3-37
When waves are scattered between two adjacent planes of a single crystal, constructive
wave interference will occur when the path length difference between such reflected
waves is an integer multiple of wavelengths. This condition is expressed by the Bragg
equation for constructive interference, !!" " #$% " = ## where d is the distance between
adjacent crystalline planes, " is the angle of incidence of the x-ray beam of photons, n is
an integer for constructive interference, and " is the wavelength of the photon beam
which is in this case, 0.062 6 nm. Ignoring the incident beam that is not scattered, the first
three angles for which maxima of x-ray intensities are found are !!"" = # " $%& #" or
% %'
#
!*
'()&) $ %'
=
%%'
&"
+ $ %'
!*
" = '(',+ - !./0 #/$" = 1(12 °
!! %
"#$ "% =
or
!!" " = " " #$% # "
# &'(%( $ )& !*
"#$ " % =
=
= &'),( , , !!" = "#$$ °
%)&
"
+'& $ )& !*
!!
%)&
or
!!" " = # " $%& # "
(
)
% # % '()&) $ *' !+
"#$ " % =
=
= '(&%- ./ , !!"" = #"$%°
% *'
&"
, $ *'
!+
!!
!
!
!
!
%*'
4-1
F corresponds to the charge passed to deposit one mole of monovalent element at
a cathode. As one mole contains Avogadro’s number of atoms,
e=
4-2
96 500 C
6.02 " 10
(a)
23
= 1.60 " 10 #19 C .
Total charge passed = i * t = (1.00 A) (3 600 s) = 3 600 C . This is
3 600 C
1.60 " 10
#19
C
= 2.25 " 1022 electrons.
As the valence of the copper ion is two, two electrons are required to deposit
each ion as a neutral atom on the cathode.
The number of Cu atoms =
number of electrons
= 1.125 " 10 22 Cu atoms.
2
(b)
So the weight (mass) of a Cu atom is:
(c)
m=q
molar weight
(2 ) or
96 500
molar weight = m( 96 500)
4-3
1.185 g
= 1.05 " 10 #22 g .
1.125 " 1022 atoms
2
2
= (1.185 g )(96 500 C )
= 63.53 g .
q
3 600 C
Thomson’s device will work for positive and negative particles, so we may apply
q
V#
" 2 .
m B ld
(a)
(
)
(b)
As the particle is attracted by the negative plate, it carries a positive charge and is
$ q
'
1.60 " 10 #19 C
=
= 9.58 " 107 C kg)
a proton. &
#27
kg
% mp 1.67 " 10
(
(c)
vx =
(d)
As vx ~ 0.01c there is no need for relativistic mechanics.
!
4-8
q
V#
0.20 radians
7
"
= (2 000 V)
2 ( 0.10 m )(0.02 m ) = 9.58 $ 10 C kg
m B2 ld
%2
4.57 $ 10 T
(a)
E V 2 000 V
=
=
4.57 " 10 #2 T = 2.19 " 106 m s
B dB 0.02 m
(
)
% sin $ (
From Equation 4.16 we have " n#'
& 2 *)
+4
or " n2 = " n1
sin# 1 4
2
sin# 4
( )
( )
. Thus the
2
2
number of α’s scattered at 40 degrees is given by
4
4
sin 20
(
# sin 10 &
2 )
(
)
" n2 = (100 cpm)
=
100
cpm
= 6.64 cpm .
%
(
4
$ sin 20 '
( sin 402 )
Similarly
" n at 60 degrees = 1.45 cpm
" n at 80 degrees = 0.533 cpm
" n at 100 degrees = 0.264 cpm
(b)
" 1%
2
From 4.16 doubling $ ' m( v( reduces " n by a factor of 4. Thus " n at
# 2&
" 1%
20 degrees = $ ' (100 cpm) = 25 cpm .
# 4&
(c)
From 4.16 we find
2
" nCu ZCu
N
= 2 Cu , ZCu = 29 , ZAu = 79 .
" nAu ZAu NAu
NCu = number of Cu nuclei per unit area
= number of Cu nuclei per unit volume * foil thickness
)
# 6.02 " 1023 nuclei & ,
22
= + 8.9 g cm 3 %
( .t = 8.43 " 10 t
63.54 g
+*
$
' .)
# 6.02 " 10 23 nuclei & ,
22
= + 19.3 g cm 3 %
( .t = 5.90 " 10 t
197.0 g
+*
$
' .-
(
NAu
)
(
So " nCu = " n Au ( 29)
)
2
8.43 # 10 22
(79 )2
2
(
$ 29 ' $ 8.43 '
5.90 # 10 2 = (100)& ) &
= 19.3 cpm .
% 79 ( % 5.90 )(
)
4-9
The initial energy of the system of α plus copper nucleus is 13.9 MeV and is just
the kinetic energy of the α when the α is far from the nucleus. The final energy of the
system may be evaluated at the point of closest approach when the kinetic energy is zero
and the potential energy is k( 2e)
!
!
Ze
r
where r is approximately equal to the nuclear radius
of copper. Invoking conservation of energy Ei = E f , K" = ( k )
2
(
)(
2Ze 2
or
r
)
)
(2)( 29) 1.60 # 10 $19 8.99 # 109
2Ze 2
r = (k )
=
= 6.00 # 10 $15 m .
K"
13.9 # 106 eV 1.60 # 10 $19 J eV
!
!
(
)(