THÔNG TIN TÀI LIỆU
39.1: a)
.smkg1037.2
)m1080.2(
)sJ1063.6(
λ
λ
24
10
34
h
p
p
h
b)
eV.3.19J1008.3
)kg1011.9(2
)smkg1037.2(
2
18
31
224
2
m
p
K
39.2:
mE
h
p
h
2
λ
m.1002.7
V)eJ1060.1()eV1020.4()kg1064.6(2
)sJ1063.6(
15
19627
34
39. 3: a)
m.1055.1
)sm1070.4(kg)1011.9(
sJ1063.6
λ
10
631
34
vm
h
e
e
b) m.1046.8m1055.1
kg1067.1
kg1011.9
λλ
1410
27
31
e
p
e
p
m
m
39.4: a) keV.2.6
m)1020.0(
)sm1000.3(s)eV10136.4(
λ
9
815
hc
E
b)
kg)1011.9(2
))m1020.0()sJ10626.6((
2
)
λ(
2
31
29342
2
m
h
m
p
K
eV.37J100.6
18
Note that the kinetic energy found this way is much smaller than the rest energy, so the
nonrelativistic approximation is appropriate.
c)
J103.8
kg)1064.6(2
))m1020.0(s)J10626.6((
2
)
λ(
2
22
27
293422
m
h
m
p
K
5.2 meV. Again, the nonrelativistic approximation is appropriate.
39.5: a) In the Bohr model .
2
π
nh
rmv
n
The de Broglie wavelength is
m.1032.3)m1029.5(2λm1029.5:1for
2
λ
1011
1
11
01
πarn
n
r
π
mv
h
p
h
n
This equals the orbit circumference.
b)
,λ4
4
)16(2
λ16)4(:4
0
400
2
4
aπ
aarn
.m1033.1λ
9
4
The de Broglie wavelength is a quarter of the circumference of the orbit, .2
4
πr
39.6: a) For a nonrelativistic particle, so,
2
2
m
p
K
.
2
λ
Km
h
p
h
b) m.1034.4)Kg1011.9()J/eV1060.1()eV800(2)sJ1063.6(
11311934
39.7:
m.1090.3
)sm340()kg005.0(
sJ1063.6
λ
34
34
mv
h
p
h
We should not expect the bullet to exhibit wavelike properties.
39.8: Combining Equations 37.38 and 37.39 gives
.1
2
γmcp
a) m.1043.41)(λ
122
mch
p
h
(The incorrect nonrelativistic calculation
gives m.)1005.5
12
b) m.1007.71)(
132
mch
39.9: a) photon
nm0.62
V)eJ10602.1()eV0.20(
)sm10998.2()sJ10626.6(
λso
λ
19
834
E
hchc
E
electron
V)eJ10602.1(eV)0.20()kg10109.9(22so)2(
19312
mEpmpE
smkg10416.2
24
nm274.0λ ph
b) photon eV96.4J10946.7λ
19
hcE
electron smkg10650.2λsoλ
27
hpph
eV1041.2J10856.3)2(
5242
mpE
c) You should use a probe of wavelength approximately 250 nm. An electron with
250λ
nm has much less energy than a photon with 250λ
nm, so is less likely to damage the
molecule.
39.10:
λ
λ
m
h
v
mv
h
2
2
2
2
λ2
λ2
1
2
1
m
h
m
h
mmvK
They will not have the same kinetic energy since they have different masses.
4
27
31
p
e
2
e
2
2
p
2
e
p
1046.5
kg1067.1
kg1011.9
λ2
λ2
m
m
m
h
m
h
K
K
39.11: a) nm10.0λ
keV12λ)c
eV150)b
sm103.7)
λ(soλ
2
2
1
6
hcE
mvE
mhvhmvp
d) The electron is a better probe because for the same
λ
it has less energy and is less
damaging to the structure being probed.
39.12: (a) λλ mhvmvh
Energy conservation:
2
2
1
mvVe
V9.66
)m1015.0()kg1011.9()C1060.1(2
)sJ10626.6(
λ22
)(
2
293119
234
2
2
2
λ
2
em
h
e
m
e
mv
V
m
h
)b( J1033.1
m1015.0
)sm100.3()sJ10626.6(
λ
15
9
834
photon
hc
hfE
V8310
C106.1
J1033.1
19
15
photon
photon
e
E
V
EKVe
39.13: For m =1,
eV.432.0J1091.6
)6.28(sin)m1010.9()kg10675.1(2
)sJ1063.6(
sin2
2
sin
λ
20
221127
234
22
2
E
θmd
h
E
mE
h
θd
39.14: Intensity maxima occur when .λsin mθd
.
ME
mh
θd
ME
h
p
h
2
sinso
2
λ
(Careful! Here, m is the order of the maxima, whereas M is the mass of the incoming
particle.)
a)
)6.60sin()VeJ1060.1()eV188()kg1011.9(2
s)J1063.6()2(
sin2
1931
34
θME
mh
d
nm.206.0m1006.2
10
b) m = 1 also gives a maximum.
one.otheronlytheisThis.8.25
m)1006.2()VeJ1060.1()eV188()kg1011.9(2
)sJ1063.6()1(
arcsin
101931
34
θ
If we let ,3
m then there are no more maxima.
c)
)6.60(sin)m1060.2(kg)1011.9(2
)sJ1063.6()1(
sin2
221031
2342
22
22
θMd
hm
E
eV.8.46J1049.7
18
Using this energy, if we let 2noisthereThus,.1sinthen,2
mθm maximum in this
case.
39.15: Surface scattering implies .λsin mθd
.341.0
m1034.8
m1096.4
arcsinSo
m.1096.4
λ
J/eV)1060.1()eV840()kg1064.6(2
sJ1063.6
2
λBut
]λarcsin[:1If
11
13
13
1927
34
θ
mE
h
p
h
d
θm
39.16: The condition for a maximum is
.arcsinso,λ.λsin
dMv
mh
θ
Mv
h
p
h
m
θd
(Careful! Here, m is the order of the maximum, whereas M is the incoming particle mass.)
a)
dMv
h
θm arcsin1
1
)sm1026.1()kg1011.9()m1060.1(
)sJ1063.6()2(
arcsin2
.07.2
)sm1026.1(kg)1011.9()m1060.1(
sJ1063.6
arcsin
4316
34
2
4316
34
θm
.14.4
b) For small angles (in radians!) so,Dθy
cm.81.1cm81.1cm61.3
cm61.3
180
radians
)14.4()cm0.50(
cm.81.1
180
radians
)07.2()cm0.50(
12
2
1
yy
π
y
π
y
39.17: a)
m)1000.1()kg1200(2
)sJ1063.6(
222
6
34
πxπm
h
v
π
h
xvm
π
h
xp
xx
.sm1079.8
32
b) Knowing the position of a macroscopic object (like a car) to within 1.00
m
μ
is, for all
practical purposes, indistinguishable from knowing “exactly” where the object is. Even with
this tiny position uncertainty of 1.00
m
μ
, the velocity uncertainty is insanely small by our
standards.
39.18: a) .
22
yuncertaintminimumfor
2
yπm
h
v
π
h
yvm
π
h
yp
yyy
sm102.3
)m100.2(kg)1067.1(2
)sJ1063.6(
4
1227
34
π
b) For minimum uncertainty,
m.1063.4
)sm502.0(kg)1011.9(2
s)J1063.6(
22
4
31
34
πvπm
h
p
π
h
z
zz
39.19: Heisenberg’s Uncertainty Principles tells us that:
.
2
π
h
px
x
We can treat the standard deviation as a direct measure of uncertainty.
.isclaimtheso
2
Therefore
sJ1005.1
2
butsJ106.3)smkg100.3(m)102.1(Here
34352510
validnot
π
h
px
π
h
px
x
x
39.20: a) ,2)()( πhvmx
x
and setting
xx
vv )010.0( and the product of the uncertainties
equal to
πh 2/ (for the minimum uncertainty) gives .sm57.9)(0.010)2( xπmhv
x
b) Repeating with the proton mass gives 31.6 s.mm
39.21: s.mkg1082.9
2
m10215.0
2
)sJ1063.6(
22
25
9
34
π
xπ
h
p
π
h
xp
m.18.5
)sm81.9(kg)00.1(
J8.50
)d
J.50.8J)1095.4(
Nikg1075.9
kg00.1
)c
eV.1009.3J1095.4
)kg1075.9(2
)m1082.9(
2
)b
2
total
total
24
26
total
524
26
2252
mg
K
hKmgh
NKK
m
p
K
e) One is claiming to know both an exact momentum for each atom (giving rise to an exact
kinetic energy of the system) and an exact position of each atom (giving rise to an exact potential
energy of the system), in violation of Heisenberg’s uncertainty principle.
39.22: .J1030.3eV1006.2
2
210292
ccmmcE
π
h
tE
s.1020.3
)J1030.3(2
sJ1063.6
2
25
10
34
2
πmcπ
h
t
39.23:
69.8J1039.1
s)106.7(2
s)J1063.6(
22
14
21
34
πtπ
h
E
π
h
tE
MeV.0869.0eV10
4
.1081.2
MeV3097
MeV0869.0
5
2
2
c
c
E
E
39.24:
eV.1027.1J1003.2
s)102.5(2
s)J1063.6(
2
1332
3
34
πtπ
h
E
39.25: To find a particle’s lifetime we need to know the uncertainty in its energy.
.s1008.1
)J1081.9(2
sJ1063.6
2
J1081.9
s)m1000.3()kg1067.1()5.4()145.0()(
24
11
34
11
28272
πEπ
h
t
cmE
39.26: a) V.419
2
)λ(
so,
2
)λ(
2
222
me
h
V
m
h
m
p
KeV
b) The voltage is reduced by the ratio of the particle masses,
V.229.0
kg1067.1
kg1011.9
V)419(
27
31
39.27: a) We recall .
2
λ
mE
h
p
h
But the energy of an electron accelerated through a
potential is just
VeE
m.1034.4λ
)V800(C)1060.1()kg1011.9(2
)sJ1063.6(
2
λ
11
1931
34
e
e
Vm
h
b) For a proton, all that changes is the mass, so
.m1001.1λ
m1034.4.
kg1067.1
kg1011.9
λλ
12
11
27
31
e
e
p
e
p
m
m
39.28: ,sin ωtψ
so
ωtψωtψψ
2
2
2**
2
sinsin .
2
is not time-independent, so
is not the wavefunction for a stationary state.
39.29: a)
kxAxψ sin . The probability density is ,sin
22
2
kxAψ and this is greatest when
5,3,1,
2
1sin
2
n
n
π
kxkx
5,3,1,
4
λ
)λ22(2
n
n
π
nπ
k
n
π
x
b) The probability is zero when
0
2
ψ
, which requires
2,1,0,
2
λ
0sin
2
n
n
k
nπ
xnπkxkx
39.30: a) The uncertainty in the particle position is proportional to the width of
xψ , and is
inversely proportional to
. This can be seen by either plotting the function for different
values of
, finding the expectation value
dxxψx
222
for the normalized wave function or
by finding the full width at half-maximum. The particle’s uncertainty in position decreases with
increasing
. The dependence of the expectation value
2
x on
may be found by considering
dxe
dxex
x
x
x
2
2
2
22
2
=
dxe
x
2
2
ln
2
1
,
4
1
2
1
ln
2
1
2
due
u
where the substitution xu
has been made. (b) Since the uncertainty in position decreases,
the uncertainty in momentum must increase.
39.31:
iyx
iyx
yxf
iyx
iyx
yxf
),(and),(
*
.1*
2
iyx
iyx
iyx
iyx
fff
39.32: The same.
),,(),,(),,(
*
2
zyxψzyxψzyxψ
)),,()(),,((),,(
*
2
iii
ezyxψezyxψezyxψ
).,,(),,(
*
zyxψzyxψ
The complex conjugate means convert all
i
’s to
i
’s and vice-versa. .1
ii
ee
39.33: Following the hint:
If we Taylor expand sin(
ax
) about a point
0
x , we get
))(()(
000
xxxfxf
.))(cos()sin(
000
xxaxaax
0
If xx is small we can even ignore the first order term and
sin(
ax
)
sin (
0
ax ).
For us
Lxx 01.0
0
which is small compared to
L
πx
L
zyx
ψL
0
23
sin
2
),,(so
.sinsin
00
L
πz
L
πy
a)
4
000
L
zyx
.101.00)0.01(
2
22
4
sin
2
63
6
3
6
3
2
L
L
V
π
L
dVψP
b)
2
1
000
zyx
.108.00)(0.01
2
2
sin
2
63
3
6
3
2
L
L
V
π
L
dV
ψP
39.34: Eq. (39.18): EψUψ
dx
ψd
m
2
22
2
Let
21
BψAψψ
)()()(
2
212121
2
22
BψAψEBψAψUBψAψ
dx
d
m
.0
22
22
2
2
22
11
2
1
22
EψUψ
dx
ψd
m
BE
ψUψ
dx
ψd
m
A
But each of
1
ψ and
2
ψ satisfy Schröedinger’s equation separately so the equation still
holds true, for any A or B.
39.35: .
2
2211
2
22
ψCEψBEUψ
dx
ψd
m
If
ψ
were a solution with energy E, then
212211
CEψBEψψCEψBE or
.)()(
2211
ψEECψEEB
This would mean that
1
ψ is a constant multiple of
212
andand, ψψψ would be wave functions
with the same energy. However,
21
EE , so this is not possible, and
ψ
cannot be a solution to
Eq. (39.18).
39.36: a)
eV)J10eV)(1.60kg)(40102(9.11
s)J10(6.63
2
λ
1931
34
mK
h
m.101.94
10
b) s.106.67
eV)J1062(40eV)(1.
kg)101(2.5m)(9.1
2
7
19
2131
mE
R
v
R
c) The width ,and'
λ
2is mtptvw
a
Rww
yy
where t is the time found in part (b)
and
a is the slit width. Combining the expression for s.mkg102.65
λ2
,
28
at
Rm
pw
y
d) ,m0.40
2
μ
pπ
h
y
y
which is the same order of magnitude.
39.37: a)
12eVλ hcE
b) Find E for an electron with
m.100.10λ
6
smkg106.626λsoλ
27
hpph
eV101.5)2(
42
mpE
V101.5so
4
VVqE
sm107.3kg)10(9.109s)mkg106.626(
33127
mpv
c) Same
λ
so same p.
.V108.2andeV108.2sokg101.673nowbut)2(
88272
VEmmpE
sm4.0kg)10(1.673s)mkg10(6.626
2727
mpv
39.38: (a) Single slit diffraction: λsin mθa
m105.13m)sin2010(150sinλ
89
θa
λλ mhvmvh
sm101.42
m)10kg)(5.1310(9.11
sJ106.626
4
831
34
v
(b) 2λsin
2
θa
0.684
m10150
m105.13
2
λ
2sin
9
8
2
a
θ
43.2
2
θ
39.39: a) The first dark band is located by aθ λsin
355nm
sin25.0
150nm
sin
λ
θ
a
b) Find
λ
for the electrons
J101.324
λ
18
photon
hc
E
smkg101.5532so)(
24
mEpm2pE
2
m104.266λ
10
ph
No electrons at locations of minima in the diffraction pattern. The angular position of these
minima are given by:
,32,1,),0.00120()m10355()m104.266(λsin
910
mmmamθ
;0.207,3;0.138,2;0.0689,1
mmm
39.40: According to Eq. 35.4
nm.600
2
00)m)sin(0.0310(40.0sin
λ
6
m
θd
The velocity of an electron with this wavelength is given by Eq. 39.1
.sm101.21
)m10600)(kg109.11(
)sJ106.63(
λ
3
931
34
m
h
m
p
v
Since this velocity is much smaller than c we can calculate the energy of the electron classically
.eV4.19J106.70)sm10kg)(1.2110(9.11
2
1
2
1
2523312
μmvK
39.41: The de Broglie wavelength of the blood cell is
.m101.66
)sm10kg)(4.0010(1.00
s)J10(6.63
λ
17
314
34
mv
h
We need not be concerned about wave behavior.
39.42: a)
mv
c
v
h
p
h
21
2
2
1
λ
2
22
2
2
2
2222
1λ
c
vh
h
c
v
hvm
2
2
2
2222
λ h
c
v
hvm
1
λ
λ
2
222
2
2
2
22
2
2
h
cm
c
c
h
m
h
v
.
λ
1
21
2
h
mc
c
v
b) .)1(
λ
2
1
1
)(
λ
1
2
21
2
c
h
mc
c
mch
c
v
.
2
λ
2
222
h
cm
c) .1000.1λ
15
mc
h
m
So
8
234
21528231
1050.8
s)J102(6.63
m)10(1.00s)m10(3.00kg)10(9.11
.)108.501()Δ1(
8
ccv
39.43: a) Recall .
22
λ
Vmq
h
mE
h
p
h
So for an electron:
.m101.10λ
V)C)(12510kg)(1.60102(9.11
sJ106.63
λ
10
1931
34
b) For an alpha particle:
.m109.10
V)C)(12510kg)2(1.60102(6.64
sJ106.63
λ
13
1927
34
[...]... 2.82 103 MeV h (6.63 10 34 J s) λ 15mc 15 (1.67 10 27 kg)(3.00 108 m s) 3.42 10 16 m (6.626 10 34 J s) h 39. 46: p ~ 2.0 10 24 kg m s , which is comparable to 10 2πa0 2π (0.5292 10 m) mv1 2.0 1024 kg m s 39. 47: x 0.40λ 0.40 39. 48: a) h h h p But xp x p x (min) 0.40 p 2π p 2πx 2π (0.4) (6.626 10 34 J s) 2.1 10 20 kg m s 2π (5.0... hc 39. 54: a) E 2.58 eV 4.13 10 19 J, with a wavelength of λ 4.82 10 7 m E 482 nm 39. 53: a) λ b) E (6.63 10 34 J s) h 6.43 10 28 J 4.02 10 9 eV 7 2πt 2π (1.64 10 s) c) λE hc, so (λ ) E λE 0, and E E λ λ , so λ λ E E 6.43 10 28 J 16 7 (4.82 10 m) 4.13 10 19 J 7.50 10 m 7.50 10 nm 22 6.63 10 J s h 39. 55:... ), or x 2 The minimum energy is mk 39. 58: a) Using the given approximation, E then h k m b) They are the same dU For x 0, x x F A For x 0, x dx Ax x F A So F ( x) for x 0 x p2 h2 b) From Problem 39. 58, E K U A x , and px h E A x 2m 2mx 2 h2 dE dE For x 0; E Ax The minimum energy occurs when 0 0 2 2mx dx dx 39. 59: a) U A x but F 13 13 h2... magnitude larger than the potential energy of an electron in a hydrogen atom 6.63 10 34 J s h 39. 49: a) p (min) 2.1 10 20 kg m s 2πx 2π (5.0 10 15 m) b) E ( pc) 2 (mc 2 ) 2 [(2.1 10 20 kg m s)(3.0 10 8 m s)]2 [(9.11 10 31 kg)(3.0 10 8 m s) 2 ]2 6.3 10 12 J 39. 5MeV K E mc 2 38.8 MeV c) The coulomb potential energy is U q1q2 (1.60 10 19 C) 2 U... uncertainty is much larger than the real uncertainty as compared to the 4.50 eV λ 39. 56: sin θ sin θ , and λ (h p) (h 2mE ) , and so λ h θ arcsin sin θ λ 2mE 7 (6.63 10 34 J s) sin 35.8 arcsin (3.00 10 11 m) 2(9.11 10 31 kg)(4.50 10 3 )(1.60 10 19 J eV) 20.9 39. 57: a) The maxima occur when 2d sin θ mλ as described in Section 38.7 b)... in the nucleus 39. 50: a) Take the direction of the electron beam to be the x-direction and the direction of motion perpendicular to the beam to be the y-direction Then, the uncertainty r in the position of the point where the electrons strike the screen is p y x r v y t m vx h 2πmy x 2K m 9.56 10 10 m, which is (b) far too small to affect the clarity of the picture 39. 51: E (6.63... 2 A2 2 m 39. 60: For this wave function, ψ1e iω1t ψ 2 e iω2t , so 2 (ψ1 e iω1t ψ 2 e iω2t )(ψ1e iω1t ψ 2 e iω2t ) * ψ1ψ1 ψ 2ψ 2 ψ1 ψ 2 e i ( ω1 ω2 )t ψ 2ψ1e i ( ω2 ω1 ) t 2 The frequencies ω1 and ω2 are given as not being the same, so is not time-independent, and is not the wave function for a stationary state 39. 61: The time-dependent... dx Since ψ is a solution of the time-independent solution with energy E , the term in parenthesis is Eψ , and so ω E , and ω ( E ) 39. 62: a) 2π E E h 2π 2π p p k λ h 2 p (k)2 ω E KE 2m 2m 2 k ω 2m ω 2π f b) From Problem 39. 61 the time-dependent Schrödinger’s equation is 2 2ψ ( x, t ) 2m x 2 ψ ( x, t ) U ( x) 0 for a free particle, so t 2ψ ( x,... t m 0.25 kg 39. 64: a) ψ 2 A2 x 2 e 2 ( αx ue 2u f ( y, z ), and 2 βy 2 γz 2 ) 2 To save some algebra, let u x 2 , so that ψ 1 2 2 , ψ (1 2u ) ψ ; the maximum occurs at u 0 2 u 1 2 b) ψ vanishes at x 0, so the probability of finding the particle in the x 0 plane is zero The wave function vanishes for x x0 2 2 2 39. 65: a) ψ ( x, y, z ) Ae ... The uncertainty principle states that w p wx k 0 is, w p w x h, which is greater than h For us, no matter what 2π h 2π 39. 68: a) For a standing wave, nλ 2 L, and p 2 (h λ ) 2 n 2 h 2 2m 2m 8mL2 b) With L a0 0.5292 10 10 m, E1 2.15 10 17 J 134 eV En 39. 69: Time of flight of the marble, from free-fall kinematic equation is just t 2(25.0 m) 2.26 s 9.81 m s 2 ht px . b)
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