1.24 "103
1.24 "103
nm =
= 0.016nm
3-34. Equation 3-24: !m =
V
80 "103V
3-36.
!2 # !1 =
6.63 $10#34 J s )(1 # cos110° )
(
h
1
#
cos
"
=
= 3.26 $10#12 m
(
)
#31
8
mc
(9.11$10 kg )(3.00 $10 m / s )
6.63 #10"34 J s )(3 #108 m / s )
(
hc
!1 =
=
= 2.43 #10"12 m
6
"19
E1 (0.511#10 eV )(1.60 #10 J / eV )
!2 = !1 + 3.26 #10"12 m = (2.43 + 3.26 )#10"12 m = 5.69 #10"12 m
E2 =
hc 1240eV nm
=
= 2.18 #105 eV = 0.218MeV
!2 5.69 #10"3 nm
Electron recoil energy Ee = E1 ! E2 (Conservation of energy)
Ee = 0.511MeV ! 0.218MeV = 0.293MeV . The recoil electron momentum makes
an angle θ with the direction of the initial photon.
PE
θ
h/λ1
110°
h/λ2
20°
2
h
cos 20° = pe sin ! = (1/ c ) E 2 # (mc 2 ) sin !
"2
(3.00 #10 m / s )(6.63 #10 J s )cos 20°
m )$(0.804 MeV ) " (0.511MeV ) % (1.60 #10
&
'
8
sin ! =
(5.69 #10
"12
(Conservation of momentum)
= 0.330 or ! = 19.3°
"34
2
2
1/ 2
"13
J / MeV )
3-37.
#! = !2 $ !1 = #! =
!1 = (100 )
3-38. (a) E1 =
h
(1 # cos " ) = (100 )(0.00243nm )(1 # cos 90° ) = 0.243nm
mc
hc 1240eV nm
=
= 1.747 "104 eV
!1
0.0711nm
(b) !2 = !1 +
(c) E2 =
h
(1 $ cos " ) = 0.01!1 Equation 3-25
mc
h
(1 # cos " ) = 0.0711 + (0.00243nm )(1 # cos180° ) = 0.0760nm
mc
hc 1240eV nm
=
= 1.634 "104 eV
!2
0.0760nm
(d) Ee = E1 ! E2 = 1.128 "103 eV
3-41. (a) Compton wavelength =
electron:
proton:
(b) E =
h
mc
h
6.63 "10!34 J s
=
= 2.43 "10!12 m = 0.00243nm
mc (9.11"10!31 kg )(3.00 "108 m / s )
h
6.63 "10!34 J s
=
= 1.32 "10!15 m = 1.32 fm
!27
8
mc (1.67 "10 kg )(3.00 "10 m / s )
hc
!
(i) electron: E =
(ii) proton: E =
1240eV nm
= 5.10 !105 eV = 0.510 MeV
0.00243nm
1240eV nm
= 9.39 "108 eV = 939 MeV
!6
1.32 "10 nm
3-50. (a) !mT = 2.898 # 10"3 m K
(b) Equation 3-18:
$ T=
2.898 # 10"3 m K
= 3.50 # 104 K
"9
82.8 # 10 m
!5
(70nm ) / (ehc /(70 nm)kT ! 1)
=
u (82.8nm ) (82.8nm )!5 / (e hc / (82.8 nm )kT ! 1)
u (70nm )
(
)(
)
6.63 " 10!34 J s 3.00 " 108 m / s
hc
=
= 5.88 and
where
(70nm )kT 70 " 10!9 m 1.38 " 10!23 J / K 3.5 " 104 K
(
)(
)
!5
(70nm ) / (e5.88 ! 1)
=
= 0.929
u (82.8nm ) (82.8nm )!5 / (e 4.97 ! 1)
u (70nm )
hc
= 4.97
(82.8nm )kT
Similarly,
)(
!5
(100nm ) / (e4.12 ! 1)
=
= 0.924
u (82.8nm ) (82.8nm )!5 / (e 4.97 ! 1)
u (100nm )
8" hc! #5
Letting C = 8" hc and a = hc / kT
3-53. (a) Equation 3-18: u (! ) = hc / ! kT
e
#1
C ! "5
gives u (! ) = a / !
e "1
(b)
# ! "5 ("1)e a / ! "a! "2
$
du
d # C ! "5 $
5! "6 &
%
=
" a/!
%
& =C%
2
a/!
d ! d ! ' ea / ! " 1(
e " 1&
e
"
1
'
(
"6
"6 a / !
C!
# a a/!
$ C! e
#a
$
=
e " 5 ea / ! " 1 & =
" 5 1 " ea / ! & = 0
2 %
2 %
a/!
a
/
!
(
(
e "1 '!
e "1 '!
(
)
(
(
(
)
)
)
(
(
)
)
The maximum corresponds to the vanishing of the quantity in brackets.
Thus, 5! 1 " e " a / ! = a
(
)
(c) This equation is most efficiently solved by trial and error; i.e., guess at a value
(
)
for a / ! in the expression 5! 1 " e " a / ! = a , solve for a better value of a / ! ;
substitute the new value to get an even better value, and so on. Repeat the
process until the calculated value no longer changes. One succession of values
is 5, 4.966310, 4.965156, 4.965116, 4.965114, 4.965114. Further iterations
repeat the same value (to seven digits), so we have
(
a
hc
= 4.965114 =
!m
!m kT
)(
6.63 # 10"34 J s 3.00 # 108 m / s
hc
=
(d) !mT =
(4.965114 )k (4.965114 ) 1.38 # 10"23 J / K
(
Therefore, !mT = 2.898 # 10"3 m K
Equation 3-5
)
)
4-1.
1
1 $
# 1
= R & 2 " 2 ' where R = 1.097 % 107 m "1 (Equation 4-2)
!mn
n )
(m
The Lyman series ends on m =1, the Balmer series on m =2, and the Paschen
series on
1
m =3. The series limits all have n = !, so = 0 .
n
1
1
# $
= R & 2 ' = 1.097 % 107 m "1
!L
(1 )
!L (limit ) = 1.097 # 107 m "1 = 91.16 # 10"9 m = 91.16nm
1
# 1 $
= R & 2 ' = 1.097 % 107 m "1 / 4
!B
(2 )
!B (limit ) = 4 / 1.097 # 107 m "1 = 3.646 # 10"7 m = 364.6nm
1
#1$
= R & 2 ' = 1.097 % 107 m "1 / 9
!P
(3 )
!P (limit ) = 9 / 1.097 # 107 m "1 = 8.204 # 10"7 m = 820.4nm
4-2.
1
1 #
" 1
= R % 2 $ 2 & where m = 2 for Balmer series (Equation 4-2)
!mn
n (
'm
1
1.097 " 107 m !1 # 1
1 $
=
! 2&
%
9
2
379.1nm
10 nm / m ' 2
n (
1 1
109 nm / m
! 2 =
= 0.2405
4 n
379.1nm 1.097 " 107 m !1
(
)
1
= 0.2500 ! 0.2405 = 0.0095
n2
n2 =
1
0.0095
!
1/ 2
n = (1 / 0.0095 )
= 10.3
!
n = 10
n = 10 ! n = 2
4-3.
1
1 #
" 1
= R % 2 $ 2 & where m = 1 for Lyman series (Equation 4-2)
!mn
n (
'm
1
1.097 " 107 m !1 #
1 $
=
1! 2 &
%
9
164.1nm
10 nm / m ' n (
1
109 nm / m
=
1
!
= 1 ! 0.5555 = 0.4445
n2
164.1nm 1.097 " 107 m !1
(
1/ 2
n = (1 / 0.4445 )
)
= 1.5
No, this is not a hydrogen Lyman series transition because n is not an
integer.
4-4.
1
1 #
" 1
= R% 2 $ 2 &
!mn
n (
'm
(Equation 4-2)
For the Brackett series m = 4 and the first four (i.e., longest wavelength lines have
n = 5,
6, 7, and 8.
1
1$
# 1
= 1.097 % 107 m "1 & 2 " 2 ' = 2.468 % 105 m "1
!45
(4 5 )
!45 =
1
= 4.052 # 10"6 m = 4052nm. Similarly,
5
"1
2.68 # 10 m
!46 =
1
= 2.625 # 10"6 m = 2625nm
3.809 # 105 m "1
!47 =
1
= 2.166 # 10"6 m = 2166nm
4.617 # 105 m "1
!48 =
1
= 1.945 # 10"6 m = 1945nm
5
"1
5.142 # 10 m
These lines are all in the infrared.
4-7.
"N #
1
A
(From Equation 4-6), where A is the product of
=
4
sin (! / 2 ) sin (! / 2 )
4
the two
quantities in parentheses in Equation 4-6.
(a)
"N (10° )
"N (1° )
=
A / sin4 (10° / 2 )
4
A / sin (1° / 2 )
=
sin4 (0.5° )
4
sin (5° )
= 1.01 # 10!4
(b)
4-9.
rd =
"N (30° )
"N (1° )
sin4 (15° )
kq! Q
ke 2 2 79
=
Ek!
(1 / 2 )m! v 2
(1.44MeV
rd =
For Ek! = 7.7 MeV :
rd = 29.5 fm
rd =
(1.44MeV
fm )(2 )(79 )
5.0 MeV
= 45.5 fm
rd = 19.0 fm
kq! Q
ke 2 2 79
=
Ek!
(1 / 2 )m! v 2
Ek! =
= 1.29 # 10!6
(Equation 4-11)
For Ek! = 5.0 MeV :
For Ek! = 12 MeV :
4-10.
=
sin4 (0.5° )
fm )(2 )(13)
4 fm
(Equation 4-11)
= 9.4 MeV
4-40. Those scattered at ! = 180° obeyed the Rutherford formula. This is a head-on
collision where the α comes instantaneously to rest before reversing direction. At
that point its kinetic energy has been converted entirely to electrostatic potential
energy, so
k (2e )(79e )
1
m! v 2 = 7.7 MeV =
where r = upper limit of the nuclear radius.
2
r
r=
k (2 )(79 )e 2
7.7 MeV
=
2 (79 )(1.440 MeV fm )
7.7 MeV
= 29.5 fm