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
m
could
be canceled from each term.)
6-54. (a) The requirement is that ! 2 (x ) = ! 2 (" x ) = ! (" x )! (" x ). This can only be
true if:
! (" x ) = ! (x ) or ! (" x ) = "! (x ).
(b) Writing the Schrödinger equation in the form
d 2!
2mE
= " 2 ! , the general
2
dx
h
solutions
of this 2nd order differential equation are:
! (x ) = A sin kx and ! (x ) = A cos kx
where k = 2mE h. Because the boundaries of the box are at x = ± L / 2,
both
solutions are allowed (unlike the treatment in the text where one boundary
was at
x = 0). Still, the solutions are all zero at x = ± L / 2 provided that an integral
number
of half wavelengths fit between x = ! L / 2 and x = + L / 2. This will occur
for:
1/ 2
! n (x ) = (2 / L ) cos n" x / L when n = 1, 3, 5,L . And for
1/ 2
! n (x ) = (2 / L ) sin n" x / L when n = 2, 4, 6,L .
The solutions are alternately even and odd.
(c)
(
6-55. ! 0 = Ae " x
(a)
)
The allowed energies are: E = h 2 k 2 / 2m = h 2 n! L2 / 2m = n 2 h 2 / 8mL2 .
2
/ 2 L2
2
2
2
2
d! 0
d!
= " x / L2 Ae " x / 2 L and ! 1 = L 0 = L " x / L2 Ae " x / 2 L = (" x / L )! 0
dx
dx
(
So,
)
(
)
d! 1
= " (1/ L )! 0 " (x / L )d! 0 / dx
dx
d 2! 1
= " (1 / L )d! 0 / dx " (1 / L )d! 0 / dx " (x / L )d 2! 0 / dx 2
And
2
dx
(
)
(
)
(
)
= 2 x / L3 ! 0 + x / L3 ! 0 + x3 / L5 ! 0
Recalling from Problem 6-3 that V (x ) = h 2 x 2 / 2mL4 , the Schrödinger
equation
(
)(
)
(
)
becomes "h 2 / 2m 3m / L3 + x3 / L5 ! 0 + h 2 x3 / 2mL5 ! 0 = E (" x / L )! 0 or,
(
)
simplifying: "3h 2 x / 2mL3 ! 0 = E (" x / L )! 0 . Thus, choosing E
appropriately
will make ! 1 a solution.
(b) We see from (a) that E = 3h 2 / 2mL2 , or three times the ground state energy.
(c) ! 1 plotted looks as below. The single node indicates that ! 1 is the first
excited state.
(The energy value in [b] would also tell us that.)
L
6-56.
x2 = "
0
2 2
n! x
x sin2
dx
L
L
2
x2 =
Letting u = n! x / L, du = (n! / L )dx
n!
2" L # " L # 2
u sin2 udu
$
%
$
%
(
L & n! ' & n! ' 0
n!
3
2 $ L % " u3 $ u 2 1 %
u cos 2u #
= )
&
&
sin
2
u
&
'
(
)
*
L + n! *, - 6 + 4 8 ,
4
.0
3
3
# L2
2 $ L % " (n! )
n!
L2
&
'
= )
(
0
(
(
0
=
(
2 2
L + n! *, & 6
4
'. 3 2n !
-
6-58. (a) For " (x, t ) = A sin (kx # !t )
d 2"
#"
= $k 2 " and
= $! A cos (kx $ !t ) so the Schrödinger equation
2
dx
#t
becomes:
"
h2k 2
A sin (kx " !t ) + V (x ) A sin (kx " !t ) = "ih! cos (kx " !t )
2m
Because the sin and cos are not proportional, this ! cannot be a solution.
Similarly,
for " (x, t ) = A cos (kx # !t ), there are no solutions.
i kx "!t
(b) For % (x, t ) = A #&cos (kx " !t ) + i sin (kx " !t )$' = Ae ( ), we have that
d 2"
#"
= $k 2 " and
= $i!". And the Schrödinger equation becomes:
2
dx
#t
h2k 2
"
# + V (x )# = "h!# for h! = h 2 k 2 / 2m + V .
2m