5-2
12
The issue is: Can we use the simpler classical expression p = (2mK )
relativistic expression p =
12
p = (2mK )
(
K 1 + 2mc
K
2
instead of the exact
12
)
? As the relativistic expression reduces to
c
2
for K << 2 mc , we can use the classical expression whenever K << 1 MeV
2
because mc for the electron is 0.511 MeV.
12
Here 50 eV << 1 MeV , so p = (2mK )
(a)
"=
=
h
=
p
h
[(2)(
0.511 MeV
c2
=
1 2
= 0.173 nm
)( 50 eV) ]
1 240 eV nm
( 2)(0.511 # 10 6 )( 50)( eV)2
[
hc
12
]
[(2 )(0.511 MeV )( 50 eV) ]1 2
12
As 50 eV << 1 MeV , p = (2mK )
(b)
"=
hc
[(2)(
0.511 MeV
c2
)(50 # 10
3
eV
= 5.49 # 10 $3 nm .
12
)]
As this is clearly a worse approximation than in (a) to be on the safe side use the
(1 +
relativistic expression for p: p = K
!
"=
!
5-7
h
hc
=
p
2
K + 2Kmc2
(
= 5.36 # 10 *3
12
)
c
so
1 240 eV nm
=
$
&% 50 # 10
nm = 0.005 36 nm
(
3 2
)
(
+ (2) 50 # 10
3
)(0.511 # 10 6 eV)')(
12
2
12
"=
. (See Problem 5-2)
1 240 MeV fm
h
hc
=
12 =
p [( 2)(938.3 MeV)(10 MeV)]
(2) (938.3 )(10)( MeV) 2
[
12
]
= 9.05 fm = 9.05 # 10 $15 m
#
&
h
h
h
h
%
(V )1 2
=
=
=
p (2 mK )1 2 (2meV )1 2 %$ (2 me )1 2 ('
%
(
6.626 # 10 $34 Js
'
* $1 2
" ='
V
12*
$31
$19
kg # 1.602 # 10
C
'& 2 # 9.105 # 10
*)
%1.226 # 10 $9 kg1 2 m 2 ( $1 2
" ='
*V
sC 1 2
'&
*)
As " = 2a 0 = 2( 0.052 9) nm = 0.105 8 nm the energy of the electron is nonrelativistic, so
we can use
"=
(
5-10
)
2
A 10 MeV proton has K = 10 MeV << 2mc = 1 877 MeV so we can use the classical
expression p = (2mK )
5-8
12
2mc
K
)
p=
h
p2
with K =
;
"
2m
(
2
)
$34
2
6.626 # 10
J %s
h
K=
2 =
2m"
2 9.11 # 10 $31 kg 1.058 # 10 $10 m
(
5-11
)(
)
2
= 21.5 # 10 $18 J = 134 eV
This is about ten times as large as the ground-state energy of hydrogen, which is 13.6 eV.
(a)
In this problem, the electron must be treated relativistically because we must use
2
relativity when pc " mc . (See problem 5-5). the momentum of the electron is
p=
h 6.626 # 10 $34 J % s
=
= 6.626 # 10 $20 kg % m s
"
10 $14 m
2
and pc = 124 MeV >> mc = 0.511 MeV . The energy of the electron is
(
E = p2 c2 + m2 c 4
12
)
2
%
= ' 6.626 " 10 #20 kg $ m s 3 " 10 8 m s
&
= 1.99 " 10 #11 J = 1.24 " 108 eV
(
)(
2
) (
+ 0.511 " 10 6 eV
2
12
2(
1.602 " 10 #19 J eV *
)
)(
)
2
so that K = E " mc # 124 MeV .
(b)
The kinetic energy is too large to expect that the electron could be confined to a
region the size of the nucleus.
$34
5-12
h
6.626 # 10
J %s
h
6
=
= 7.27 # 10 m s .
= mv , we find that v =
$31
m"
"
9.11 # 10
kg 1 # 10 $10 m
From the principle of conservation of energy, we get
Using p =
(
(
)(
)(
)
9.11 " 10 #31 kg 7.27 " 106 m s
mv2
eV =
=
2
2
)
2
= 2.41 " 10
#17
J = 151 eV .
Therefore V = 151 V .
5-15
me v02
p2
. As the wavenumber and angular
=
2
2me
frequency of the electron’s de Broglie wave are given by p = hk and E = h" , substituting
For a free, non-relativistic electron E =
these results gives the dispersion relation " =
5-17
2
( )
12
2%
"
E = $p2 c2 + (me c2 ) ' . As E = h" and p = hk
#
&
E 2 = p2 c2 + me c 2
d" hk
p
hk 2
=
=
= v0 .
. So vg =
dk me me
2me
!
2
h" = #%h2 k 2 c 2 + me c 2 &(
$
'
(
)
#
me c 2
%
2 2
" ( k) = % k c +
h2
%$
(
)
12
or
2 &1 2
(
(
('
2&
# 2 2
k c + me c 2 h (
%
" $
'
vp = =
k
k
(
d"
vg =
dk
k0
)
12
12
2
#
) m c2 , &
= %% c2 + + e . ((
* hk $
'
2&
#
1 % 2 2 ) me c 2 , (
= k c ++
.
2%
* h - ('
$
/1 2
2kc 2 =
kc 2
12
2&
# 2 2
2
%$ k c + me c h ('
(
)
0# 2 2
2 &1 2 4
2
2 %k c + m e c h ( 2 0
12
2
' 2 2#k 2 c2 + m c2 h 2 & 42 = c 2
vp vg = 1 $
5 1%
5
e
(
k
' 26
2
2 23$
23
26
Therefore, vg < c if vp > c .
(
5-23
(a)
"p"x = m"v"x #
"v #
(b)
5-24
(a)
)
(
)
h
2
h
2$ J % s
=
= 0.25 m s
4$ m"x 4$ (2 kg )(1 m)
The duck might move by (0.25 m s)( 5 s) = 1.25 m . With original position
uncertainty of 1m, we can think of "x growing to 1 m + 1.25 m = 2.25 m .
h
"x"p = h so if "x = r , "p #
r
2
(b)
( #p)
p2
h2
K=
"
=
2me
2me
2mer 2
U="
E=
(c)
ke
r
h2
2
2me r2
"
ke 2
r
To minimize E take
dE
h2
ke 2
h2
="
= Bohr radius = a 0 . Then
3 + 2 =0#r =
dr
mer
r
me ke 2
2
2%
"
" h %" me ke 2 %
me k 2 e 4
2 m e ke
E=$
(
ke
=
= (13.6 eV .
$ 2 '
# 2me '&$# h2 '&
# h &
2h2
5-34
(a)
g(" ) = ( 2# )
$1 2
%
& V (t )(cos " t $ i sin " t )dt , V( t) sin " t is an odd function so this
$%
integral vanishes leaving g(" ) = 2 (2# )
sketch of g(" ) is given below.
!
$1 2
%
'2*
& V0 cos " tdt = )( # ,+
0
12
V0
sin " %
.A
"
!
2
h" = #%h2 k 2 c 2 + me c 2 &(
$
'
(
)
#
me c 2
%
2 2
" ( k) = % k c +
h2
%$
(
)
12
or
2 &1 2
(
(
('
2&
# 2 2
k c + me c 2 h (
%
" $
'
vp = =
k
k
(
d"
vg =
dk
k0
)
12
12
2
#
) m c2 , &
= %% c2 + + e . ((
* hk $
'
2&
#
1 % 2 2 ) me c 2 , (
= k c ++
.
2%
* h - ('
$
/1 2
2kc 2 =
kc 2
12
2&
# 2 2
2
%$ k c + me c h ('
(
)
0# 2 2
2 &1 2 4
2
2 %k c + m e c h ( 2 0
12
2
' 2 2#k 2 c2 + m c2 h 2 & 42 = c 2
vp vg = 1 $
5 1%
5
e
(
k
' 26
2
2 23$
23
26
Therefore, vg < c if vp > c .
(
5-23
(a)
"p"x = m"v"x #
"v #
(b)
5-24
(a)
)
(
)
h
2
h
2$ J % s
=
= 0.25 m s
4$ m"x 4$ (2 kg )(1 m)
The duck might move by (0.25 m s)( 5 s) = 1.25 m . With original position
uncertainty of 1m, we can think of "x growing to 1 m + 1.25 m = 2.25 m .
h
"x"p = h so if "x = r , "p #
r
2
(b)
( #p)
p2
h2
K=
"
=
2me
2me
2mer 2
U="
E=
(c)
ke
r
h2
2
2me r2
"
ke 2
r
To minimize E take
dE
h2
ke 2
h2
="
= Bohr radius = a 0 . Then
3 + 2 =0#r =
dr
mer
r
me ke 2
2
2%
"
" h %" me ke 2 %
me k 2 e 4
2 m e ke
E=$
(
ke
=
= (13.6 eV .
$ 2 '
# 2me '&$# h2 '&
# h &
2h2
5-34
(a)
g(" ) = ( 2# )
$1 2
%
& V (t )(cos " t $ i sin " t )dt , V( t) sin " t is an odd function so this
$%
integral vanishes leaving g(" ) = 2 (2# )
sketch of g(" ) is given below.
!
$1 2
%
'2*
& V0 cos " tdt = )( # ,+
0
12
V0
sin " %
.A
"
g(!)
2
V, t
$
!
0
#
(b)
2$
$
#
%
%
2$
%
$
%
As the major contribution to this pulse comes from " ’s between "
"# $
#
"
and , let
$
#
%
and since "t = # .
&
&$ )
"#"t = ( +% = $
'% *
(c)
Substituting "t = 0.5 µs in "# =
$
"1
1
6
we find
=
= 1 # 10 Hz .
"t
2 "t 2 0.5 # 10 $6 s
(
)
6
As the range is 2 "f , the range is 2 " 10 Hz . For "t = 0.5 ns , the range is
2"f = 2 # 109 Hz .
5-35
(a)
1 +$
A +$ #& 2 ( k # k0 ) 2 ikx
A #& 2 k02 +$ #& 2 (k 2 #( 2k 0 +ix & 2 )k )
a ( k )eikx dk =
e
e dk =
e
dk
%
%
%e
2" #$
2" #$
2"
#$
. Now complete the square in order to get the integral into the standard form
f ( x) =
"
+#
2
"az
$ e dz :
"#
"
e
"
"#
2
( k 2 "(2k 0 +ix # 2 ) k) = e +# 2 (k 0 +ix
2#
2 2
) e "# 2 ( k "( k 0 +ix 2# 2 )) 2
A "# 2 k 20 # 2 ( k 0 +ix 2# 2 ) 2 +% "# 2 ( k "( k0 +ix
e
e
& e
2$
k = "%
A "x 2 4# 2 ik0 x +% "# 2 z 2
=
e
e
dz
& e
2$
z = "%
f ( x) =
!
!
!
$
ix '
where z = k " & k0 +
) . Since
%
2# 2 (
+$
"#
% e
z ="$
real part of f ( x) , Re f ( x) is Re f ( x) =
2 2
z
A
dz =
e #x
2
2#
A #x 2
&1 2
e
, f ( x) =
#
" 2
4"
2
2
)) 2
dk
4" 2 ik 0x
e
. The
cos k 0 x and is a gaussian
" 2
envelope multiplying a harmonic wave with wave number k0 . A plot of Re f ( x)
is shown below:
!
!
Re f ( x)
A #x 2
e
" 2
4" 2
x
cos k0 x
Comparing
(c)
2
A #x 2 4" 2
"(x 2#x )
e
to Ae
implies "x = # .
" 2
By same reasoning because " 2 =
$ 1' 1
1
1
. Finally "x"k = # &
= .
2 , "k =
% 2# )( 2
2#
4 #k
2
5-36
mu h
u
1
h
. vphase = f" =
= = vphase . This is different from
mu 2 = hf and " =
2h mu 2
2
mu
the speed u at which the particle transports mass, energy, and momentum.
E=K =
!
!
!
!
!
!