Physics Week 6 Solutions

Ex 24,26,30,33,37,44,48,53,54,56,62,63,64,69,70,78

24. max

( / )hc Kφ λ= − is greatest whenmax

0.K = Thus, 1240 eV nm/900 nm 1.38 eVφ = ⋅ = (see Example 39.2).

26. The wavelengths in the Paschen series for hydrogen are given by Equation 34.9, with2

3n = and

14, 5, 6, . . . .n = Thus,

1 2 2

H 1 19 /( 9).R n nλ −= − If one substitutes 1282 nmλ = and1

H 0.01097 (nm) ,R−= one finds

2

19n = 2

1(1282 0.01097)( 9),n× − or1

9 1.41/(14.1 9) 5,n = × − = corresponding to the second line in this series.

30.For a slow electron ( ),v c<< the de Broglie wavelength is / ,h mvλ = so34

/ (6.626 10 J s)v h mλ −= = × ⋅ × 31 1 1

(9.11 10 kg) (1 mm) 72.7 cm/s.− − −× =

33. INTERPRET We want to find the uncertainty in the velocity of a proton, given its uncertainty in position.

DEVELOP To find ,v∆ we use the uncertainty principle, x p∆ ∆ ≥ h (Equation 34.15) with p m v∆ = ∆ and

1 fm.x∆ =

EVALUATE The above equation gives

7(197.3 MeV fm/ )0.21 6.3 10 m/s

(938 MeV)(1 fm)

p cv c

m m x

∆ ⋅∆ = ≥ = = = ×

∆

h

ASSESS The quantity / 197.3 MeV/p x c∆ = ∆ =h is barely small enough compared to 938 MeV/mc c= to justify

using the non-relativistic relation ,p mv= but this is good enough for approximate purposes.

37. INTERPRET The neutron is confined in the uranium nucleus with x∆ equal to the diameter of the nucleus.

We want to find the minimum energy of the neutron using the uncertainty principle.

DEVELOP Using the same reasoning as given in Example 34.6, for a neutron2

( 940 MeV)mc = confined to a

uranium nucleus ( 15 fm),x∆ ≈ the uncertainty principle requires that

22 1

2 2 2

pK

m m x

= ≥

∆

h

EVALUATE From the above equation, we find the minimum kinetic energy to be

2 2 2

min 2

1 1 (197.3 MeV fm/30 fm)23.0 keV

2 2 2 2(940 MeV)2

cK

m x xmc

⋅ = = = =

∆ ∆

h h

ASSESS This is smaller than the 5 MeV estimated for the nucleon in Example 34.6 by a factor of2

15 , since x∆ is

15 times larger. Most estimates of nuclear energies for single-particle states, based on the uncertainty principle,

give values of the order 1 MeV, consistent with experimental measurements.

44. Einstein’s equation for the photoelectric effect (Equation 34.7) givesmax max

( / ) , or /( )K hc hc Kλ φ λ φ= − = + =

1240 eV nm/(1.3 4.28) eV 222 nm.⋅ + = (We used the work function listed in Table 34.1.)

48. A scattered photon with half the initial energy ( / )hc λ has wavelength 2 ,λ λ′ = so (1 cos90 )c

λ λ λ∆ = = − ° =

2.43 pm,c

λ = is the initial wavelength of photons Compton-scattered from electrons at the given angle.

(See Equation 34.8.)

53. INTERPRET This problem is about whether or not the photoelectric effect can occur with a given wavelength of

the light source.

DEVELOP To determine whether or not the photoelectric effect can occur, we first find the work function.

Einstein’s photoelectric effect equation (Equation 34.7), and the data for the blue light, give the work function of

the photocathode material as

max max

1240 eV nm0.85 eV 2.03 eV

430 nm

hcE K Kγφ

λ

⋅= − = − = − =

EVALUATE The energy of a photon of the red light is only

1240 eV nm1.96 eV

633 nm

hc

λ

⋅= =

and therefore is insufficient to eject photoelectrons.

ASSESS A wavelength of 633 nm is greater than the cutoff wavelength ofcutoff

/ 610 nmhcλ φ= = for the

photocathode material.

54. The energy of the photon emitted in a hydrogen atom transition between adjacent states1 2 1

( 1)n n n→ = − is

/E hcγ λ= = 2 2 2 2

H 1 1 H 1 1 1[( 1) ] (2 1) ( 1)hcR n n hcR n n n− − − −− − = − − (see Equations 34.6 and 34.9 and the discussion

of the Bohr atom in the text). This is a maximum for the smallest allowed1 2

1,n n> ≥ which is1

2.n = Then

(a) 3 3,max H4 4

( ) ( )(13.6 eV) 10.2 eV,E hcRγ = = = and (b) 2 1 1 1.n n= − = (This is the Lyman alpha photon at 122

nm.)

56. The series limit 1( )n = ∞ is2

2 H/ 91.2 nmn Rλ∞ = = for the Lyman series 1( 1;n = see solution to Exercise 25), and

four times this, or 365 nm, for the Balmer series2

2( 4).n =

62. Since 0.1 nm 2.43 pm,c

λ λ= >> = non-relativistic expressions can be used to write the kinetic energy in terms

of the de Broglie wavelength,2 2 2 2 2 2 2/2 ( / ) /2 /2 ( ) /2 (1.24 keV nm) /2(511 keV)K p m h m h m hc mcλ λ λ= = = = = ⋅

2(0.1 nm) 150 eV.= Since the kinetic energy gained in an acceleration from rest equals the potential energy

difference, then ,K e V= ∆ and 150 V.V∆ =

63. INTERPRET The resolution of a microscope depends on the wavelength; a smaller wavelength implies a greater

electron speed.

DEVELOP The electron microscope is better if the wavelength is / 450 nm.h pλ = < Since ,p mv= the above

condition allows us to obtain the minimum electron speed.

EVALUATE The above inequality gives

34

31

6.626 10 J s1.62 km/s

(450 nm) (9.91 10 kg)(450 nm)

hv

m

−

−

× ⋅> = =

×

So the minimum speed is 1.62 km/s.

ASSESS Since 450 nm 0.00243 nmC

λ>> = (Compton wavelength of electron), our use of the non-relativistic

momentum was justified. The electron microscope could provide resolutions down to about 1 nm and

magnifications of6

10 .

64. The velocities are small enough compared to c so that p mv= can be used to find the de Broglie wavelength

/ .h mvλ = The ratios are (a)34 32

(6.63 10 J s)/(0.1 kg)(5 m/s)(0.1 m) 1.33 10− −× ⋅ = × for the ball, (b) 1.47 for the

alpha particle, and (c) 92.9 for the U238

nucleus. (d) The ratio is negligible for the ball, and most significant for

the U238

nucleus.

69. INTERPRET In this problem we want to show that if a photon’s wavelength is equal to a particle’s Compton

wavelength, then the photon’s energy is equal to the particle’s rest energy.

DEVELOP The Compton wavelength of a particle is / ,C

h mcλ = and its rest energy is2.E mc= On the other

hand, the energy of a photon is / .E hf hcγ λ= =

EVALUATE When the wavelength of the photon is ,C

λ λ= its energy is

2

/C

hc hcE mc

h mcγ

λ= = =

which is the same as the particle’s rest energy.

ASSESS We emphasize that photons have zero rest mass and their energy is completely kinetic in nature.

70. The frequencies of hydrogen spectral lines for transitions to the nth level

are2 2

H 1/ ( ),f c cR n nλ − −= = − where 1n =

1, 2, . . . .n n+ + The lowest and highest frequencies are2 2

1 H[ ( 1) ]f cR n n− −= − + and

2

H ,f cR n−

∞ = so the

frequency

interval covered by the series is2

1 H ( 1) .f f f cR n−

∞∆ = − = +

78. INTERPRET We show that the correspondence principle holds for the Bohr model in that the frequency of a

photon emitted in a 1n n+ → transition for large n equals the orbital frequency of the electron. We’ll do this by

taking the limit of both the transition energy and the orbital frequency as .n → ∞

DEVELOP The energy of the photon is2 2 2

0 2 1/2 (1/ 1/ ),E ke a n n∆ = − where 2n n= and 1 1.n n= + The frequency of

the photon is .E

h

∆ The orbital frequency can be calculated from the orbital radius2 2 2/2 /r ke E ke mv= − = and the

orbital velocity /v n mr= h using /2 .o

f v rπ=

We will use the binomial expansion on fγ and compare the result with the orbital frequency .o

f

EVALUATE From /v n mr= h and2 2/r ke mv= we obtain the orbital radius

2 2

2 .n

e mkr = h The orbital frequency is

4 2 2

3 3 3

0

,2 2

o

v e mk e kf

r n ha nπ π= = =

hwhere

2

0 2a

e mk=

h

The frequency of the photon emitted is ( )2 2

2 2 20 0

21 1 1 1

2 2( 1)(1 (1 ) ),ke ke

ha ha nn n nfγ

−

+= − = − + so for 1 1,

n=

2 2 2

2 2 3

0 0 0

2 21 1

2 2

ke ke e kf

n nha n ha n ha nγ

≈ − − = =

ASSESS For large values of n, the optical frequency fγ is the same as the orbital frequency .o

f