:5

n

H

m104.33

sm103.00λ

14 7

)34()

34(λ,

7144

m105.890

)sm10(3.000s)J10(6.626

8 34

.eV2.107J

103.371

m105.896

)sm10(3.000s)J10(6.626λ

eV

2.109J

103.375

3 19

7

8 34

J10(4.78

C)10(1.60(184)4

1

)2()92(4

1

14 6





b) The above result may be substituted into Coulomb’s law, or, the relation between the magnitude of the force and the magnitude of the potential energy in a Coulombic field is

N

13.8m)

10(5.54

)evJ10(1.6eV)10(4.78

14

19 6

C)1060.1(1644

)82()2(

0

2 19

0 0

2 1

ε

e e r ε

q q U

103.63J

K

kg106.64

J)102(5.812

mv

K

λandnm97.0λ

so,)4(

11λ

38.23: a) Following the derivation for the hydrogen atom we see that for Be all we 3 

need do is replace e2 by4e2.Then

.eV13.6016

)(Be)

H(168

)4(1)

2 2

2 2 2

E h

n

e m ε

So for the ground state, (Be3 ) 218eV

E

b) The ionization energy is the energy difference between the nandn1 levels

So it is just 218 eV for Be , which is 16 times that of hydrogen.3

8

)4

(

λ

1

2 2

2 1

1 8 2

2

2 1 3 2 0

2 2

n c h

e m



4

11)m10(1.74λ

1,1to

2 2 0 3

n

e m

h n ε



38.24: a), b) For either atom, the magnitude of the angular momentum is 

2

h

s

mkg

L so for this state L33.1610 34kgm2 s

38.26: a) We can find the photon’s energy from Eq 38.8

1)m10(1.097)

sm10(3.00s)J10(6.63

12

1

19

2 2 1 7 8

34 2

E

The corresponding wavelength is λ hc E 434nm

b) In the Bohr model, the angular momentum of an electron with principal quantum

number n is given by Eq 38.10

2

h n

L

Thus, when an electron makes a transition from n = 5 to n = 2 orbital, there is the

following loss in angular momentum (which we would assume is transferred to the photon):

s

J103.172

s)J103(6.632

)52

However, this prediction of the Bohr model is wrong (as shown in Chapter 41)

38.27: a) 2.18 10 m/s

)sJ1063.6(2

C)1060.1(1

:2

34 0

2 19 1

sm1009.122

5 1

3

6 1

v v h

3 3 2 0 2

0

2 2 2

21

22

me

h n ε nh e

me h n v

3

s1022.1)2(:

2

s1053.1C)1060.1(kg)1011.9(

)sJ1063.6(41

15 3

1 3

15 3

1 2

16 4

19 31

3 34 2

0 1

T T n

T

c) number of orbits 8.2 10

s1022.1

s100

b) Using the eight-figure value for 2.1798741 10 18

λgivesE hc hcR  

,m100967758

R

19

18 3



m10890.5

s)m10000.3(J10626.6(λ

19 7

8 34

1 2

/ 3

.J10371.3m

10896.5

s)m10000.3(J10626.6(λ

22 19

19 2

8 34

2 2

s)m1000.3(sJ1063.6(λ

20 5

8 34

J1050

38.32: 20.66 eV18.70eV1.96eV3.1410 19J,andλ  632nm,

E

hc f

c

in good agreement

min

hc hf

m1011.3)

V4000)(

C1060.1(

)sm10s)(3.00J

1063.6(

19

8 34

This is the same answer as would be obtained if electrons of this energy were used Electron beams are much more easily produced and accelerated than proton beams

b) The shortest wavelength would correspond

to the maximum electron energy, eV, and so λ 0.0414nm

eV hc

38.35: An electron’s energy after being accelerated by a voltage V is just E eV The most energetic photon able to be produced by the electron is just:

V

λ

e

hc E

hc 



.nm0829.0m1029.8V)105.1)(

C1060.1(

)sm1000.3)(

sJ1063.6(

4 19

8 34

λ1

cos

mc h

,731.0nm002426

0

nm0042.0

1

nm002426

0

nm0021.01cos.nm0500.0nm0521

.)λmaximizeto

is180))(

1(1(λ

1065.6

38.38: a) λ 0.0691nm.b)λλ(h mc)(1cos)(2.42610 12 m)

eV hc

.nm0698.0λsom,1011.7)

mc

h

λ, 180 ; λ 2λ

λ    

6

1070.9λ

2λ

mc

h Δ

38.40: The change in wavelength of the scattered photon is given by Eq 38.23

 (1 cos ).

λ)cos1(λλ

h

Thus,

.m1065.2)11()100.0)(

m/s1000.3)(

kg1067.1

(

)sJ1063.6(

8 27

38.41: The derivation of Eq (38.23) is explicitly shown in Equations (38.24) through

(38.27) with the final substitution of

)

cos1(λ

λyieldingλ

p h

p

m10400

Km1090.2m

1090

9

3 3

K728.2

Km1090.2Km1090.2

λand,mm966.0K

00.3

Km10898

2

Note that a more precise value of the Wien displacement law constant has been used b)

A factor of 100 increase in the temperature lowers λ by a factor of 100 to 9.66 m  and mraises the frequency by the same factor, to 3.101013 Hz c) Similarly,λm 966nm

.Hz1010

3

and f   14

38.45: a) H  Aeσ T4;Ar2l

4

4 2 8

2 3

4

)KmW10671.5)(

26.0)(

m30.0(m)10(0.20

W100

K1006

Much of the emitted radiation is in the ultraviolet

38.46: (a) Wien’s law: T k

m λ

nm97m107.9K

30,000

mK1090.2

m

This peak is in the ultraviolet region, which is not visible The star is blue because the

largest part of the visible light radiated is in the blue violetpart of the visible spectrum (b) PσAT4 (Stefan-Boltzmann law)

m102.8

)K000,30)(

4(Km

W105.67W)

1086.3)(

2 8 26

m108.2

8

9 sun

(c) The visual luminosity is proportional to the power radiated at visible wavelengths Much of the power is radiated nonvisible wavelengths, which does not contribute to the visible luminosity

21

but)1(

λ

2)

(

2 5

2

x

x x e

e

hc λ

2)λ(λ

2)

(

kT hc

hc λ

I

(38.31), which is Rayleigh’s distribution.

38.48: a) As in Example 38.10, using four-place values for the physical constants,

b) With T 2000 K and the same values for λ andλ,

37.14

λ 

kT hc

and so

.1054.7λ)

c) With 4.790and (λ) λ 1.36 10

λ,K

hc T

d) For these temperatures, the intensity varies strongly with temperature, although for even higher temperatures the intensity in this wavelength interval would decrease From the Wien displacement law, the temperature that has the peak of the corresponding distribution in this wavelength interval is 5800 K (see Example 38.10), close to that used

in part (c)

38.49: a) To find the maximum in the Planck distribution:

2 5

2 2

5

2 5

2

)1(

λ

)(

2)1(

λ

)2(50)1(

λ

2λ

e

hc e

hc d

d d

5Solve

λ55

λ)1(

5

λ λ

kT

hc x

e x e e

)sm1000.3)(

sJ1063.6()965.4(

23

8 34

38.50: Combining Equations (38.28) and (38.30),

nm

872m1072

8

)KmW1067.5W/m10(6.94

K)m1090.2((

Km1090.2

λ

7

1/4 4 2 8

2 6

3 1/4

3 m

38.51: a) Energy to dissociate an AgBrmolecule is just 

J1000.1mole1

)mole(

E E

.eV04.1eV)J

10

(1.60

J10

)sm1000.3)(

sJ1063.6(

19

8 34

m/s1000.3λ

14 6

J1063.6)Hz1000.1)(

sJ1063.6

19

26 8

38.52: a) Assume a non-relativistic velocity and conserve momentum

λ2λ2

12

1

m

h m

h m mv

λλ

2 2

2

mc

h hc m

(10.2

)sm1000.3)(

sJ1063.6(λ

.neglectedbe

canrecoilsosmallquiteisThis.1042.5eV

2.10

eV1053.5

eV

105.53J

1084.8)m1022.1)(

kg1067.1(2

)sJ1063.6(

9 8

8 27

2 7 27

2 34

38.53: Given a source of spontaneous emission photons we can imagine we have a

uniform source of photons over a long period of time (any one direction as likely as any other for emission) If a certain number of photons pass out though an area A, a distance

D from the source, then at a distance 2D, those photons are spread out over an area

38.55: a) Plot: Below is the graph of frequency versus stopping potential.

Threshold frequency is when the stopping potential is zero

Hz104.60Hz

1011.4

89

)sm1000.3(

c) The work function is just hf th (6.6310 34 Js)(4.601014Hz)3.0510 19 JeV

V

sJ1058.6

)C1060.1)(

HzV1011.4(

34

19 15

)10.0)(

W200()

(

)(

14

h hf

P dN dE

dt dE dt

So.cmsecphotons10

00.14

6930)

cmsecphotons10

(1.00

4

secphotons10

03

2 11

38.57: a) Recall  

λ

0

hc eV

0 1

2 01

1λ

1λ

λ)(

e

hc V hc hc V

V

e

b)

.V477.0m1095.2

1m

1065.2

1C)

10(1.60

)sm10s)(3.00J

1063

6

(

7 7

11

8 34

So the change in the stopping potential is an increase of 0.739 V

38.58: From Eq (38.13), the speed in the ground state is (2.19 106 m s)

v  gives Z = 13.7, or 14 as an integer b) The ionization energy is EZ2 (13.6

eV), and the rest mass energy of an electron is 0.511 MeV, and setting

smkg1099.6λ

m09485.0

2λλso

1λ

(9.11

s)J10(6.63m)

100830

0

(

)cos1(λ

λ)cos1(λ

λ

8 31

h

b) Since the collision is one-dimensional, the magnitude of the electron’s momentum must be equal to the magnitude of the change in the photon’s momentum Thus,

.smkg102smkg1065

1

)m10(0830.0

10781.0

1)sJ1063.6(λ

1λ1

23 23

1 9 34

16 16

2 e e

38.61: a) 1.69 10 kg.

207

2 1

p e r

m m

m m m

m

m m m

b) The new energy levels are given by Eq (38.18) with m e replaced by m r

.keV53.2eV

1

1 2

3

2 31

kg 28

2 2

2

4 2

n

n m

m h

n

e m E

e

r r

n 

c)

)sm10(3.00s)J1063.6(

))eVJ10(1.60eV))1053.2(4Ve1053.2()(

1

λ

1

8 34

19 3

3 2

eV,0.8eV,

0.5eV,

1055.8

)sm1000.3(s)J1063.6(λ

18 8

8 34

would expect a maximum kinetic energy of 14.5eV3.4eV11.1eV, which is exactly 10.2 eV above that measured in part (a), explaining the anomoly

38.64: a) In terms of the satellite’s mass M , orbital radius Rand orbital period T,

MR h

L h n

2 2

V  and its centripetal acceleration is 2  22 3 

R M

L R

L R

R M

ML R

GM

earth

2 3

2

2 2

2 2

h n

R

 c) R2knn, and for the next orbit, n1,andRn(h2 42GMearthM)

Insertion of numerical values from Appendix F and using n from part (a) gives

m,10

h n mvr L



2But

2

2 2 2

2 2

4

h n r

D mr

h n D

mv r

r

mv Dr



 b) The energy E KU  mv  Dr ,sinceF Dr

2

12

to

2 2 2

2

2

12

1So.2

1

Dr Dr r

m

D m E kx U

nh m

D mD

nh D

h m

D n n E E

22

)(

38.66: a) (1 ( )) 1,

d e e

d e

d e d r,

m m

values gives mr,d 0.999728me b) Let R

R m

m R

e

d r, e

p r,

,3)4(λand,Then

nm

033.0λ

13

4λ

d r,

e p r,

m R

R R

R R

38.67: a) The H line is emitted by an electron in the  n3 energy level,

eV

51.1)

i

n n hc

E E

658 nm, 103 nm, and 122 nm for the respective photons above

38.68:

kln

)(

1 2

g ex kT

) (

E E T

e n

J

101.63eV2.10

eV

6.13

eV

4.34

eV6.13

18 g

ex

g

2 ex

E

E E

ln(10)KJ10(1.38

J)1063.1(

12 23

ln(10)KJ10(1.38

J)1063.1(

8 23

ln(10)KJ10(1.38

J)1063.1(

4 23

absorption lines

38.69: The transition energy equals the sum of the recoiling atom’s kinetic energy and

the photon’s energy

.λ

k tr

E E

hc E

E

hc E

E E E

2 2

)λ(λ

11

11

11

1λ

E hc E

E hc

E

E E

hc

E E E

hc E

E E hc

k k

tr k tr

k tr

tr k tr

tr k tr

Conservation of momentum, assuming atom initially at rest, yields:

mc

h mhcλ

h

m

h m

P E P

h

k k

2

λ2

λ

λ22λ

2 2 2

2

2 2

34

106.6)sm10kg)(3.0010

2(1.67

s)J1063.6(

 K m v

38.71: a) Largest wavelength shift:

s)m10kg)(3.0010

(9.11

s)J1063.6(2))

1(1

(

8 31

Smallest energy implies largest  so,

2

eV1011.52λ

2λ1

dE mcdT

mc

pIV dt

dE mc

.0(

)V10A)(18.010

0.60)(

01.0

d) A high melting point and heat capacity—tungsten and copper, for example

2

λ

1λ

1)

(f f hc h

E

319.5J

10111.5m10100.1

1m

101.132

1)

sm10s)(3.00J

which is a loss for the photon, but which is a gain for the electron So, the kinetic energy

of the electron is 5.11110 17 J319.5eV

.sm1006.1kg1011.9

J)1011.5(2

b) If all the energy of the electron is lost in the emission of a photon, then

J105.111

)sm10s)(3.00J

1063.6(

nm

3.89m

10892

38.74: a) λ1 (h mc)(1cosθ1),λ2 (h mc)(1cosθ2), and so the overall wavelength shift is λ(h mc)(2cosθ1cosθ2).

(b) For a single scattering through angle θ,λs  (h mc)(1cosθ)

For two successive scatterings through an angle of  2, for each scattering, λt 2(h mc)(1cos 2)

t s 2

2 s

2

λλand))2cos(

1()2(cos1so1)

2

cos(

))2(cos1(2)(λand))2(cos1(2

h θ

θ

Equality holds only when  180

c) (h mc)2(1cos30.0)0.268(h mc).d)(h mc (1cos60)0.500(h mc),which is indeed greater than the shift found in part (c)

38.75: a) The wavelength of the gamma rays is

m101.24eV)

J10(1.60eV)10(1.00

)sm10(3.00s)J10(6.63

19 6

8 34

m101.24m

cos1(

mc

h mc

h mc

10

6.42

s)J10(6.63

m)10(5.00)sm10(3.00Kg)10

2(9.11λ

2

So

9 11

34

33 8

collisions10

sec103.15ears

13

26 13

6 y

.collisionsec

0.0945m

10

45

38.76: a) The final energy of the photon is ,and ,

(

1λ

1

λ

)1()λ()

λ(λ

2 2 2

mc

mc hc

hc K

hc

hc K

E

hc



)1(

(K mc2   since the relativistic expression must be used for three-figure accuracy) b)  arccos(1λ (h mc))

3.801

(0.250))

sm10(3.00kg)10(9.11m)10(5.101

mm105.10λ

34

8 31

102.43

m)103.34m10(5.101

arccos

nm

103.34

12

12 12

λ

2)

λ( 5 λ 2

)1(

2)

1(

)(

2)

hf e

f c

hc f

f

c df f I d

3 2

4 4 5 2

3

4 5 4

3 2 4

0

3 3

2 4

3

15

2240

)()2()2(240

1)(2

1

)(2)1(

2

h c

T k c

h

kT h

c kT

dx e

x h

c

kT e

c

df hf

x kT

T k

as shown in Eq (38.36) Plugging in the values for the constants we get σ 5.6710 8 W m2 K4

38.78: I σT4,PIA,andE Pt;combining,

hrs

2.45s

10

8.81

K)(473)KmW10(5.67)m10(4.00

J)(100

3

4 4

2 8 2

6 4

4

me

h n ε

T  and frequency is just

.4

1

3 3 2

0

4

h n

me f

2

)1(

11

11then,1and

n n

n

.4

large

for

2

211

1)

11(

11

1

3 3 2 0

4 3

2 2 2

h n

me f

n n

n n

n n

(

λ

(h dN dt A The intensity is

,)λ((

)((dN dt E A dN dt hc A

and comparison of the two expressions gives the pressure as ( c I )

38.81: Momentum:

)( p P P

p

P p P

)(

2)(

(2)(

)(pc pc 2 E2 E2  pc pc 2  Pc2 p p  Ec p p  pc2  Ec p

0 )

hc Pc E

Pc E

hc Pc

E

Pc E Pc

E

Pc E hc

Pc E pc

Pc E p Pc Ec pc

Pc Ec p

p

Pc Ec p Ec pc Pc

λ(λ

2λ

)(

λ2λλ

)(

22

)(

)2

(

2 2

2

2 2

2

If

2 2 2

2 2

mc E

Pc mc E

c2

11

E

m E

E

mc Pc

2 1

)m10(10.6kg)

1011.9(1J1060.1

6 4

2 31

c hc

(1.2410 16 m)(156.0)7.0810 15 m

c) These photons are gamma rays We have taken infrared radiation and converted it into gamma rays! Perhaps useful in nuclear medicine, nuclear spectroscopy, or high energy physics: wherever controlled gamma ray sources might be useful