2
5-3.
(hc )
p2
Ek = eVo =
=
2m! 2 2mc 2 ! 2
5-4.
!=
h
=
p
h
(b) For a proton: ! =
1240eV nm
(
"(2 ) 0.511 $ 106 eV
%
(
5-5.
)(
3
)
4.5 $ 10 eV #&
1/ 2
1240eV nm
(
6
$(2 ) 983.3 # 10 eV
&
(c) For an alpha particle:
1240eV nm
$(2 ) 3.728 # 109 eV
&
)
(from Equation 5-2)
2mc 2 Ek
(a) For an electron: ! =
!=
(
hc
=
2mEk
2
(1240eV nm )
1
Vo =
= 940V
e 2 5.11 " 105 eV (0.04nm )2
)(4.5 # 10 eV )%'
3
1/ 2
)(4.5 # 10 eV )%'
3
1/ 2
= 0.0183nm
= 4.27 # 10"4 nm
= 2.14 # 10"4 nm
! = h / p = h / 2mEk = hc / "$ 2mc 2 (1.5kT )#%
1/ 2
(from Equation 5-2)
Mass of N2 molecule =
2 ! 14.0031u 931.5MeV / uc 2 = 2.609 ! 104 MeV / c 2 = 2.609 ! 1010 eV / c 2
(
!=
5-11.
!=
)
1240eV nm
(
)
(
)
#(2 ) 2.609 % 1010 eV (1.5 ) 8.617 % 10"5 eV / K (300 K )$
&
'
h
=
p
h
2m p Ek
1/ 2
= 0.0276nm
= 0.25nm
Squaring and rearranging,
2
2
(hc ) =
(1240eV nm )
h2
Ek =
=
= 0.013eV
2
2
2
2
2m p !
2 mpc !
2 938 " 106 eV (0.25nm )
(
)
n! = D sin "
#
sin ! = 0.822
" ! = 55°
(
)
sin " = n! / D = (1)(0.25nm )/ (0.304nm )
5-12. (a) n! = D sin " # D =
=
(b) sin " =
n!
nhc
=
sin " sin " 2mc 2 Ek
(1)(1240eV
nm )
(
(sin 55.6° )!$ 2 5.11 # 105 eV
)
1/ 2
(50eV )"%
= 0.210nm
(1)(1240eV nm )
n!
=
= 0.584
D (0.210nm )# 2 5.11 % 105 eV (100eV )$1 / 2
&
'
(
)
! = sin"1 (0.584 ) = 35.7°
5-17. (a) y = y1 + y2
= 0.002m cos (8.0 x / m ! 400t / s ) + 0.002m cos (7.6 x / m ! 380t / s )
1
!1
"
= 2 (0.002m )cos $ (8.0 x / m # 7.6 x / m ) # (400t / s # 380t / s )%
2
&2
'
1
!1
"
# cos % (8.0 x / m + 7.6 x / m ) $ (400t / s + 380t / s )&
2
'2
(
= 0.004m cos (0.2 x / m ! 10t / s )" cos (7.8 x / m ! 390t / s )
(b) v =
! 390 / s
=
= 50m / s
k 7.8 / m
(c) vs =
"!
20 / s
=
= 50m / s
"k 0.4 / m
(d) Successive zeros of the envelope requires that 0.2"x / m = ! , thus
!
2!
#x =
= 5! m with #k = k1 " k2 = 0.4m "1 and #x =
= 5! m.
0.2
#k
h
h
hc
1240eV nm
5-22. (a) ! = =
=
=
= 0.549nm
1/ 2
2
p
2mEk
2mc Ek " 2 0.511 $ 106 eV (5eV )#
%
&
(
)
d sin! = " / 2 For first minimum (see Figure 5-17).
d=
!
0.549nm
=
= 3.15nm slit separation
2 sin"
2 sin 5°
(b) sin 5° = 0.5cm / L where L = distance to detector plane L =
0.5cm
= 5.74cm
2 sin 5°
5-23. (a) The particle is found with equal probability in any interval in a force-free
region.
Therefore, the probability of finding the particle in any interval ∆x is
proportional to
∆x. Thus, the probability of finding the sphere exactly in the middle, i.e.,
with
∆x = 0 is zero.
(b) The probability of finding the sphere somewhere within 24.9cm to 25.1cm is
proportional to ∆x =0.2cm. Because there is a force free length L = 48cm
available
to the sphere and the probability of finding it somewhere in L is unity, then
the
probability that it will be found in ∆x = 0.2cm between 24.9cm and 25.1cm
(or any
interval of equal size) is: P!x = (1 / 48 )(0.2cm ) = 0.00417.
L
5-24. Because the particle must be in the box
L
2
2
#! * ! dx = 1 = # A sin (" x / L )dx = 1
0
Let u = ! x / L;
x = 0 " u = 0;
!
0
x = L " u = ! and dx = (L / ! )du , so we have
!
2
" A (L / ! )sin
2
0
udu = A2 (L / ! )" sin2 udu = 1
0
!
!
u sin 2u #
= (L / ! ) A2 (! / 2 ) = LA2 / 2 = 1
(L / ! ) A ) sin udu = (L / ! )A "% $
&
4 (0
'2
0
2
!
2
( )
2
1/ 2
A2 = 2 / L " A = (2 / L )
5-25. (a) At x = 0 :
(b) At x = ! :
(c) At x = 2! :
2
2
Pdx = ! (0, 0 ) dx = Ae0 dx = A2 dx
Pdx = Ae "!
2
/ 4! 2
Pdx = Ae "4!
2
2
/ 4! 2
2
dx = Ae "1 / 4 dx = 0.61A2 dx
2
2
dx = Ae "1 dx = 0.14 A2 dx
(d) The electron will most likely be found at x = 0, where Pdx is largest.
5-27. #E #t $ h % #E $ h / #t =
1.055 " 10!34 J s
$ 6.6 " 10!9 eV
!7
!19
10 s 1.609 " 10 J / eV
(
)
5-35. The size of the object needs to be of the order of the wavelength of the 10MeV
neutron.
! = h / p = h / " mu. " and u are found from:
Ek = mn c 2 (! " 1) or ! " 1 = 10 MeV / 939 MeV
1/ 2
! = 1 + 10 / 939 = 1.0106 = 1 / (1 " u 2 / c 2 )
or u = 0.14c
h
hc
1240eV nm
=
=
= 9.33 fm
2
" mu #&" mc (u / c )$' #(1.0106 ) 939 % 106 eV (0.14 )$
&
'
Nuclei are of this order of size and could be used to show the wave character of
10MeV neutrons.
Then, ! =
5-39.
(
)
h
h
# !t =
2
2!E
!16
6.58 " 10 eV s
#t =
= 1.32 " 10!24 s
2 " 250 " 106 eV
!E !t "
5-40. (a) For a proton or neutron:
!x!p "
#v =
h
and !p = m!v assuming the particle speed to be non-relativistic.
2
h
1.055 " 10!34 J s
=
= 3.16 " 107 m / s $ 0.1c (non!27
!15
2m#x 2 1.67 " 10 kg 10 m
(
)(
)
relativistic)
(
)(
1.67 " 10!27 kg 3.16 " 107 m / s
1 2
(b) Ek # mv =
2
2
2
)
= 8.34 " 10!13 J = 5.21MeV
(c) Given the proton or neutron velocity in (a), we expect the electron to be
relativistic,
in which case, Ek = mc 2 (! " 1) and
"p =
h
# ! mv
2"x
$
!v #
h
2m"x
For the relativistic electron we assume v ! c
! $
h
1.055 # 10"34 J s
=
= 193
2mc%x 2 9.11 # 10"31 kg 3.00 # 108 m / s 10"15 m
(
(
)(
)(
Ek = mc 2 (! " 1) = 9.11 # 10"31 kg 3.00 # 108 m / s
)(
2
)
) (192 ) = 1.58 # 10
"11
J = 98MeV