Note that the kinetic energy found this way is much smaller than the rest energy, so the nonrelativistic approximation is appropriate.. Again, the nonrelativistic approximation is approp
Trang 139.1: a) 2.37 10 kg m s.
) m 10 80 2 (
) s J 10 63 6 ( λ
10
34
h p p h
) kg 10 11 9 ( 2
) s m kg 10 37 2 ( 2
18 31
2 24
2
m
p
K
39.2:
mE
h p
h
2
λ
V) e J 10 60 1 ( ) eV 10 20 4 ( ) kg 10 64 6 ( 2
) s J 10 63 6
19 6
27
34
) s m 10 70 4 ( kg) 10 11 9 (
s J 10 63 6
6 31
34
v m
h
e e
kg 10 67 1
kg 10 11 9 λ
27
31
p
e p
m
m
m) 10 20 0 (
) s m 10 00 3 ( s) eV 10 136 4 (
8 15
E
b)
kg) 10 11 9 ( 2
)) m 10 20 0 ( ) s J 10 626 6 ((
2
) λ (
2 9 34
2 2
m
h m
p
K
6.010 18 J37eV
Note that the kinetic energy found this way is much smaller than the rest energy, so the
nonrelativistic approximation is appropriate
J 10 3 8 kg)
10 64 6 ( 2
)) m 10 20 0 ( s) J 10 626 6 ((
2
) λ ( 2
22 27
2 9 34
2 2
m
h m
p K
5.2 meV Again, the nonrelativistic approximation is appropriate
39.5: a) In the Bohr model
2π
nh r
mv n The de Broglie wavelength is
m 10 32 3 ) m 10 29 5 ( 2 λ m 10 29 5 :
1 for
2
1
11 0
n
r π mv
h
p
This equals the orbit circumference
4
) 16 ( 2 λ 16
) 4 ( :
4 0 0
2
n
λ 1.33 10 9 m 4
The de Broglie wavelength is a quarter of the circumference of the orbit, 2πr4
39.6: a) For a nonrelativistic particle, ,so
2
2
m
p
2
λ
Km
h p
h
b) (6.6310 34 Js) 2(800eV)(1.6010 19 J/eV)(9.1110 31Kg)4.3410 11m
Trang 239.7: 3.90 10 m.
) s m 340 ( ) kg 005 0 (
s J 10 63 6
mv
h p
h
We should not expect the bullet to exhibit wavelike properties
39.8: Combining Equations 37.38 and 37.39 gives
pmc γ2 1 a) λ (h mc) 2 14.4310 12 m
p
h
(The incorrect nonrelativistic calculation gives 5.0510 12 m.)
b) (h mc) 2 17.0710 13 m
39.9: a) photon
V) e J 10 602 1 ( ) eV 0 20 (
) s m 10 998 2 ( ) s J 10 626 6 ( λ
so
8 34
E
hc hc
E
electron
p2 (2m) so p 2mE 2(9.109 10 31 kg)(20.0eV)(1.602 10 19 J eV)
E
s m kg
10
416
2 24
nm 274 0
λh p
b) photon Ehc λ7.94610 19 J4.96eV
electron λh p so ph λ2.65010 27 kgm s
eV 10 41 2 J 10 856 3 ) 2
E
c) You should use a probe of wavelength approximately 250 nm An electron with
250
λ nm has much less energy than a photon with λ250 nm, so is less likely to damage the molecule
39.10:
λ
λ
m
h v mv
h
2 2 2
λ 2 λ 2
1 2
1
m
h m
h m mv
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
10 46 5 kg 10 67 1
kg 10 11 9
λ 2
λ 2
m m m
h m h K
K
Trang 339.11: a) λ0.10nm
keV 12 λ )
c
eV 150 )
b
s m 10 3 7 ) λ ( so
λ 2 2
1
6
hc
E
mv
E
m h v h
mv
p
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) λh mvvh mλ
Energy conservation: 2
2
1mv V
V 9 66
) m 10 15 0 ( ) kg 10 11 9 ( ) C 10 60 1 ( 2
) s J 10 626 6 (
λ 2 2
) ( 2
2 9 31
19
2 34 2
2 2
λ 2
em
h e
m e
mv
m 10 15 0
) s m 10 0 3 ( ) s J 10 626 6 ( λ
15 9
8 34
photon
hf hc E
V 8310 C
10 6 1
J 10 33 1
19
15 photon
photon
e
E V
E K V e
39.13: For m =1,
eV
432 0 J 10 91 6
) 6 28 ( sin ) m 10 10 9 ( ) kg 10 675 1 ( 2
) s J 10 63 6 ( sin
2
2 sin
λ
20
2 2 11 27
2 34 2
2 2
E
θ md
h E
mE
h θ d
39.14: Intensity maxima occur when dsinθmλ
.
ME
mh θ
d ME
h p
h
2 sin
so 2
(Careful! Here, m is the order of the maxima, whereas M is the mass of the incoming
particle.)
a)
) 6 60 sin(
) V e J 10 60 1 ( ) eV 188 ( ) kg 10 11 9 ( 2
s) J 10 63 6 ( ) 2 ( sin
34
θ ME
mh d
2.0610 10 m0.206nm
b) m = 1 also gives a maximum.
one
other only the is This 8
25
m) 10 06 2 ( ) V e J 10 60 1 ( ) eV 188 ( ) kg 10 11 9 ( 2
) s J 10 63 6 ( ) 1 ( arcsin
10 19
31
34
θ
If we let m3, then there are no more maxima
Trang 4c)
) 6 60 ( sin ) m 10 60 2 ( kg) 10 11 9 ( 2
) s J 10 63 6 ( ) 1 ( sin
2 34 2
2 2
2 2
θ Md
h m E
7.4910 18 J46.8eV
Using this energy, if we let m2, thensinθ 1.Thus, thereisnom2 maximum in this case
39.15: Surface scattering implies d sinθmλ
341 0 m 10 34 8
m 10 96 4 arcsin So
m
10 96 4 λ
J/eV) 10
60 1 ( ) eV 840 ( ) kg 10 64 6 ( 2
s J 10 63 6 2
λ
But
] λ arcsin[
: 1 If
11 13 13
19 27
34
θ
mE
h p
h
d θ
m
39.16: The condition for a maximum is sin λ λ ,so arcsin
dMv
mh θ
Mv
h p
h m
θ d
(Careful! Here, m is the order of the maximum, whereas M is the incoming particle mass.)
dMv
h θ
m 1 1 arcsin
) s m 10 26 1 ( ) kg 10 11 9 ( ) m 10 60 1 (
) s J 10 63 6 ( ) 2 ( arcsin
2
07 2 ) s m 10 26 1 ( kg) 10 11 9 ( ) m 10 60 1 (
s J 10 63 6 arcsin
4 31
6
34 2
4 31
6
34
θ m
14
4
b) For small angles (in radians!) yDθ,so
cm
81 1 cm 81 1 cm 61 3
cm 61 3 180
radians )
14 4 ( ) cm 0 50
(
cm
81 1 180
radians )
07 2 ( ) cm 0 50
(
1 2 2
1
y y
π y
π y
39.17: a)
m) 10 00 1 ( ) kg 1200 ( 2
) s J 10 63 6 ( 2
2
34
π x πm
h v
π
h x v m π
h x
s m 10 79
8 32
b) Knowing the position of a macroscopic object (like a car) to within 1.00 μm is, for all practical purposes, indistinguishable from knowing “exactly” where the object is Even with this tiny position uncertainty of 1.00 μm, the velocity uncertainty is insanely small by our standards
Trang 539.18: a)
2 2
y uncertaint minimum
for
h v
π
h y v m π
h y
) m 10 0 2 ( kg) 10 67 1 ( 2
) s J 10 63 6
12 27
34
π
b) For minimum uncertainty,
m
10 63 4 ) s m 50 2 0 ( kg) 10 11 9 ( 2
s) J 10 63 6 ( 2
2
4 31
34
π v πm
h p
π
h z
z z
39.19: Heisenberg’s Uncertainty Principles tells us that:
2π
h
p
x x
We can treat the standard deviation as a direct measure of uncertainty
is
claim the so 2 Therefore
s J 10 05 1 2 but s J 10 6 3 ) s m kg 10 0 3 ( m) 10 2 1 (
valid not π
h p x
π
h p
x
x
x
39.20: a) (x)(mv x)h 2π, and setting v x (0.010)v xand the product of the uncertainties equal to h 2/ π (for the minimum uncertainty) gives v x h (2πm(0.010)x)57.9m s
b) Repeating with the proton mass gives 31.6 mm s
2
m 10 215 0 2
) s J 10 63 6 ( 2
2
25 9
34
π x π
h p π
h x
p
m
18 5 ) s m 81 9 ( kg) 00 1 (
J 8 50 )
d
J
50.8 J) 10 95 4 ( Ni kg 10 75 9
kg 00 1 )
c
eV
10 09 3 J 10 95 4 ) kg 10 75 9 ( 2
) m 10 82 9 ( 2
)
b
2
total total
24 26
total
5 24
26
2 25 2
mg
K h K
mgh
NK K
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
2
2 10 2
9
mc E
π
h t
s
10 20 3 ) J 10 30 3 ( 2
s J 10 63 6 2
25 10
34 2
π mc π h t
Trang 639.23:
69 8 J 10 39 1 s) 10 6 7 ( 2
s) J 10 63 6 ( 2
2
14 21
34
π t π
h E π
h t
E
MeV
0869
0
eV
104
MeV 3097
MeV 0869
2
2
c
c E
E
s) 10 2 5 ( 2
s) J 10 63 6 ( 2
13 32
3
34
π t π
h E
39.25: To find a particle’s lifetime we need to know the uncertainty in its energy.
s 10 08 1 ) J 10 81 9 ( 2
s J 10 63 6 2
J 10 81 9
s) m 10 00 3 ( ) kg 10 67 1 ( ) 5 4 ( ) 145 0 ( ) (
24 11
34 11
2 8 27
2
π E π
h t
c m E
2
) λ ( so , 2
) λ ( 2
2 2
2
me
h V m
h m
p K eV
b) The voltage is reduced by the ratio of the particle masses,
V
229 0 kg 10 67 1
kg 10 11 9 V) 419
31
2
λ
mE
h p
h
But the energy of an electron accelerated through a potential is just EeV
m
10 34
4
λ
) V 800 ( C) 10 60 1 ( ) kg 10 11 9 ( 2
) s J 10 63 6 ( 2
λ
11
19 31
34
e
e
V m h
b) For a proton, all that changes is the mass, so
m 10 01 1 λ
m 10 34 4 kg 10 67 1
kg 10 11 9 λ λ
12
11 27
31
e
e p
e p
m m
39.28: ψ sin ωt, so
2 * ψ*ψsin2ωt ψ2sin2ωt.
2 is not time-independent, so is not the wavefunction for a stationary state
39.29: a) ψ x Asinkx The probability density is ψ2 A2 sin2kx, and this is greatest when
5 , 3 , 1 , 2 1
sin2kx kx nπ n
, 1,3,5
4
λ ) λ 2 2(
π
nπ k
nπ x
Trang 7b) The probability is zero when ψ2 0, which requires
2 , 1 , 0 , 2
λ 0
sin2 n n
k
nπ x nπ kx kx
39.30: a) The uncertainty in the particle position is proportional to the width of ψ x , and is inversely proportional to This can be seen by either plotting the function for different values of , finding the expectation value x2 ψ2x2dx 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 x on 2 may be found by considering
dx e
dx e x x
x
x
2 2
2
2 2 2
dx
e 2 x2
ln 2
, 4
1 2
1 ln 2
du
e u
where the substitution u xhas been made (b) Since the uncertainty in position decreases, the uncertainty in momentum must increase
iy x
iy x y x f iy
x
iy x y
x
1
* 2
iy x
iy x iy x
iy x f f f
39.32: The same.
) , , ( ) , , ( )
, , (x y z 2 ψ* x y z ψ x y z
ψ(x,y,z)e i 2 (ψ*(x,y,z)ei)(ψ(x,y,z)ei)
ψ*(x,y,z)ψ(x,y,z)
The complex conjugate means convert all i’s to i’s and vice-versa e i ei 1
39.33: Following the hint:
If we Taylor expand sin(ax ) about a point x , we get 0 f(x0) f(x0)(xx0)
) )(
cos(
)
sin(ax0 a ax0 xx0 If x is small we can even ignore the first order term and x0
sin(ax ) sin (ax ).0
For us xx0 0.01L which is small compared to
L
πx L
z y x ψ
2 sin
2 ) , , ( so
sin
L
πz L
πy
Trang 8a)
4 0 0
0
L z y
10 1.00 )
0.01 ( 2
2 2 4
sin
2
L V
π L
dV ψ P
b)
2
1 0 0
0 y z
x
2 sin
2
L L
V
π L
dV ψ P
39.34: Eq (39.18): Uψ Eψ
dx
ψ d
2
2 2 2
Let ψ Aψ1Bψ2
) (
) (
) (
2
2
Bψ Aψ E Bψ Aψ U Bψ Aψ dx
d
0 2
2 2 2 1
1 2
1 2 2
dx
ψ d m B Eψ Uψ dx
ψ d
m
But each of ψ and 1 ψ satisfy Schröedinger’s equation separately so the equation still 2
holds true, for any A or B
2 2
ψ CE ψ BE Uψ dx
ψ d
If ψ were a solution with energy E, then
2 1
2 2 1
) (
) (E1 E ψ1 C E E2 ψ2
This would mean that ψ is a constant multiple of 1 ψ2,andψ1andψ2 would be wave functions with the same energy However, E 1 E 2, so this is not possible, and ψ cannot be a solution to
Eq (39.18)
39.36: a)
eV) J 10 eV)(1.60 kg)(40
10 2(9.11
s) J 10 (6.63 2
λ
19 31
34
mK h
1.9410 10m
eV) J 10 6 2(40eV)(1
kg) 10 1 (2.5m)(9.1 2
7 19
2 31
m E
R v
R
c) The width is 2 λ'andw v t p t m,
a R w
w y y where t is the time found in part (b) and a is the slit width Combining the expression for , 2 λ 2.6510 28kgm s
at
R m p
h y
y
which is the same order of magnitude
Trang 939.37: a) E hc λ12eV
b) Find E for an electron with λ0.1010 6m
λh pso ph λ6.62610 27kgm s
eV 10 1.5 ) 2
E
V 10 1.5
so 4
E
s m 10 7.3 kg) 10 (9.109 s)
m kg 10 6.626
v
c) Same λso same p
V 10 8.2 and
eV 10 8.2 so
kg 10 1.673 now
but ) 2
E
s m 4.0 kg) 10 (1.673 s)
m kg 10
v
39.38: (a) Single slit diffraction: asinθmλ
m 10 5.13 m)sin20
10 (150 sin
λ
λh mvvh m
s m 10 1.42 m) 10 kg)(5.13 10
(9.11
s J 10
8 31
34
v
(b) asinθ2 2λ
m 10 150
m 10 5.13 2
λ 2
8
a θ
θ2 43.2
39.39: a) The first dark band is located by sinθ λ a
sin25.0
150nm sin
θ
a
b) Find λfor the electrons
λ
18 photon
hc
E
E p 2 (2 m)so p 2mE 1.55310 24kgm s
λh p4.26610 10m
No electrons at locations of minima in the diffraction pattern The angular position of these minima are given by:
, 3 2, 1, ),
0.00120 (
) m 10 355 ( ) m 10 4.266 ( λ
; 0.207 ,
3
; 0.138 ,
2
; 0.0689
,
m
39.40: According to Eq 35.4
nm
600 2
00) m)sin(0.03 10
(40.0 sin
λ
6
m
θ d
The velocity of an electron with this wavelength is given by Eq 39.1
s m 10 1.21 )
m 10 600 )(
kg 10 9.11 (
) s J 10 6.63 ( λ
3 9
31
34
m
h m
p v
Since this velocity is much smaller than c we can calculate the energy of the electron classically
eV 4.19 J 10 6.70 )
s m 10 kg)(1.21 10
(9.11 2
1 2
μ mv
Trang 1039.41: The de Broglie wavelength of the blood cell is
m 10 1.66 )
s m 10 kg)(4.00 10
(1.00
s) J 10 (6.63
3 14
34
mv h
We need not be concerned about wave behavior
39.42: a)
mv c
v h p h
2 2
2 1
2
2 2 2 2
2 2 2 2
λ
c
v h h c
v h v
2 2
2 2 2 2 2
c
v h v
1
λ
2 2 2 2
2
2 2 2
2 2
h
c m c c
h m
h v
λ 1
2 2
h mc
c v
2
1 1
) (
λ 1
2 2
h
mc c
mc h
c
2
λ 2
2 2 2
h
c m
c) λ 1.00 10 15
mc
h
2 34
2 15 2
8 2
31
10 50 8 s)
J 10 2(6.63
m) 10 (1.00 s) m 10 (3.00 kg) 10
) 10 8.50 1 ( ) Δ 1
2 2
λ
V mq
h mE
h p
h
m 10 1.10 λ V) C)(125 10
kg)(1.60 10
2(9.11
s J 10 6.63
19 31
34
b) For an alpha particle:
m 10 9.10 V)
C)(125 10
kg)2(1.60 10
2(6.64
s J 10 6.63
19 27
34