6-43. (a) For x > 0, h 2 k22 / 2m + V0 = E = h 2 k12 / 2m = 2V0
1/ 2
1/ 2
h . Because k1 = (4mV0 )
So, k2 = (2mV0 )
2
2
(b) R = (k1 ! k2 )
(
(k1 + k2 )
= 1 ! 1/ 2
2
h, then k2 = k1 / 2
(Equation 6-68)
) (
1 + 1/ 2
2
) = 0.0294,
or 2.94% of the incident particles
are
reflected.
(c) T = 1 ! R = 1 ! 0.0294 = 0.971
(d) 97.1% of the particles, or 0.971 ! 106 = 9.71 ! 105 , continue past the step in
the +x
direction. Classically, 100% would continue on.
6-44. (a) For x > 0, h 2 k22 / 2m ! V0 = E = h 2 k12 / 2m = 2V0
1/ 2
So, k2 = (6mV0 )
2
(b) R = (k1 ! k2 )
1/ 2
h . Because k1 = (4mV0 )
h, then k2 = 3 / 2k1
2
(k1 + k2 )
2
R = (k1 ! k2 )
2
(k1 + k2 )
(
= 1! 3/ 2
2
) (1 +
3/ 2
2
) = 0.0102
Or 1.02% are reflected at x = 0.
(c) T = 1 ! R = 1 ! 0.0102 = 0.99
(d) 99% of the particles, or 0.99 ! 106 = 9.9 ! 105 , continue in the +x direction.
Classically, 100% would continue on.
6-45. (a)
E = 4eV
9eV =V0
! = 2m (V0 " E ) h
(
)
= 2 0.511 ! 106 eV / c 2 (eV ) / h
0.6 nm = a
= 5.11 ! 106 eV
=
and ! a = 0.6nm # 11.46nm "1 = 6.87
eV
/h
c
2260eV
= 11.46nm !1
197.3eV nm
Since ! a is not
1, use Equation 6-75:
The transmitted fraction
"1
#
$
# % 81 &
$
sinh2 ! a
2
T = '1 +
( = '1 + ) * sinh (6.87 )(
- + 80 ,
.
-' 4 (E V0 )(1 " E V0 )(.
(
Recall that sinh x = e x ! e ! x
"1
) 2,
!1
" 81 $ e6.87 ! e !6.87 % 2 #
!6
T = &1 + )
* ' = 4.3 ( 10 is the transmitted
2
&- 80 +
, '.
fraction.
(b) Noting that the size of T is controlled by ! a through the sinh2 ! a and
increasing T
implies increasing E. Trying a few values, selecting E = 4.5eV yields
T = 8.7 " 10!6
or approximately twice the value in part (a).
6-48. Using Equation 6-76,
T % 16
E#
E $ "2! a
where E = 2.0eV , V0 = 6.5eV , and a = 0.5nm.
&1 " ' e
V0 ( V0 )
" 2.0 # " 2.0 # !2(10.87 )(0.5)
T $ 16 &
$ 6.5 % 10!5
' &1 ! 6.5 ' e
6
.
5
(
)(
)
(Equation
6-75
T = 6.6 " 10!5 .)
2
6-49.
(k ! k )
R= 1 2 2
(k1 + k2 )
and T = 1 ! R
(Equations 6-68 and 6-70)
(a) For protons:
k1 = 2mc 2 E / hc = 2 (938MeV )(40 MeV ) / 197.3MeV fm = 1.388
k2 = 2mc 2 (E ! V0 ) / hc = 2 (938MeV )(10 MeV ) / 197.3MeV fm = 0.694
yields
2
2
" 1.388 ! 0.694 # " 0.694 #
R=$
% =$
% = 0.111 And T = 1 ! R = 0.889
& 1.388 + 0.694 ' & 2.082 '
(b) For electrons:
1/ 2
! 0.511 "
k1 = 1.388 #
$
% 938 &
1/ 2
= 0.0324
! 0.511 "
k2 = 0.694 #
$
% 938 &
= 0.0162
2
" 0.0324 ! 0.0162 #
R=$
% = 0.111 And T = 1 ! R = 0.889
& 0.0324 + 0.0162 '
No, the mass of the particle is not a factor. (We might have noticed that
could
be canceled from each term.)
7-1.
En1 n2 n3 =
E311 =
h 2! 2 2
n1 + n22 + n32
2mL2
(
)
(Equation 7-4)
h 2! 2 2 2 2
h 2! 2
3
+
1
+
1
=
11
E
where
E
=
0
0
2mL2
2mL2
(
)
(
)
(
)
E222 = E0 22 + 22 + 22 = 12 E0 and E321 = E0 32 + 22 + 12 = 14 E0
The 1st, 2nd, 3rd, and 5th excited states are degenerate.
E
(×E0)
14
321
12
222
311
10
221
8
6
211
4
111
2
s = 3/ 4
m
7-2.
En1 n2 n3 =
h 2! 2 " n12 n22 n32 # h 2! 2 " 2 n22 n32 #
+ %=
+ %
$ +
$ n1 +
2m & L12 L22 L23 ' 2mL12 &
4
9 '
n1 = n2 = n3 = 1 is the lowest energy level.
E111 = E0 (1 + 1 / 4 + 1 / 9 ) = 1.361E0 where E0 =
The next nine levels are, increasing order,
h 2! 2
2mL12
(Equation 7-5)
(Problem 7-2 continued)
7-3.
n1
n2
n3
E (! E0 )
1
1
2
1.694
1
2
1
2.111
1
1
3
2.250
1
2
2
2.444
1
2
3
3.000
1
1
4
3.028
1
3
1
3.360
1
3
2
3.472
1
2
4
3.778
(a) " n1 n2 n3 (x, y, z ) = A cos
n!z
n1! x
n!y
sin 2 sin 3
L
L
L
(b) They are identical. The location of the coordinate origin does not affect the
energy
level structure.
7-4.
" 111 (x, y, z ) = A sin
!x
!y
!z
sin
sin
L1
2 L1
3L1
" 112 (x, y, z ) = A sin
!x
!y
2! z
sin
sin
L1
2 L1
3L1
!x
!y
!z
sin
sin
L1
L1
3L1
!x
!y
2! z
" 122 (x, y, z ) = A sin sin
sin
L1
L1
3L1
" 121 (x, y, z ) = A sin
" 113 (x, y, z ) = A sin
!x
!y
!z
sin
sin
L1
2 L1
L1
7-5.
En1 n2 n3 =
n32 # h 2! 2 " 2 n22 n32 #
n22
h 2! 2 " n12
$ 2+
%=
+
+ %
$ n1 +
2m $ L1 (2 L1 )2 (4 L1 )2 % 2mL12 &
4 16 '
&
'
(from Equation 7-5)
"
n2 n2 #
h 2! 2
E0 = $ n12 + 2 + 3 % where E0 =
4 16 '
2mL12
&
(Problem 7-5 continued)
(a)
n1
n2
n3
E (! E0 )
1
1
1
1.313
1
1
2
1.500
1
1
3
1.813
1
2
1
2.063
1
1
4
2.250
1
2
2
2.250
1
2
3
2.563
1
1
5
2.813
1
2
4
3.000
1
1
6
3.500
(b) 1,1,4 and 1,2,2
7-7.
(
)(
2
)
)(
1.055 # 10"34 J s ! 2
h 2! 2
E0 =
=
= 37.68eV
2mL2 2 9.11 # 10"31 kg 0.10 # 10"9 m 2 1.609 # 10"19 J / eV
(
E311 ! E111 = "E = 11E0 ! 3E0 = 8 E0 = 301eV
(Problem 7-7 continued)
E222 ! E111 = "E = 12 E0 ! 3E0 = 9 E0 = 339eV
)
E321 ! E111 = "E = 14 E0 ! 3E0 = 11E0 = 415eV
7-8.
(a) Adapting Equation 7-3 to two dimensions (i.e., setting k3 = 0), we have
" n1 n2 = A sin
n1! x
n!y
sin 2
L
L
(b) From Equation 7-5, En1 n2 =
h 2! 2 2
n1 + n22
2mL2
(
)
(c) The lowest energy degenerate states have quantum numbers n1 = 1, n2 = 2,
and n1 = 2,
n2 = 1.