n 2 a0
4-13. (a) rn =
Z
r6 =
(Equation 4-18)
62 (0.053nm )
1
( )=
(b) r6 He
4-15.
+
= 1.91nm
62 (0.053nm )
2
" 1
1
1 #
= Z 2R % 2 $ 2 &
% n f ni &
!
'
(
= 0.95nm
(Equation 4-22)
"1 1 #
" n i2 $ 1 #
1
= R % 2 $ 2 & =R % 2 &
!ni
' 1 ni (
' ni (
!ni =
ni2
(
=
n i2
" n i2 #
=
91
.
17
nm
(
)$ 2 %
1.0968 ' 107 m n i2 & 1
( ni & 1 )
) (
R ni2 & 1
)(
)
4
9
(91.17nm ) = 121.57nm !3 = (91.17nm ) = 102.57nm
3
8
16
!4 = (91.17 nm ) = 97.25nm
!" = 91.17 nm
15
None of these are in the visible; all are in the ultraviolet.
!2 =
∞
4
3
2
λ, nm
80
90
100
110
120
130
4-19. (a)
au =
µe h 2
µe
h2
9.11 " 10!31 kg
=
=
a
=
(0.0529nm ) = 2.56 " 10!4 nm
0
µ µ ke 2 µ µ µe ke 2 µ µ
1.69 " 10!28 kg
(b) Eµ =
µµ k 2e4
2h 2
=
µ µ µe k 2 e 4 µ µ
1.69 " 10!28 kg
=
E
=
(13.6eV ) = 2520eV
0
µ e 2h 2
µe
9.11 " 10!31 kg
(c) The shortest wavelength in the Lyman series is the series limit
( ni = !, n f = 1 ). The photon energy is equal in magnitude to the ground
state energy ! Eµ .
!" =
hc 1240eV nm
=
= 0.492nm
Eµ
2520eV
(The reduced masses have been used in this solution.
4-21.
Energy (eV)
0
-2
n=∞
n=4
n=3
(d)
-4
(b)
n=2
(c)
-6
-8
-10
-12
-14
n=1
(a)
(a) Lyman limit, (b) H ! line, (c) H ! line, (d) longest wavelength line of Paschen
series
4-24. (a) The reduced mass correction to the Rydberg constant is important in this case.
1
#
R = R" &
(1+ m / M
En = !hcR / n 2
$
#1$
6
!1
' = R" & 2 ' = 5.4869 % 10 m
)
( )
(from Equation 4-26)
(from Equations 4-23 and 4-24)
(
)(
)
2
E1 = ! (1240eV nm ) 5.4869 " 106 m !1 10!9 m / nm / (1) = !6.804eV
Similarly, E2 = !1.701eV and E3 = !0.756eV
(b) Lyman α is the n = 2 ! n = 1 transition.
hc
= E2 # E1
"
$
"! =
hc
1240eV nm
=
= 243nm
E2 # E1 #1.701eV # (#6.804eV )
Lyman β is the n = 3 ! n = 1 transition.
"! =
hc
1240eV nm
=
= 205nm
E3 # E1 #0.756eV # (#6.804eV )
4-25. (a) The radii of the Bohr orbits are given by (see Equation 4-18)
r = n 2 a0 / Z where a0 = 0.0529nm and Z = 1 for hydrogen.
2
For n = 600, r = (600 ) (0.0529nm ) = 1.90 ! 104 nm = 19.0 µ m
This is about the size of a tiny grain of sand.
(b) The electron’s speed in a Bohr orbit is given by
v 2 = ke 2 / mr with Z = 1
Substituting r for the n = 600 orbit from (a), then taking the square root,
(
)(
v 2 = 8.99 " 109 N m 2 1.609 " 10!19 C
v 2 = 1.33 ! 107 m 2 / s 2
"
2
) / (9.11" 10
!31
)(
kg 19.0 " 10!6 m
)
v = 3.65 ! 103 m / s
For comparison, in the n = 1 orbit, v is about 2 ! 106 m / s
1
1 #
2" 1
= R (Z $ 1) % 2 $ 2 &
!
' n1 n2 (
4-26. (a)
"1
#
1 &$
2% 1
!3 = * 1.097 ' 107 m "1 (42 " 1) ( 2 " 2 ) + = 6.10 ' 10"11 m = 0.0610nm
, 1 3 -/
.
(
)
"1
#
1 &$
2% 1
!4 = * 1.097 ' 107 m "1 (42 " 1) ( 2 " 2 ) + = 5.78 ' 10"11 m = 0.0578nm
, 1 4 -/
.
(
)
"1
(b) !lim it
#
2% 1
&$
= * 1.097 ' 107 m "1 (42 " 1) ( 2 " 0 ) + = 5.42 ' 10"11 m = 0.0542nm
,1
-/
.
(
)
4-27.
1
1 $
1 $
2# 1
2# 1
= R (Z % 1) & 2 % 2 ' = R (Z % 1) & 2 % 2 ' for K!
"
(1 2 )
( n1 n2 )
1/ 2
#
$
1/ 2
%
&
#
$
1
1
& =%
&
Z "1 = %
% ! R '1 " 1 ( &
%* (0.0794nm ) 1.097 ) 10"2 / nm (3 / 4 )&+
%* ,. 4 -/ &+
Z = 1 + 39.1 ! 40 Zirconium
(
4-29.
)
n 2 a0
(Equation 4-18)
Z
The n =1 electrons “see” a nuclear charge of approximately Z ! 1 , or 78 for Au.
rn =
(
)(
)
r1 = 0.0529nm / 78 = 6.8 " 10!4 nm 10!9 m / nm 1015 fm / m = 680 fm , or about 100
times
the radius of the Au nucleus.
4-36.
"E =
hc 1240eV nm
=
= 1.610eV . The first decrease in current will occur when
!
790nm
the voltage reaches 1.61V.
4-43.
# % 1
1 &$
! = ( R * 2 " 2 +)
(. *, n f ni +- )/
Because R % µ,
µH =
"1
d!
'! =
'µ = " R "2
dµ
(
(
dR / d µ = R / µ. &! ' " R
me m p
me + m p
µD =
"2
"1
% 1
1 & dR
'µ
** 2 " 2 ++
n
n
d
µ
f
i
,
-
)
# 1
1 $
(( 2 " 2 ))
* n f ni +
)
"1
(R / µ )&µ = "! (&µ / µ )
me md
me + md
(
)
me md ! m p
m m / (me + md )
m / (me + md )
"µ µ D ! µ H µ D
=
=
!1 = e d
!1 = d
!1 =
µ
µH
µH
m p (me + md )
me m p / me + m p
m p / me + m p
(
)
If we approximate md = 2m p and me
(
md , then
)
m
!µ
" e and
µ
2m p
"! = #! ("µ / µ ) = # (656.3nm )
4-45. (a) En = ! E0 Z 2 / n 2
0.511MeV
= #0.179nm
2 (938.28MeV )
(Equation 4-20)
For Li++, Z = 3 and En = !13.6eV (9 )/ n 2 = !122.4 / n 2 eV
The first three Li++ levels that have the same (nearly) energy as H are:
n = 3, E3 = !13.6eV
n = 6, E6 = !3.4eV
n = 9, E9 = !1.51eV
Lyman α corresponds to the n = 6 → n = 3 Li++ transitions. Lyman β
corresponds
to the n = 9 → n = 3 Li++ transition.
(b) R (H ) = R" (1 / (1 + 0.511MeV / 938.8MeV )) = 1.096776 # 107 m !1
R (Li ) = R" (1 / (1 + 0.511MeV / 6535MeV )) = 1.097287 # 107 m !1
For Lyman α:
1
1 $
#
= R (H )'1 " 2 ( = 1.096776 % 107 m "1 10"9 m / nm (3 / 4 ) & 121.568nm
!
) 2 *
(
)
For Li++ equivalent:
1
#1 1 $
#1 1 $ 2
= R (Li )& 2 " 2 ' Z 2 = 1.097287 % 107 m "1 10"9 m / nm & " ' (3)
!
(3 6 )
( 9 36 )
(
! = 121.512nm
"! = 0.056nm
4-50. For He: En = !13.6eV Z 2 / n 2 = !54.4eV / n 2
(Equation 4-20)
(a)
0
5
3
-10
Energy (eV)
2
-20
-30
-40
-50
1
∞
4
E1 = -54.4 eV
E2 = -13.6 eV
E3 = -6.04 eV
E4 = -3.04 eV
E5 = -2.18 eV
E∞ = 0 eV
)
(b) Ionization energy is 54.5eV.
(c) H Lyman α: ! = hc / "E = 1240eV nm / (13.6eV # 3.4eV ) = 121.6nm
H Lyman β: ! = hc / "E = 1240eV nm / (13.6eV # 1.41eV ) = 102.6nm
He+ Balmer α: ! = hc / "E = 1240eV nm / (13.6eV # 6.04eV ) = 164.0nm
He+ Balmer β: ! = hc / "E = 1240eV nm / (13.6eV # 3.40eV ) = 121.6nm
#! = 42.4nm #" = 19.0nm
(The reduced mass correction factor does not change the energies calculated
above
to three significant figures.)
(d) En = !13.6eV Z 2 / n 2 because for He+, Z = 2, then Z 2 = 22. Every time n is an
even number a 22 can be factored out of n 2 and cancelled with the Z 2 = 22 in
the numerator; e.g., for He+,
E2 = !13.6eV 22 / 22 = !13.6eV
2
2
(H ground state)
E4 = !13.6eV 2 / 4 = !13.6eV / 2
2
(H ! 1st excited state)
E6 = !13.6eV 22 / 62 = !13.6eV / 32
M
(H ! 2nd excited state)
etc.
Thus, all of the H energy level values are to be found within the He+ energy
levels, so
He+ will have within its spectrum lines that match (nearly) a line in the H
spectrum.
ke 2
ke 2
=! 2
4-52. (a) En = !
2rn
2n ro
hf = En ! En !1 = !
En !1 = !
ke 2
ke 2 "
ke 2
$
!
!
2n 2 ro $ 2 (n ! 1)2 ro
&
(
2
2 (n ! 1) ro
#
%
%
'
)
2
2
ke 2 ! 1
1 " ke 2 n # n # 2n + 1
$
f =
# %=
2
2hro $ (n # 1)2 n 2 % 2hro
n 2 (n # 1)
&
'
ke 2 2n ! 1
ke 2
=
"
for n
2hro n 2 (n ! 1)2 ro hn3
(b) f rev =
v
2! r
2
f rev
=
"
1
v2
1 mv 2
1 ke 2
ke 2
=
=
=
4! 2 r 2 4! 2 mr r
4! 2 mr r 2
4! 2 mro3 n 6
(c) The correspondence principle implies that the frequencies of radiation and
revolution
are equal.
2
" ke 2 #
ke 2
2
f =$
=
= f rev
%
3
2
3 6
4! mro n
& ro hn '
2
2
ke 2 " hn3 #
h2
h2
ro =
=
=
$
%
4! 2 mn 6 & ke 2 '
4! 2 mke 2 mke 2
which is the same as ao in Equation 4-19.
4-53.
2
kZe 2 mv 2
=
r
r
"
1/ 2
# kZe 2 $
!v = %
&
( mr )
=
kZe 2 (! mv )
=
r2
mr
(from Equation 4-12)
v
1' " 2
" 2 $ kZe 2 % # $ kZe 2 %
c 2 ! 2 $ kZe 2 %
2
=
Therefore,
!
&c + (
(
)
)' = (
)
1 * ! 2 + mr ,
mr
+
, . + mr ,
-
!2 $
1 " kZe 2 #
'
(
c 2 ) mao *
%
! = 0.0075Z 1 / 2
E = KE % kZe 2 / r = mc 2 (! % 1) %
%
v = 0.0075cZ 1 / 2 = 2.25 & 106 m / s & Z 1 / 2
# 1
$ kZe 2
kZe 2
= mc 2 &
% 1' %
r
r
&( 1 % " 2
')
And substituting ! = 0.0075 and r = ao
!
"
1
#
E = 511 % 10 eV
& 1$ & 28.8Z eV
2
#
$
#' 1 & (0.0075 )
$(
3
= 14.4eV ! 28.8Z eV = !14.4 Z eV
4-57. Refer to Figure 4-16. All possible transitions starting at n = 5 occur.
n = 5 to n = 4, 3, 2, 1
n = 4 to n = 3, 2, 1
n = 3 to n = 2, 1
n = 2 to n = 1
Thus, there are 10 different photon energies emitted.
(Problem 4-57 continued)
ni
nf
fraction
no. of photons
5
4
1/4
125
5
3
1/4
125
5
2
1/4
125
5
1
1/4
125
4
3
1/ 4 !1/ 3
42
4
2
1/ 4 !1/ 3
42
4
1
1/ 4 !1/ 3
42
3
2
1 / 2 !#1 / 4 + 1 / 4 (1 / 3)"$
83
3
1
1 / 2 !#1 / 4 + 1 / 4 (1 / 3)"$
83
2
1
!(1 / 2 (1 / 4 + 1 / 4 )(1 / 3))+ 1 / 4 (1 / 3) + 1 / 4 "
#
$
250
Total = 1,042
Note that the number of electrons arriving at the n = 1 level (125+42+83+250) is 500,
as it should be.