The spacing between energy levels is so small that the energy appears continuous and the balls particle-like as opposed to wave-like... The uncertainty is seen to increase with n.. c The
Trang 140.1: a) 2 22 1 22 34 22
m)kg)(1.520
.0(8
s)J1063.6(8
h n
E n
J
1022
J)102.1(22
mv
E
sm101.1
m5
67 67
2
2 2
2 1
h E
E
d) No The spacing between energy levels is so small that the energy appears
continuous and the balls particle-like (as opposed to wave-like)
kg)(5.010
8(1.673
s)J10626.6
19 6
8
38
3)14(
2 2
2 1 2
E E m h L mL
h mL
h E
108(9.11
3s)
J1063.6
m/s)10(3.00s)J1063.6(λ
18 9
11.9(8
)12(s)J1063.6(8
)(
)(
2 2 2 34 2
2 2 2
2 2
m n h L m
1043
m) 10 kg)(3.33 10
2 34 2
photon energy), which does not correspond to the 13.6eV ground state energy of the hydrogen atom Note that the energy levels for a particle in a box are proportional to n ,2whereas the energy levels for the hydrogen atom are proportional to 1 2
n
Trang 240.5: (5.00 10 3 kg)(0.010m s)2 2.5 10 7 J
2 1 2 2
2 1
8
so
h L
x ψ
2,1,0,2
.4)12(2
)12
(
m x
π m L
L L
xc) These answers are consistent with the zeros and maxima of Fig 40.5
40.8: a) The third excited state is n4,so
2
2 2
8)14
2 9 31
2 34
m/s)10s)(3.0J
1063.6(
8 34
Trang 3h
8/2
λ8
10 10
2 2 1
2
2 1
h mL
h
E
m106.0
s)J1063.6(λ
24 10
34 1
2 2
3 2
2 3
ψ
d and for ψ to be a solution of Eq.(40.3), 8 2 22
2 2
m E h
m π E
2 2
2
Eψ dx
ψ d m π
kx Ak
kx Ak dx
d kx A dx
d ψ
dx
d
cos)
sin(
)cos
2
2 2
8
coscos
8
2
2 2
2 2 2
2 2
mE h
mE π k
m π
h k E
kx EA kx m π
h Ak
U dx
ψ d
22
ψ E ECψ CEψ
Uψ dx
ψ d m C
Trang 440.13: a) Uψ Eψ
dx
ψ d
2
2 22:Eq.(40.1)
2 2
ψ U m
Eq for this case
40.14: According to Eq.40.17, the wavelength of the electron inside of the square well is
given by
.)3(2λ
2
0 in
U m
h mE
By an analysis similar to that used to derive Eq.40.17, we can show that outside the box
.)2(2)(
2
λ
0 0
out
U m
h U
E m
Thus, the ratio of the wavelengths is
.2
3)2(2
)3(2λ
λ
0
0 in
U m
U m
2625.0625
1 2
2 2 1
m1043.3J)
10kg)(3.2010
109.9
(
2
625
19 31
h
hc E
U
hc
)375.5(
8)8)(
375.5(
1 0
m/s)1000.3(m)10kg)(1.5010
11.9(
34
8 2
9 31
Trang 540.17: Since U0 E6 , we can use the results from Section 40.3, E10.625E,E3 5.09and
2 15 27
2 34 2
2
2 2
m)10kg)(4.010
67.1(2
s.)J10054.1(
109.15
m/s)10s)(3.00J
1063.6(
12
8 34
1 3
40.18: Since U0 E6 , we can use the results from Section 40.3
m1099.4)
sm10kg)(3.0010
8(9.11
m)10s)(4.55J
1063.6)(
805.1(8
λ)805.1
(
So
.8
805.1)625.043.2(λ
.625.0and43
2
.8
10 8
31
7 34
2
2 1
2
1 2
mL
h E
hc E
E
E
E E
2sin:
.Eq.(40.15))
(2
2cos2
2sin2
2
2 2
x mE mE
B x mE mE
A dx
Ce dx
(2[or,2
2 / 1 0 0
Trang 640.21: e L.
U
E U
E
0 0
01
16
eV11.0
eV0.60
J) 10 kg)(8.0 10 11 9 ( 2
34 19 31
eV11.0
ev0.61eV11.0
eV0.6
e U
E U
E
0 0
01
16
with E 5.0eV,L0.6010 9 m,and
kg10
1
λ K K and K1are for xLwhereK12U0 and λ2 and K2 are for
0 0 2
where
0 xL K EU U
2
12
λ
λ
0
0 1
K
40.24: Using Eq 40.21
56.2eV15.0
eV0.121eV15.0
eV0.12161
16
0 0
E G
nm
26.00.025
2.56ln)m109.8(2
1)
ln(
21
J/eV)10
eV)(1.6012.0
kg)(15.010
11.9(2)(
2
1 8 2
1
9
34
19 31
Ge
T
π
E U
Trang 740.25: a) Probability of tunneling is T Ge 2L
41
32141
32161
16
0 0
E G
and
sJ10054.1
J/eV)
10eV)(1.6032
kg)(41eV10
11.9(2)(
2
34
9 31
.102.174
.274
.2So
.m1054
1
3 7
7 m)
10 5 2 )(
m 10 54 1 ( 2
1 10
10 1
T e
1011.9
1067.1
zero toequalpurposespractical
allfor andis
The.10
74.274
.2
m1059.6
143
330 m)
10 5 2 )(
m 10 59 6 ( 2
1 11
10 1
11
small T
e e
2
E U m U
E U
E G
16Giving
) ( 2 2 0 0
0 L e
U
E U
E T
E U m
eV106.91J
1011.1
kg0.250
N/m1102
s)J10055.1(2
12
1
14
33 0
1
15 33
0
34 0
E
m
k ω
δ x dx
ψ d ψ x dx
dψ δ
k
ω.
/m k δ
m
2
12
1if
Trang 840.29: The photon’s energy is
E E
30
1
2 4
26 2
8 2
2
2 2
m)10(5.25
kg)104.9()sm1000.3(4λ
4λ
.mN2
40.30: According to Eq.40.26, the energy released during the transition between two
adjacent levels is twice the ground state energy
eV
2.11
2 02
10eV)(1.60(11.2
)sm10s)(3.00J
1063.6(
8 34
c hf
E
For n1,E ω k m 1.54410 21 J
m129)
(λso
k m ψ
A
ψ
This is more or less what is shown in Fig (40.19)
)0
(
)2
k m ψ
A ψ
This figure cannot be read this precisely, but the qualitative decrease in amplitude with distance is clear
Trang 940.33: For an excited level of the harmonic oscillator
22
12
1
A k ω n
k
ω n
1
mv ω
12
(
)12()
12()(
)12(
max max
m ω n
ωm n
k
ω n
mv A p x
m
ω n
v
,)12(
Soxp n which agrees for the ground state (n0)withxp
The uncertainty is seen to increase with n.
4 / 0
cos12
12sin
dx L
πx L
dx L
πx L
,2
141
2sin2
x L
14
12
sin2
L x L
about 0.0409 c) The particle is much likely to be nearer the middle of the box than the edge d) The results sum to exactly 1 , which means that the particle is as likely to be 2between x0andL 2as it is to be between xL 2andxL e) These results are represented in Fig (40.5b)
Trang 102 2
1 4 / 3 4 / | | L 2sin
dx ψ P
.818.0
121
2sin2
11sin
2
Let
4 3 4
4 / 3 4 / 2
π z d z P
L
π dx dz L
πx z
2 4
/ 3 4 /
2 2
2sin
dx ψ P
2
1
2sin2
12
1sin
2
2 / 3 2 / 2
P
c) This is consistent with Fig 40.5(b) since more of ψ is between1 than
4
3and
L L
xand the proportions appear correct
and2
)2(sin)2(b),4
sin)2()
43
x
dx L
πx L
π L
dx ψ
L
2sin
2
|:|
4
2 2
L dx ψ
π L
dx ψ
L
2
3sin
2
|:|
n n n
n n
n n
This is never larger than it is forn1,andR3. b) R approaches zero; in the classical
limit, there is no quantization, and the spacing of successive levels is vanishingly small compared to the energy levels
Trang 1140.39: a) The transition energy (4 1)
8 2
2 1
m
1092.1λ
s)J1063.6(3
m)1018.4(sm10(3.00kg)1011.9(83
8λ
5
34
2 9 8
31 2
hc
s)J1063.6)(
23(
m)1018.4(sm10(3.00kg)1011
2
2 9 8
π dx
dψ
)cos(
2
dx
dψ L
0for2
2
11
2cos
2
3
2 1
2 3
2 1
π dx
dψ
L
πx L
π L
πx L
π L dx
b)
L
πx L
π L dx
dψ L
πx L
.0for
83
x
L x L
π dx
2 1
d) 2 8 32 cos 2 8 32for ,sincecos2 1
πx L
Trang 12nπ k
At x0 the initial momentum at the wall is p iˆ
2initial
and the final momentum,
2
ˆ2So
.ˆ2final
L
hn L
hn L
hn L
.ˆ2
p
L
hn L
hn L
hn L
Trang 1340.43: a) For a free particle, U(x)0 so Schrodinger'sequationbecomes 2 (2 )
dx
x ψ d
(
.)
(
k 2 2
k
x
x
e k dx
x ψ
d
ke dx
x dψ
ke dx
x
kx
e k dx
x ψ
m
k
d) Note
Trang 1440.44: b) ( ) ( ) ( ) ( )
2 2
x Eψ x ψ x U dx
x ψ d
)())((2)(
2 2
2
x ψ x U E
m dx
x ψ
2where
)()
(
2 2
2
x U E h
m A
x Aψ dx
x ψ d
has the opposite sign of ψ (x)
If EU (x), we are outside the well and A0 From the above equation, this implies
has the same sign as ψ (x)
40.45: a) We set the solutions for inside and outside the well equal to each other at the
well boundaries, x0and L
,)
0sin(
:
x since we must have D0for x0
L x for C
De L
mE B
L mE A
where
cossin
mE k
De kL B
kA k
kB k
kA dx
dψ
.sin
(2and
116
0 0
E G
Ge
.0010.0and,kg1011.9,eV0.10,
eV5.5If
m1009.1s)
J10054.1(
eV)J1060.1)(
eV5.4)(
kg1011.9(
eV5.51eV0.10
eV5.516
0010.0ln)m1009
Trang 1540.47: a)
1 0
2 0
)(
4
)hsin(1
L κ U
0
2 2 0
)(
16)
(16
12
1sinh
e U E
U E
e U T
e κL κL
κL κL
κL
.116where
0 0
E G
Ge
b) κL1 implies either κ is large or L is large (or both are large) If L
is large, the barrier is wide If is large, U0 E is big, which implies E is small
compared to U0
c) AsEU0,κ0sinhκLκL
1 2
2 2 0 1
0
2 2 2 0
4
21)
(41
U E
L κ U T
since 2 2 ( 02 )
E U m
since2
12
21But
1 2 1
2 2
2 0 0
U
ψ d k
.)
(2or
,)(
2gives
π
oscillator energy is given by
J1062.6)srad6.12)(
sJ10054.1(2
12
J1062
12
315
33 0
Trang 1640.50: a) ( (12)) ( (12)) ,andsolvingfor ,
2
1 2
n hf
n ω n
mv
.103.12
1Hz)50.1)(
sJ1063.6(
s)m360.0(kg)020.0)(
21(2
12
1
30 34
2 2
b) The difference between energies is
.J1095.9)Hz50.1)(
sJ1063
detected with current technology
dx
ψ d
2
2 2
2
12
:)21.40(
Cxe dx
.2
3but
2
12
2
3:)25-42(
Eq
3)(
4
42
42
2)(
22
2 2
2 2
2
2 2
2
3 2 2 2
2 2
2
2 2
2 2
ω E m
k ω ψE x
k x mω ω
ψ
x mω mω
Cxe
x ω m x m xmω Ce
e x Cmω Ce
dx
d
x mω
x mω
x mω x
Trang 1740.52: With u(x) x2,amω ,|ψ|2 isof theformC*Cu(x)e -au(x),whichhasamaximum
decreasingrapidly
aand0atminimuma
showsalsofigureThe)
19
thisand),26.40(.Eqin1fromfound,3is
(a)partofresultthe
0
asand,0at vanishhence
andb).,
1or
,
1
at
2 2
2 2
n A
ψ
x x
ψ ψ
m x
x
mω
Trang 1840.53: a) Eq (42-32): , and areall
2 2
2 2
2 2
z y
ψ Eψ Uψ z
ψ y
ψ x
,then
)()()(if
Now.andfor
similarly
and,2
12
so,equationr
SchrodingeD
1the
to
solutions
2
2 2
2 2
2
2
2 2
2 2
2 2
2
y x x z
x
y
z y x z
y x z
y
x x x x
n n
n n
n
n
n n
n n
n n n
n
n n n n
ψ ψ dz
ψ d dz
ψ ψ
ψ d x
ψ z
ψ y ψ x ψ ψ ψ
ψ
ψ E ψ x k dx
ψ d m
ψ y
ψ x
ψ
12
2 2 2 2
2 2
2 2
2
12
12
1
][
][
2
12
2
12
2
12
2 2
2 2
2 2
2 2
2 2
2 2
ω n
n n E
ω n
n n
E
ψ E ψ E E E
ψ ψ ψ E E E
ψ ψ ψ z k dz
ψ d m
ψ ψ ψ y k dy
ψ d m
ψ ψ ψ x k dx
ψ d m
z y x n n n
z y
x n
n n
n n n
n n
n n n n n n
n n n n
n n n n
n n n n
z y
z y
z x z
y x
z y x z y x
y x z z
z x y y
z y x x
c) As seen in b) there is just one set of quantum numbers (0, 0, 0) for the ground state and three possibilities (1, 0, 0), (0, 1, 0) and (0, 0, 1) for the first excited state
Trang 1940.54:Let1 k1 m,ω2 k2 m,ψ n x(x)beasolutionof Eq.(40.21)with E n x
)
()()(),,(formtheofsolutiona
forlook,53
40
Problem
inAs(a)2
1energy
and,ofinsteadvariable
tindependen
the
as
with
but)25-42.(
Eqofsolutiona
be)(andsolution,similar
abe)(,
2
1
2 1
z ψ y ψ x ψ z y x ψ
ω n
E x
z
z ψ y
ψ ω
n
z y x z
z x
n n n
z n
n n
2 2
2
12
with similar relations for and 2 Adding,
2 2
2
z
ψ y
E E z
ψ y
ψ x
2 1 2
2 2
2 2
2
2
2
12
12
12
ψ U E
ψ U E E E
z y
n
)(
)(
1)
state
excitedfirst
theandstateground
the
both
for
numbersquantum
ofsetoneonlyisThere)
)23((and
,as
,
1
and0to
scorrespondlevel
excitedfirst
The)
2(
and
,
0
toscorrespondlevel
groundThe
b)integers
enonnegativall
and
,
2 2
2 1
2 1
2
2
2 2
2 1
ω ω
E ω
ω
n
n n ω
ω E
n
n n n
n
n
z
y x z
y x z
Trang 2040.55: a) ψ(x) Asinkxandψ(L 2)0ψ(L 2)
where,8
)2(2
2λ
λ
λ
222
2sin0
2
2 2 2
2 2 2
mL
h n mL
h n m
p E L
nh n
h p n L
π L
nπ k nπ kL kL
A
n n
)12(
2
)12()
12(
2λ
λ
2)12(2
)12(22
cos0
2
2 2
h n E
L
h n p
n L
π L
π n k
π n kL kL
mL
h n
E n d) Part (a)’s wave functions are odd, and part (b)’s are even
40.56: a) As with the particle in a box, ψ(x) Asinkx,whereAisa constantand
Eq (14.17) and Eq (40.18) c) At xL,AsinkLDeκL andkAcoskLκDeκL.Dividing the second of these by the first gives
,cotkL κ
a transcendental equation that must be solved numerically for different values of the length L and the ratio E U0
Trang 2140.57: a) ( ) 2 ( ( )).
2)
m
p x U K
.))((2)(λλ
x U E m
h x
recall k0),EU(x)getssmaller,soλ(x)getslarger
b a
n dx x U E m h
x U E m h
dx x
dx
2))
((2
1))((2)
2))
((2 e) U(x)0for0xLwithclassicalturningpointsat x0and xL.So,
.82
2
12
2
),(partfromSo,.22
2))
((
2
2
2 2 2
mL
n h L
hn m E
hn
L
mE
d L
mE dx
mE dx
mE dx
x U
f) Since U(x)0 in the region between the turning points atx0and xL,theresults
is the same as part (e) The height U never enters the calculation WKB is best used 0
with smoothly varying potentials U (x)
Trang 2240.58: a) At the turning points 2
2
1
TP
2 TP
k
E x
x k E
K
E
nh dx x k E m
2
22
22
k m
mE k
m x
k m mE x
k E
k
E a
k
E A
)1(arcsin2
2
2arcsin2
222
arcsin2
2
0
2 2 2 2
m E k
E k m
k E
k E k
E k
E k
E k
E k
m
A
x A
x A x k m dx
x A k m
b b
a
2so,Recall
.2so
,
h E m
k ω
hn π k
m E
2
1recall
2ω E ω n
However, our approximation isn’t bad at all!
Trang 2340.59: a) At the turning points TP TP
A
E x
x A
x A E m
0 2 ( ) Let 2 ( )2
(2
So.2,0whenand,0,when
A
E x dx
dx Ax E m
0
0 2 21
)(
22
2toequalisthisWKB,Using
.)2(3
23
2
mE mA
12
)2(3
n
mAh m
E
hn mE
,
49.023,59.012,10
1 3 3 3 3 2 2
• A sharp step gave ever-increasing level differences (~n2)
• A parabola (~x2)gaveevenlyspacedlevels(~n)
• Now, a linear potential (~x)giveseverdecreasingleveldifferences(~n 3).Roughly: If the curvature of the potential (~ second derivative) is bigger than that of a parabola, then the level differences will increase If the curvature is less than a parabola, the differences will decrease