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.
(a)
12
Here 50 eV << 1 MeV , so p = (2mK )
"=
=
(b)
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 )
"=
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
!
"=
!
5-5
(a)
h
hc
=
p
2
K + 2Kmc2
(
= 5.36 # 10 *3
"=
12
)
2mc
K
2
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
h
h hc 1 240 eV nm 124 eV
or p = =
. As
=
=
p
(10 nm) ( c)
" "c
c
2&
#
K = E " mc2 = % p2 c 2 + mc 2 (
$
'
( )
12
" mc2 ,
2
we must use the relativistic expression for K, when pc " mc . Here
pc = 124 eV << mc2 = 0.511 MeV , so we can use the classical expression for K,
2
p
.
K=
2m
2
(124 eV )
p2
p2 c2
K=
=
=
= 0.150 eV
2m 2 mc2 2(0.511 MeV )
(b)
Electrons with " = 0.10 nm p =
hc 12 400 eV
as in (a). As
=
"c
c
2
pc << mc 2 = 0.511 MeV , use K =
(
Electrons with " = 10 fm = 10 # 10
(c)
2
(12 400) ( eV)
p2
= p2 c 2 =
= 150 eV .
2m
(2) 0.511 " 106 eV
$15
m, p=
)
hc 1.24 # 103 MeV
. As
=
"c
c
pc >> mc 2 = 0.511 MeV , use
2%
"
K = $p2 c 2 + mc2 '
#
&
(
)
12
( mc2 = pc ( mc2 = 1 240 MeV ( 0.511 MeV = 1 239 MeV .
2
For alphas with mc = 3 726 MeV :
2
(a)
p still is
p
124 eV
. As pc << 3 726 MeV , we use K =
:
2m
c
K=
p2 c2
2mc
2
(124 eV )2
#6
=
= 2.06 " 10
(2)( 3 726 MeV)
For alphas with " = 0.10 nm , p =
(b)
eV .
12 400 eV
2
. As pc << mc = 3 726 MeV ,
c
2
(12 400 eV)
p2
p2 c2
K=
=
=
= 0.020 6 eV .
2m 2mc2 (2) (3 726 MeV)
p=
(c)
1.24 " 103 MeV
2
and pc = 1 240 MeV ~ mc = 3 726 MeV . We use
c
2
K = "$p2 c 2 + mc2 %'
#
&
= 201 MeV .
(
)
12
1 2
2
2
( mc2 = "$(1 240 MeV) + (3 726 MeV) %'
#
&
!
5-7
2
A 10 MeV proton has K = 10 MeV << 2mc = 1 877 MeV so we can use the classical
12
expression p = (2mK )
"=
5-8
!
. (See Problem 5-2)
1 240 MeV fm
h
hc
=
=
p [( 2)(938.3 MeV)(10 MeV)]1 2
(2) (938.3 )(10)( MeV) 2
[
#
&
h
h
h
h
%
( )1 2
=
12 =
12 =
12 V
p (2 mK )
%
(2meV )
$ (2 me ) ('
%
(
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
'&
*)
"=
(
!
( 3 726 MeV
)
12
]
= 9.05 fm = 9.05 # 10 $15 m
5-11
(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
$14
"
10
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
2
) + (0.511 " 10 6 eV) (1.602 " 10 #19
12
2(
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 .
2
5-15
these results gives the dispersion relation " =
5-17
2
me v0
p
. 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 =
2
( )
12
2%
"
E = $p2 c2 + (me c2 ) ' . As E = h" and p = hk
#
&
E 2 = p2 c2 + me c 2
2
d" hk
p
hk
=
=
= 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 .
(
5-18
)
(
h
#3
#3
where "p = m"v = (0.05 kg) 10 $ 30 m s = 1.5 $ 10 kg % m s . Therefore,
2
h
6.626 # 10 $34 J % s
$32
"x =
=
= 3.51 # 10
m.
2"p 4 & 1.5 # 10 $3 kg % m s
(
"x"p #
(
5-19
)
K=
)
)
mv2
p2
:
=
2
2m
(1" 10 6 eV)(1.6 " 10 #19
)
J eV =
p2
(
2 1.67 " 10
#27
kg
"p = 0.05p = 1.160 # 10 $21 kg % m s and "x"p =
"x =
)
$ p = 2.312 " 10
kg % m s ,
h
. Thus
4#
6.63 # 10 $34 J %s
(1.16 # 10 $21
#20
)
kg % m s ( 4& )
= 4.56 # 10
$14
m.
Note that non-relativistic treatment has been used, which is justified because the kinetic
energy is only
(1.6 " 10 #13 ) " 100% = 0.11%
1.50 " 10 #10
of the rest energy.
5-23
(a)
"p"x = m"v"x #
"v #
(b)
h
2
h
2$ 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 .
5-24
(a)
"x"p = h so if "x = r , "p #
(b)
K=
h
r
2
( #p)
p2
h2
"
=
2me
2me
2mer 2
ke 2
r
h2
ke 2
E=
"
r
2me r2
U="
(c)
To minimize E take
dE
h2
ke 2
h2
="
+
=
0
#
r
=
= Bohr radius = a 0 . Then
dr
mer 3
r2
me ke 2
2
2%
"
" h %" me ke 2 %
me k 2 e 4
2 m e ke
E=$
= (13.6 eV .
'
$
2 ' ( ke $
2 ' =
# 2me &# h &
# h &
2h2
5-25
To find the energy width of the " -ray use "E"t #
"E #
h
or
2
h
6.58 $ 10 %16 eV & s
%6
#
# 3.29 $ 10 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.
5-26
The full width at half-maximum (FWHM) is 110 MeV. So "E = 55 MeV and using
h
"Emin "t min = ,
2
"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 x1 "
x
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
(
!
!
)
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# )
$1 2
%
'2*
0
sketch of g(" ) is given below.
g(!)
(b)
sin " %
.A
"
2
V, t
$
!
0
#
12
& V0 cos " tdt = )( # ,+ V0
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 .
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 =
"
"
"
!
!
!
!
!