5-25
To find the energy width of the " -ray use "E"t #
"E #
h
or
2
h
6.58 $ 10 %16 eV & s
#
# 3.29 $ 10 %6 eV .
2"t (2 ) 0.10 $ 10 %9 s
(
)
#6
As the intrinsic energy width of ~ ±3 " 10 eV is so much less than the experimental
resolution of ±5 eV , the intrinsic width can’t be measured using this method.
The full width at half-maximum (FWHM) is 110 MeV. So "E = 55 MeV and using
h
"Emin "t min = ,
2
5-26
"t min =
h
6.58 # 10 $16 eV % s
=
& 6.0 # 10 $24 s
2 "E
2 55 # 106 eV
(
)
' = lifetime ~ 2 "t min = 1.2 # 10 $23 s
5-27
For a single slit with width a, minima are given by sin " =
sin " # tan " =
n#
where n = 1, 2 , 3 , K and
a
x
x x1 "
2"
x $ x1 "
,
= and 2 =
# 2
= or
L L
a
L
a
L
a
"=
a#x 5 Å $ 2.1 cm
=
= 0.525 Å
L
20 cm
(
)
2
1.24 $ 10 4 eV %Å
( hc) 2
p2
h2
E=
=
=
=
= 546 eV
2m 2m"2 2mc 2 "2 2 5.11 $ 10 5 eV (0.525 Å) 2
(
!
2
)
2
2
With one slit open P1 = "1 or P2 = "2 . With both slits open, P = "1 + "2 . At a
maximum, the wavefunctions are in phase so
5-29
(
Pmax = "1 + "2
)2 .
At a minimum, the wavefunctions are out of phase and
!
(
Pmin = "1 # "2
Now
"1
P1
=
P2
"2
5-32
!
2
= 25 or
"1
= 5 , and
"2
(
(
"1 + "2
Pmax
=
Pmin
"1 # "2
!
!
2
(a)
f =
(
)2 .
)
) 2 = (5 "2 + "2 ) 2 = 62
)2 ( 5 "1 # "2 )2 42
#19
J
E (1.8 ) 1.6 " 10
=
= 4.34 " 1014 Hz
#34
h
6.63 " 10
J $s
=
36
= 2.25 .
16
5-25
To find the energy width of the " -ray use "E"t #
"E #
h
or
2
h
6.58 $ 10 %16 eV & s
#
# 3.29 $ 10 %6 eV .
2"t (2 ) 0.10 $ 10 %9 s
(
)
#6
As the intrinsic energy width of ~ ±3 " 10 eV is so much less than the experimental
resolution of ±5 eV , the intrinsic width can’t be measured using this method.
The full width at half-maximum (FWHM) is 110 MeV. So "E = 55 MeV and using
h
"Emin "t min = ,
2
5-26
"t min =
h
6.58 # 10 $16 eV % s
=
& 6.0 # 10 $24 s
2 "E
2 55 # 106 eV
(
)
' = lifetime ~ 2 "t min = 1.2 # 10 $23 s
5-27
For a single slit with width a, minima are given by sin " =
sin " # tan " =
n#
where n = 1, 2 , 3 , K and
a
x
x x1 "
2"
x $ x1 "
,
= and 2 =
# 2
= or
L L
a
L
a
L
a
"=
a#x 5 Å $ 2.1 cm
=
= 0.525 Å
L
20 cm
(
)
2
1.24 $ 10 4 eV %Å
( hc) 2
p2
h2
E=
=
=
=
= 546 eV
2m 2m"2 2mc 2 "2 2 5.11 $ 10 5 eV (0.525 Å) 2
(
!
2
)
2
2
With one slit open P1 = "1 or P2 = "2 . With both slits open, P = "1 + "2 . At a
maximum, the wavefunctions are in phase so
5-29
(
Pmax = "1 + "2
)2 .
At a minimum, the wavefunctions are out of phase and
!
(
Pmin = "1 # "2
Now
"1
P1
=
P2
"2
5-32
!
2
= 25 or
"1
= 5 , and
"2
(
(
"1 + "2
Pmax
=
Pmin
"1 # "2
!
!
2
(a)
f =
(
)2 .
)
) 2 = (5 "2 + "2 ) 2 = 62
)2 ( 5 "1 # "2 )2 42
#19
J
E (1.8 ) 1.6 " 10
=
= 4.34 " 1014 Hz
#34
h
6.63 " 10
J $s
=
36
= 2.25 .
16
c
= 691 nm
f
(b)
"=
(c)
"E #
h 6.63 $ 10 %34 J &s
=
"t
2' 2 $ 10 %6 s
(
"E # 5.276 $ 10
6-2
(a)
)
%29
J = 3.30 $ 10 %10 eV
Normalization requires
L
' A2 * L4 '
' 2& x *
' 4& x * *
$
2
1 = %#$ " dx = A2 % #4 L cos2 )
, dx = )
L ) 1 + cos)
, dx
,
%
#
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 ) 0 &
& L *) *)
0
0
2
2
2%
L
8
% 2 (% L ( % 2 ( % L (
% 4 $ x(
= ' *' * + ' * ' * sin '
& L )& 8 ) & L ) & 4$ )
& L *)
6-3
(a)
0
1 1
+
= 0.409
4 2$
$ 2" x '
$ 2" '
10
+1
10
Asin&
) = Asin 5 * 10 x so & ) = 5* 10 m ,
% # (
% # (
2#
"=
= 1.26 $ 10 %10 m .
5 $ 1010
(
)
(b)
p=
h 6.626 # 10 $34 Js
=
= 5.26 # 10 $24 kg m s
" 1.26 # 10 $10 m
(c)
K=
p2
2m
m = 9.11 " 10 #31 kg
5.26 " 10 #24 kg m s)
(
K=
( 2 " 9.11 " 10 #31 kg)
K=
6-6
=
2
= 1.52 " 10 #17 J
1.52 " 10 #17 J
= 95 eV
1.6 " 10 #19 J eV
" ( 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 '
Therefore E =
2 2
6-9
En =
h2 k 2
.
2m
2
n h
3h
2 , so "E = E2 # E1 =
8mL
8mL2
"E = (3)
(1240 eV nm c)2
(
)(
)
8 938.28 # 106 eV c2 10 $5 nm
2
= 6.14 MeV
hc
1240 eV nm
=
= 2.02 $ 10 %4 nm
# E 6.14 $ 106 eV
This is the gamma ray region of the electromagnetic spectrum.
"=
2 2
6-10
En =
n h
8mL2
2
(
)
6.63 " 10 #34 Js
h2
=
8mL2 8 9.11 " 10 #31 kg 10 #10 m
(
(a)
)(
2
)
= 6.03 " 10 #18 J = 37.7 eV
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)
hc
= En i # En f
"
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
hf =
6-12
2
#( 3 8 ) h" &
hc $ h ' 2 2
%
(
L
=
"E =
=&
2
*1
and
)
# % 8mL2 (
%$ 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
#10
2
(
)
6.626 " 10 #34 J $ s
h2
E1 =
=
8mp L2 8 1.67 " 10 #27 kg 2 " 10 #10 m
(
=
(b)
)(
8.22 " 10 #22 J
= 5.13 " 10 #3 eV
#19
1.60 " 10
J eV
Electron in the same box:
m
)
2
= 8.22 " 10 #22 J
2
(
)
6.626 " 10 #34 J $s
h2
=
E1 =
8me L2 8 9.11 " 10 #31 kg 2 " 10 #10 m
(
(c)
6-16
(a)
)(
2
)
= 1.506 " 10 #18 J = 9.40 eV .
The electron has a much higher energy because it is much less massive.
$ # 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&
L3
P=
#
0
(b)
12
$2'
" dx = & )
% L(
2
L3
* 3
12
* x'
2
2 -/ * ( 3) 02
2
# sin &% L )( dx = * # sin +d+ = * / 6 , 8 2 = 0.195 5 .
0
0
.
1
2$
12
$ 100# x '
" 2%
" = Asin&
) , A= $ '
# L&
% L (
L3
P=
=
(c)
6-18
# 100 " x&
2
2# L &
) sin2 %$ L (' dx = L %$ 100" ('
L 0
100" 3
1 100" 1
# 200" &
) sin2 *d* = 50" . 6 + 4 sin%$ 3 (' 1
0
,
/
-
0
1 , 1 / # 2" & 1
3
+.
= 0.331 9
(= +
1 sin%
3 - 200" 0 $ 3 ' 3 400"
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"
5"
2
3"
n" x
L
From this sketch we see that P( x) is a maximum when
n" x " 3" 5"
#
= ,
,
, K = "% m +
$
L
2 2
2
1&
(
2'
or when
x=
L"
1%
$ m + '&
n#
2
Likewise, P( x) is a minimum when
x=
6-29
(a)
m = 0, 1, 2, 3, K, n .
n" x
= 0 , " , 2" , 3" , K = m" or when
L
Lm
n
m = 0, 1, 2, 3, K, n
Normalization requires
$
$
2
$
2
(
)
(
)
1 = % " dx = C 2 % e #2x 1 # e #x dx = C 2 % e #2x # 2e #3x + e #4x dx . The integrals are
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 2 % xe #2x 1 # e #x dx = C 2 % x e #2x # 2e #3x + e #4x dx .
#$
!
0
0
#
The integrals are readily evaluated with the help of the formula
1
"ax
$ xe dx = a 2 to
0
)1 # 1& 1 ,
2
"1
2 ) 13 ,
get x = C * " 2% ( + - = C *
- . Substituting C = 12 nm gives
$
'
4
9
16
144
+
.
+
.
2
x =
!
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.
!
!
!
13
nm = 1.083 nm .
12
!
6-30
The possible particle positions within the box are weighted according to the probability
2
2
2 $ n# x '
density " = sin &
) The position is calculated as
L
% L (
L
2
x = # x " dx =
0
% n$ x (
2L
nx
xsin 2 '
(so that
* dx . Making the change of variable " =
#
L0
& L )
L
# dx
2L "
2
) gives x = 2 $ # sin n# d# . Using the trigonometric identity
d" =
L
" 0
"
*(
L &("
2 sin2 " = 1 # cos2" , we get x = 2 ' $ # d# % $ # cos 2n# d# + . An integration by parts
(,
" () 0
0
"2
L
. Thus, x = ,
2
2
, there is an extra factor of x in the
shows that the second integral vanishes, while the first integrates to
2
independent of n. For the computation of x
integrand. After changing variables to " =
#x
we get
L
"
*(
L2 &(" 2
"3
2
, the second may
3 ' $ # d# % $ # cos2 n# d# + . The first integral evaluates to
(,
3
" () 0
0
be integrated by parts twice to get
#
1#
& 1 )
#
#
2
"
cos2n
"
d
"
=
%
" sin 2n" d" = ( 2 + " cos 2n" 0 = 2 .
$
$
' 2n *
n0
2n
0
x2 =
Then x
6-31
2
2
=
L
"3
2
$& " 3
" (& L2
L
#
=
#
.
%
)
&' 3 2n 2 &* 3 2 (n" ) 2
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
#
x = $ x2 " dx = 2C 2 $ x2 e %2x
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
#
&
# x0 &
2
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 .
x )
dx = #2C ( 0 +e #2x
' 2*
2&
$x
x0
= 1 # e#
0
2
= 0.757