6-2
(a)
Normalization requires
L
' A2 * L4 '
' 2& x *
' 4& x * *
$
2
1 = %#$ " dx = A2 % #4 L cos2 )
, dx = )
, dx
, % # L ) 1 + cos)
4
( L +
( L + ,+
( 2 + 4(
so A =
(b)
L
8
2
.
L
L
8
L
2$ x (
% 4 (% 1( 8 %
% 4$ x ( (
P = # " dx = A # cos '
dx = ' * ' * # ' 1 + cos'
dx
*
&
L
)
&
L
)
&
2
)
& L *) *)
0&
0
0
2
2
2%
% 2 (% L ( % 2 ( % L (
% 4 $ x(
= ' *' * + ' * ' * sin '
& L )& 8 ) & L ) & 4$ )
& L *)
6-3
(a)
L
8
=
0
1 1
+
= 0.409
4 2$
$ 2" x '
$ 2" '
10
+1
10
Asin&
) = Asin 5 * 10 x so & ) = 5* 10 m ,
% # (
% # (
2#
%10
"=
m.
10 = 1.26 $ 10
5 $ 10
(
)
(b)
p=
h 6.626 # 10 $34 Js
=
= 5.26 # 10 $24 kg m s
" 1.26 # 10 $10 m
(c)
K=
p
2m
2
m = 9.11 " 10 #31 kg
( 5.26 " 10 #24 kg m s)
K=
( 2 " 9.11 " 10 #31 kg)
2
= 1.52 " 10 #17 J
1.52 " 10 #17 J
= 95 eV
1.6 " 10 #19 J eV
Solving the Schrödinger equation for U with E = 0 gives
K=
6-5
(a)
" d 2( %
" h2 % # dx 2 &
U=$
.
'
# 2m& (
If " = Ae
(b)
#x 2 L2
then
d2"
3
2 $ 1 ' #x 2
e
2 = 4Ax # 6AxL &
% L4 )(
dx
(
)
U( x) is a parabola centered at x = 0 with U( 0) =
L2
"3h2
mL2
" h 2 % " 4 x2
%
,U=$
2 ' $ 2 ( 6' .
# 2 mL & # L
&
<0 :
U
2
3h
mL2
6-6
x
" ( x) = Acos kx + Bsin kx
#"
= $kA sin kx + kB cos kx
#x
#2"
= $k 2 Acos kx $ k 2 Bsin kx
# x2
% $2m (
% $2mE (
' 2 * (E $U )" = '
* ( Acoskx + Bsin kx)
& h )
& h2 )
The Schrödinger equation is satisfied if
"2# % $2m (
='
* (E $U )# or
" x2 & h 2 )
# "2mE &
"k2 ( Acos kx + Bsin kx) = % 2 (( Acos kx + Bsin kx) .
$ h '
2 2
Therefore E =
6-9
En =
n 2 h2
!
, so "E = E2 # E1 =
8mL2
"E = (3)
h k
.
2m
3h2
8mL2
(1240 eV nm c)2
(
)(
)
8 938.28 # 106 eV c2 10 $5 nm
2
= 6.14 MeV
hc
1240 eV nm
%4
=
= 2.02 $ 10 nm
# E 6.14 $ 106 eV
This is the gamma ray region of the electromagnetic spectrum.
"=
6-10
En =
n 2 h2
8mL2
2
(
)
6.63 " 10 #34 Js
h2
=
8mL2 8 9.11 " 10 #31 kg 10 #10 m
(
(a)
)(
E1 = 37.7 eV
E2 = 37.7 " 22 = 151 eV
E3 = 37.7 " 32 = 339 eV
E 4 = 37.7 " 4 2 = 603 eV
(b)
hf =
hc
= En i # En f
"
2
)
= 6.03 " 10 #18 J = 37.7 eV
1 240 eV $ nm
hc
=
En i # En f
E n i # En f
For n i = 4 , nf = 1 , E ni " E n f = 603 eV " 37.7 eV = 565 eV , " = 2.19 nm
n i = 4 , nf = 2 , " = 2.75 nm
n i = 4 , nf = 3 , " = 4.70 nm
n i = 3 , nf = 1 , " = 4.12 nm
n i = 3 , nf = 2 , " = 6.59 nm
n i = 2 , nf = 1 , " = 10.9 nm
"=
6-11
In the present case, the box is displaced from (0, L) by
wavefunctions by replacing x with x "
L
. Accordingly, we may obtain the
2
L
in the wavefunctions of Equation 6.18. Using
2
*# n" x & n" * # n" & #
# n" x &
# n" x &
L&# n" &
# n" &
/ = sin%
sin ,%
(' %$ x ) (' / = sin ,%
()
( cos%$
(' ) cos%
( sin%$
(
$
L
2
$
L
'
2
$
L
'
2
$
L
'
2 '
,+
/.
+
.
we get for "
L
L
# x#
2
2
# 2&
" 1 ( x) = % (
$ L'
# ) x&
" 2%
2" ( x %
cos%
'
( ; P1 (x) = $# '& cos $
L
# L &
$ L '
12
# 2) x &
"2%
2 " 2( x %
sin %
'
( ; P2 ( x) = $# '& sin $
L
# L &
$ L '
12
# 3 ) x&
"2%
2 " 3( x %
cos %
'
( ; P3 ( x) = $# '& cos $
L
# L &
$ L '
#2&
" 2 ( x) = % (
$ L'
#2&
" 3 ( x) = % (
$ L'
12
!
6-12
2
# ( 3 8 ) h" &
hc $ h ' 2 2
(
=&
"E =
and L = %
2 ) 2 *1
# % 8mL (
%$ mc ('
6-13
(a)
[
12
]
= 7.93 ) 10 *10 m = 7.93 Å.
Proton in a box of width L = 0.200 nm = 2 " 10
=
(b)
2
#22
J
8.22 " 10
= 5.13 " 10 #3 eV
#19
1.60 " 10
J eV
Electron in the same box:
2
(
)
6.626 " 10 #34 J $s
h2
E1 =
=
8me L2 8 9.11 " 10 #31 kg 2 " 10 #10 m
(
!
!
m
6.626 " 10 #34 J $ s)
(
E1 =
=
= 8.22 " 10 #22 J
2
2
#27
#10
8mp L
8(1.67 " 10
kg )( 2 " 10
m)
h2
!
!
#10
(c)
)(
2
)
= 1.506 " 10 #18 J = 9.40 eV .
The electron has a much higher energy because it is much less massive.
6-16
(a)
$ # x'
" ( x) = Asin & ) , L = 3 Å. Normalization requires
% L(
L
L
% $ x(
LA2
2
1 = # " dx = # A2 sin2 '
dx
=
*
& L )
2
0
0
" 2%
so A = $ '
# L&
12
L3
P=
(b)
$2'
L3
$ * x'
L3
=
6-18
* 3
( 3)1 2 0
2 -*
12
$ 100# x '
" 2%
" = Asin&
,
A
=
)
$# '&
% L (
L
P=
(c)
2
2
2
2
# " dx = &% L )( # sin &% L )( dx = * # sin +d+ = * // 6 , 8 22 = 0.195 5 .
0
0
0
.
1
# 100 " x&
2
2# L &
sin2 %
(
( dx = %$
)
L 0
$ L '
L 100" '
100" 3
# 200" & /
1 , 100" 1
2
) sin *d* = " . 6 + sin%$ 3 (' 1
50 4
0
0
1 , 1 / # 2" & 1
3
+.
= 0.331 9
(' = +
1 sin%$
3 400"
3 - 200" 0
3
Yes: For large quantum numbers the probability approaches
1
.
3
Since the wavefunction for a particle in a one-dimension box of width L is given by
$ n# x '
2
2
2 $ n # x'
" n = Asin&
) it follows that the probability density is P( x) = " n = A sin &
),
% L (
% L (
which is sketched below:
P(x)
"
2
"
3"
2
2"
From this sketch we see that P( x) is a maximum when
5"
2
3"
n" x
L
n" x " 3" 5"
1&
#
= ,
,
, K = "% m + (
$
L
2 2
2
2'
or when
x=
!
!
L"
1%
$ m + '&
n#
2
m = 0, 1, 2, 3, K, n .
Likewise, P( x) is a minimum when
x=
6-20
n" x
= 0 , " , 2" , 3" , K = m" or when
L
Lm
n
m = 0, 1, 2, 3, K, n
d2"
# 2m &
= % 2 ( {U (x) ) E}" (x) . In the well
$h '
2mE
interior, U( x) = 0 and solutions to this equation are sin kx and cos kx , where k 2 = 2 .
h
The waves symmetric about the midpoint of the well ( x = 0 ) are described by
The Schrödinger equation, after rearrangement, is
dx2
" ( x) = Acos kx
"L < x < +L
In the region outside the well, U( x) =U , and the independent solutions to the wave
# 2m &
±" x
2
equation are e
with " = % 2 ( (U ) E ) .
$h '
(a)
The growing exponentials must be discarded to keep the wave from diverging at
infinity. Thus, the waves in the exterior region, which are symmetric about the
midpoint of the well are given by
" ( x) = Ce #$ x
x > L or x < "L .
At x = L continuity of " requires AcoskL = Ce
"# L
. For the slope to be
"# L
continuous here, we also must require "Ak sin kL = "Ce
. Dividing the two
equations gives the desired restriction on the allowed energies: k tan kL = " .
(b)
!
The dependence on E (or k) is made more explicit by noting that k 2 + " 2 =
12
#2 mU
2&
which allows the energy condition to be written k tan kL = $ 2 " k '
% h
(
2
!
!
!
(
2
)
2
(2 ) 511 " 10 3 eV c2 ( 5 eV )( 0.1 nm )
2mUL
=
= 1.312 7 .
2
h2
(197.3 eV # nm c)
With this value, the equation for " = kL
!
.
Multiplying by L, squaring the result, and using tan " + 1 = sec " gives
2mUL2
2
2
from which the desired form follows immediately,
(
kL) sec ( kL) =
!
h2
2mU
. The ground state is the symmetric waveform having the
k sec( kL) =
h
lowest energy. For electrons in a well of height U = 5 eV and width 2L = 0.2 nm ,
we calculate
!
!
2
2mU
2 ,
h
"
1
= (1.312 7)
cos "
!
!
!
!
!
2
= 1.145 7
!
can be solved numerically employing methods of varying sophistication. The
simplest of these is trial and error, which gives " = 0.799 From this, we find
k = 7.99 nm "1 , and an energy
2
(
#1
h2 k 2 (197.3 eV" nm c) 7.99 nm
E=
=
2m
2 511$ 10 3 eV c2
(
6-29
(a)
)
2
)
= 2.432 eV .
Normalization requires
$
$
2
$
2
(
)
(
)
2
#2x
1 # e #x dx = C 2 % e #2x # 2e #3x + e #4x dx . The integrals are
1 = % " dx = C % e
0
#$
0
# 1 & 1 , C2
2 )1
elementary and give 1 = C * " 2% ( + - =
. The proper units for C are those
$ 3 ' 4 . 12
+2
"1 2
of (length )
(b)
12
thus, normalization requires C = (12)
nm
"1 2
.
The most likely place for the electron is where the probability "
2
is largest. This
d"
is also where " itself is largest, and is found by setting the derivative
equal
dx
zero:
0=
d"
= C #e #x + 2e #2x = Ce #x 2 e #x # 1 .
dx
{
}
{
}
"x
The RHS vanishes when x = " (a minimum), and when 2e = 1 , or x = ln 2 nm .
Thus, the most likely position is at xp = ln 2 nm = 0.693 nm .
(c)
!
The average position is calculated from
$
2
2
$
x = % x" dx = C % xe
(1 # e #x )
#2x
0
#$
2
2
$
(
dx = C % x e
0
#2x
# 2e
#3x
+e
#4x
) dx .
#
The integrals are readily evaluated with the help of the formula
1
"ax
$ xe dx = a 2 to
0
# 1& 1 ,
2
"1
2 )1
2 ) 13 ,
get x = C * " 2% ( + - = C *
- . Substituting C = 12 nm gives
$
'
9
16 .
+ 144 .
+4
x =
!
13
nm = 1.083 nm .
12
We see that x is somewhat greater than the most probable position, since the
probability density is skewed in such a way that values of x larger than xp are
weighted more heavily in the calculation of the average.
!
!
6-31
2
The symmetry of " ( x) about x = 0 can be exploited effectively in the calculation of
average values. To find x
$
!
!
2
x = % x" (x) dx
#$
!
We notice that the integrand is antisymmetric about x = 0 due to the extra factor of x (an
odd function). Thus, the contribution from the two half-axes x > 0 and x < 0 cancel
2
exactly, leaving x = 0 . For the calculation of x
, however, the integrand is symmetric
and the half-axes contribute equally to the value of the integral, giving
#
2
#
2
2
2 %2x
x = $ x " dx = 2C $ x e
0
x0
dx .
0
3
" x0 %
Two integrations by parts show the value of the integral to be 2$ ' . Upon substituting
# 2 &
12
$ x2 '
" 1 % " x0 % 3 x20
x
2 12
2
= & 0 ) = 0 . In
for C , we get x = 2 $ ' (2 )$ ' =
and "x = x # x
2
# x0 & # 2 &
2
% (
2
calculating the probability for the interval "#x to +"x we appeal to symmetry once
again to write
2
(
2
+$x
2
2
$x
P = % " dx = 2C % e
#$x
#2x x 0
0
or about 75.7% independent of x0 .
!
!
!
!
!
!
!
2 & x0 ) #2x
dx = #2C ( +e
' 2*
)
$x
x0
=1 #e
0
# 2
= 0.757