Chapter 40 - Introduction to Quantum Physics P40.1

−

−

× ⋅= =

×

33

9

2.898 10 m K 5.18 10 K560 10 m

T ×

P40.6 (a) ( )( )( )σ − −= = × × ⋅ = ×P 44 4 2 8 2 41 20.0 10 m 5.67 10 W m K 5 000 K 7.09 10 WeA T 4

(b) ( )λ λ λ−= = × ⋅ ⇒ =3max max max5 000 K 2.898 10 m K 580 nmT

(c) We compute: ( )( )

( ) ( )

−−

−

× ⋅ ×= =

×

34 86

23B

6.626 10 J s 3.00 10 m s2.88 10 m

1.38 10 J K 5 000 Khc

k T×

The power per wavelength interval is ( ) ( )( )π

λ λλ λ

= =⎡ ⎤−⎣ ⎦

P2

5B

2exp 1

hc AAI

hc k T,

and ( )( ) ( )π π − −= × × × = × ⋅

22 34 8 4 192 2 6.626 10 3.00 10 20.0 10 7.50 10 J m shc A − 4

( )

( ) ( )μ μ

−

−

× ⋅ ×= =

−⎡ ⎤× −⎣ ⎦

= ×

P19 4 13

5 4.9739

10

7.50 10 J m s 1.15 10 J m s580 nm

1580 10 m exp 2.88 m 0.580 m 1

7.99 10 W m

e⋅

(d)–(i) The other values are computed similarly:

(d)

(e)

(f)

(c)

(g)

(h)

(i)

λ πλ λ

λ λ−

−

−

× × ×× × ×

× ×× ×

P2

/5

1251 26 1226

250 23 227

13 10

13 10

21 ( ),

1.00 nm 2882.6 7.96 10 7.50 10 9.42 105.00 nm 576.5 2.40 10 2.40 10 1.00 10400 nm 7.21 1347 7.32 10 5.44 10580 nm 4.97 143.5 1.15 10 7.99 10700 nm 4.12 60.4 4

Bhc k T

B

hc Ahc ek T

W m

−

− − − −

× ××

× × × ×

12 10

4

5 5 14

.46 10 7.38 101.00 mm 0.00288 0.00289 7.50 10 0.26010.0 cm 2.88 10 2.88 10 7.50 10 2.60 10 9

(j) We approximate the area under the ( )λP versus λ curve, between 400 nm and 700 nm, as two trapezoids:

( ) ( )

( ) ( )

−

−

⎡ ⎤ ⎡+ × − ×⎣ ⎦ ⎣=

⎡ ⎤ ⎡+ × − ×⎣ ⎦ ⎣+

P10 9

10 9

5.44 7.99 10 W m 580 400 10 m2

7.99 7.38 10 W m 700 580 10 m2

⎤⎦

⎤⎦

so the power radiated as visible light is

= ×P 42.13 10 Wapproximately 20 kW .

P40.7 (a) , so σ=P 4eA T

( ) ( )σ π −

⎡ ⎤×⎛ ⎞ ⎢ ⎥= = = ×⎜ ⎟ ⎢ ⎥⎡ ⎤⎝ ⎠ × × ⋅⎢ ⎥⎣ ⎦⎣ ⎦

P1 4

1 4 263

28 8 2 4

3.85 10 W 5.78 10 K1 4 6.96 10 m 5.67 10 W m K

TeA

(b) λ− −

−× ⋅ × ⋅= = = × =

×

3 37

max 3

2.898 10 m K 2.898 10 m K 5.01 10 m 501 nm5.78 10 KT

P40.14 (a) ( )( )

( )( )λ

φ

−

−

× ⋅ ×= = =

×

34 8

19

6.626 10 J s 3.00 10 m s296 nm

4.20 eV 1.60 10 J eVchc

λ −

×= = = ×

×

815

9

3.00 10 m s1.01 10 Hz

296 10 mcc

cf

(b) φλ

= + Δ Shc e V :

( )( )

( )( ) (−

− −−

× ×)= × + ×

×

34 819 19

9

6.626 10 3.00 104.20 eV 1.60 10 J eV 1.60 10

180 10 SVΔ

Therefore, Δ = 2.71 VSV

P40.24 (a) ( )λ θΔ = −1 cose

hm c

:

( )( ) ( )λ−

−−

×Δ = − ° = ×

× ×

3413

31 8

6.626 10 1 cos37.0 4.88 10 m9.11 10 3.00 10

(b) λ

=00

hcE :

( )( ) ( )( )λ

−−

× ×× × =

34 83 19

0

6.626 10 3.00 10 m s300 10 eV 1.60 10 J eV

λ −= × 120 4.14 10 m

and λ λ λ −′ = + Δ = × 120 4.63 10 m

( )( )

λ

−−

−

× ⋅ ×′ = = = × =

′ ×

34 814

12

6.626 10 J s 3.00 10 m s4.30 10 J 268 keV

4.63 10 mhcE

(c) ′= − = − =0 300 keV 268.5 keV 31.5 keVeK E E

P40.31 Maximum energy loss appears as maximum increase in wavelength, which occurs for

scattering angle 180°. Then ( )λ ⎛ ⎞Δ = − ° =⎜ ⎟⎝ ⎠

21 cos180 hmc mc

h where m is the mass of the

target particle. The fractional energy loss is

λ λ λ λ λ

λ λ λ λ λ′′ ′−− − Δ

= = = =′ + Δ +

00 0

0 0 0 0

22

hc hc h mcE EE hc h mc

Further, λ =00

hcE

, so′−= =

+ +0 0

20 0

2 22 2

h mcE E EE hc E h mc mc E0

.

(a) For scattering from a free electron, , so =2 0.511 MeVmc

( )( )

′−= =

+0

0

2 0.511 MeV0.667

0.511 MeV 2 0.511 MeVE E

E

(b) For scattering from a free proton, , and =2 938 MeVmc

( )( )

′−= =

+0

0

2 0.511 MeV0.00109

938 MeV 2 0.511 MeVE E

E

P40.34 ( )( )

λ−

−−

× ⋅= = = = ×

× ×

3413

27 6

6.626 10 J s 3.97 10 m1.67 10 kg 1.00 10 m s

h hp mu

*P40.40 (a) For the massive particle, ( )γ= − 21K mc and λγ

= =mh hp mu

. For the photon (which

we represent as γ), and=E K( )γλ γ

= = = =− 21

c ch ch chf E K mc

. Then the ratio is

( )γλ γ γ γ

λ γ γ= =

− − −2 2 2

1 where =1 1 1 /m

ch mu umc h c u c

. The ratio can be written

( )− − 2 21 1 /

u

c u c

(b) ( )

( )γλ

λ= =

⎡ ⎤− − −⎢ ⎥⎣ ⎦2 2

1 0.91.60

1 0.9 1 1 0.9 1m

(c) The ratio for a particular particle speed does not depend on the particle mass: There would be no change.

(d) ( )

( ) ( )( )γλ

λ= =

⎡ ⎤− − −⎢ ⎥⎣ ⎦

3

2 2

1 0.0012.00 10

1 0.001 1 1 0.001 1m

×

(e) As → 1uc

, γ → ∞ and γ −1 becomes nearly equal to γ. Then γλ γλ γ

→ =1 1m

.

(f) As → 0uc

, −

⎛ ⎞ ⎛ ⎞− − − − − =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠≈

1 22 2

2 2

1 11 1 1 12 2

u uc c

2

2

uc

and( )( )

γλλ

→ = → ∞2 2

211 2m

u c cuu c

.

P40.47 (a) ( )( )

λ−

−−

× ⋅= = = ×

×

347

27

6.626 10 J s 9.92 10 m1.67 10 kg 0.400 m s

hmu

(b) For destructive interference in a multiple-slit experiment, θ λ⎛ ⎞= +⎜ ⎟⎝ ⎠

1sin2

d m ,

with for the first minimum. Then = 0m

λθ − ⎛ ⎞= =⎜ ⎟⎝ ⎠

1sin 0.028 42d

°

θ= tanyL

( )( )θ= = ° =tan 10.0 m tan 0.028 4 4.96 mmy L

(c) We cannot say the neutron passed through one slit. If its detection forms part

of an interference pattern, we can only say it passed through the array of slits. If we test to see which slit a particular neutron passes through, it will not form part of the interference pattern.

P40.49 For the electron,

( )( )( )− −Δ = Δ = × × = × ⋅31 4 329.11 10 kg 500 m s 1.00 10 4.56 10 kg m sep m u −

( )π π

−

−

× ⋅Δ = = =

Δ × ⋅

34

32

6.626 10 J s 1.16 mm4 4 4.56 10 kg m s

hxp

For the bullet,

( )( )( )− −Δ = Δ = × = × ⋅4 30.020 0 kg 500 m s 1.00 10 1.00 10 kg m sp m u

π

−Δ = = ×Δ

325.28 10 m4

hxp

P40.50 (a) Δ Δ = Δ Δ ≥h

2p x m u x so

( )( )

ππ π

⋅Δ ≥ = =

Δ2 J s 0.250 m s

4 4 2.00 kg 1.00 mhum x

(b) The duck might move by ( )( ) =0.25 m s 5 s 1.25 m . With original position

uncertainty of 1.00 m, we can think of Δx growing to + =1.00 m 1.25 m 2.25 m .

P40.52 With , the uncertainty principle

requires

−Δ = × 152 10 mx−Δ ≥ = × ⋅

Δh 202.6 10 kg m s

2xpx

.

The average momentum of the particle bound in a stationary nucleus is zero. The uncertainty in momentum measures the root-mean-square momentum, so we take

−≈ × ⋅203 10 kg m srmsp . For an electron, the non-relativistic approximation = ep m u would predict ≈ × 103 10 m su , while u cannot be greater than c.

Thus, a better solution would be ( ) ( ) γ⎡ ⎤= + ≈ =⎣ ⎦1 22 22 256 MeVe eE m c pc m c

γ ≈ =− 2 2

11101 v c

so ≈ 0.999 96u c

For a proton, =p

um

gives = × 71.8 10 m su , less than one-tenth the speed of light.

P40.53 (a) At the top of the ladder, the woman holds a pellet inside a small regionΔ ix . Thus, the uncertainty principle requires her to release it with typical horizontal

momentum Δ = Δ =Δh

2x xi

p m ux

. It falls to the floor in a travel time given by

= + 2102

H gt as =2Htg

, so the total width of the impact points is

( ) ⎛ ⎞Δ = Δ + Δ = Δ + = Δ +⎜ ⎟Δ Δ⎝ ⎠

h 22f i x i i

i i

H Ax x u t x xm x g x

where =h 2

2HA

m g

To minimizeΔ fx , we require ( )( )Δ

=Δ

0f

i

d x

d x or − =

Δ 21 0i

Ax

so Δ =ix A

The minimum width of the impact points is

( )Δ =

⎛ ⎞⎛ ⎞Δ = Δ + = = ⎜ ⎟⎜ ⎟Δ⎝ ⎠ ⎝ ⎠

h1 4

min

2 22i

f ii x A

A Hx x Ax m g

(b) ( ) ( ) ( )−−

−

⎡ ⎤× ⋅ ⎡ ⎤Δ = = ×⎢ ⎥ ⎢ ⎥×⎢ ⎥ ⎣ ⎦⎣ ⎦

1 2 1 43416

4 2min

2 1.054 6 10 J s 2 2.00 m5.19 10 m

5.00 10 kg 9.80 m sfx