Chapter 28
Atomic Physics
Answers to Even Numbered Conceptual
6.
Classically, the electron can occupy any energy state. That is, all energies would be
allowed. Therefore, if the electron obeyed classical mechanics, its spectrum, which
originates from transitions between states, would be continuous rather than discrete.
8.
The de Broglie wavelength of macroscopic objects such as a baseball moving with a typical
speed such as 30 m/s is very small and impossible to measure. That is, λ = h m v , is a very
small number for macroscopic objects. We are not able to observe diffraction effects
because the wavelength is much smaller than any aperture through which the object could
pass.
10. In both cases the answer is yes. Recall that the ionization energy of hydrogen is 13.6 eV. The electron can absorb a photon of energy less than 13.6 eV by making a transition to some intermediate state such as one with n = 2 . It can also absorb a photon of energy greater than 13.6 eV, but in doing so, the electron would be separated from the proton and have some residual kinetic energy. Problem Solutions
28.1
The Balmer equation is
4 ⎛ n2 ⎞
⎛ 1 1⎞
= R H ⎜ 2 − 2 ⎟ , or λ =
⎜
⎟
R H ⎝ n2 − 4 ⎠
λ
⎝2 n ⎠
1
When n = 3 ,
λ=
4
1.097 37 × 107 m
−1
⎛ 9 ⎞
−7
⎜
⎟ = 6.56 × 10 m = 656 nm
⎝ 9− 4 ⎠
When n = 4 ,
λ=
4
1.097 37 × 107 m
−1
⎛ 16 ⎞
−7
⎜
⎟ = 4.86 × 10 m = 486 nm
⎝ 16 − 4 ⎠
When n = 5 ,
λ=
4
1.097 37 × 107 m
−1
⎛ 25 ⎞
−7
⎜
⎟ = 4.34 × 10 m = 434 nm
⎝ 25 − 4 ⎠
423
424
CHAPTER 28
28.3
(a) From Coulomb’s law,
F=
ke q1q2
r2
( 8.99× 10
=
9
N ⋅m
2
C 2 )( 1.60 × 10−19 C )
(1.0× 10
−10
m
)
2
= 2.3× 10−8 N
2
(b) The electrical potential energy is
PE =
9
keq1q2 ( 8.99× 10 N ⋅ m
=
r
⎛
1 eV
= − 2.3× 10−18 J⎜
-19
⎝ 1.60 × 10
28.7
(a)
2
C 2 )( − 1.60× 10−19 C )( 1.60 × 10−19 C )
1.0 × 10−10 m
⎞
⎟ = − 14 eV
J⎠
rn = n2a0 yields r2 = 4( 0.0529 nm ) = 0.212 nm
(b) With the electrical force supplying the centripetal acceleration,
m evn2 kee2
kee2
m ekee2
and pn = m evn =
= 2 , giving vn =
rn
rn
m ern
rn
Thus,
m ekee2
p2 =
=
r2
( 9.11× 10
−31
kg )( 8.99× 109 N ⋅ m
2
C 2 )( 1.6 × 10−19 C )
0.212× 10−9 m
= 9.95× 10−25 kg ⋅ m s
(c)
⎛ h ⎞
⎛ 6.63× 10−34 J⋅ s⎞
−34
Ln = n ⎜
⎟ → L2 = 2⎜
⎟ = 2.11× 10 J⋅ s
π
π
2
2
⎝
⎠
⎝
⎠
9.95× 10−25 kg ⋅ m s)
(
p22
1
2
=
= 5.44× 10−19 J= 3.40 eV
(d) KE2 = m ev2 =
−31
2
2m e
2( 9.11× 10 kg )
2
(e)
8.99× 109 N ⋅ m 2 C 2 )( 1.60 × 10−19 C )
(
ke ( − e) e
=−
PE2 =
r2
( 0.212× 10−9 m )
= −1.09× 10−18 J= − 6.80 eV
2
2
Atomic Physics
(f)
28.10
E2 = KE2 + PE2 = 3.40 eV − 6.80 eV = − 3.40 eV
⎛ 1
1⎞
= RH ⎜ 2 − 2 ⎟
⎜ n f ni ⎟
λ
⎝
⎠
1
(b) From
1
RH
or λ =
λ=
28.15
(a)
E=
hc
(c)
f=
c
λ
λ
⎛ ni2n2f ⎞
⎜ 2
⎟ with ni = 6 and n f = 2
⎜ ni − n2f ⎟
⎝
⎠
1
1.097 37 × 107 m
( 6.63× 10
−34
=
=
−1
⎡ ( 36)( 4) ⎤
−7
⎢
⎥ = 4.10× 10 m = 410 nm
36
4
−
⎣
⎦
J⋅ s)( 3.00× 108 m s)
410× 10− 9 m
= 4.85× 10− 19 J= 3.03 eV
3.00 × 108 m s
= 7.32× 1014 H z
−9
410 × 10 m
⎛ 1
1⎞
= R H ⎜ 2 − 2 ⎟ , it is seen that (for a fixed value of n f ) λm ax occurs when
⎜ n f ni ⎟
λ
⎝
⎠
ni = n f + 1 and λm in occurs when ni → ∞ .
From
1
(
)
(a) For the Lyman series n f = 1 ,
1
λm ax
= ( 1.097 37 × 107 m
−1
= ( 1.097 37 × 107 m
−1
) ⎛⎜⎝ 11 − 21 ⎞⎟⎠ → λ
= 1.22× 10−7 m = 122 nm
) ⎛⎜⎝ 11 − ∞1 ⎞⎟⎠ → λ
= 9.11× 10−8 m = 91.1 nm
2
m ax
2
and
1
λm in
(b)
(
)
For the Paschen series n f = 3 ,
1
λm ax
and
m in
2
= ( 1.097 37 × 107 m
−1
) ⎛⎜⎝ 31 − 41 ⎞⎟⎠ → λ
2
2
m ax
= 1.87 × 10−6 m = 1.87 × 103 nm
425
426
CHAPTER 28
28.17
The batch of excited atoms must make these six transitions to get back to the ground
state: ni = 2 → n f = 1, also ni = 3 → n f = 2 and ni = 3 → n f = 1, and also ni = 4 → n f = 3
and ni = 4 → n f = 2 and ni = 4 → n f = 1 . Thus, the incoming light must have just enough
energy to produce the ni = 1 → n f = 4 transition. It must be the third line of the Lyman
series in the absorption spectrum of hydrogen. The incoming photons must have
wavelength given by
16
16
⎛ 1 1 ⎞ 15R H
=
or λ =
= RH ⎜ 2 − 2 ⎟ =
15R H 15( 1.097 37 × 107 m
16
λ
⎝1 4 ⎠
1
−1
)
= 97.2 nm
Z 2 ( 13.6 eV )
( 3) ( 13.6 eV )
, E1 = −
= − 122 eV
2
2
n
( 1)
2
28.27
(a) From En = −
n2a0
( 1) a0 0.0529× 10−9 m
=
= 1.76 × 10−11 m
gives r1 =
3
3
Z
2
(b) Using rn =
28.33
In the 3p subshell, n = 3 and l = 1 . The 6 possible quantum states are
m l = +1
m s = ± 12
l =1
ml =0
m s = ± 12
l =1
m l = −1
m s = ± 12
l =1
(a) For a given value of the principle quantum number n, the orbital quantum number
l varies from 0 to n − 1 in integer steps. Thus, if n = 4 , there are 4 possible
n= 3
n= 3
n= 3
28.34
values of l : l = 0,1,2, and 3
(b) For each possible value of the orbital quantum number l , the orbital magnetic
quantum number m l ranges from −l to + l in integer steps. When the principle
quantum number is n = 4 and the largest allowed value of the orbital quantum
number is l = 3 , there are 7 distinct possible values for m l . These values are:
m l = −3,− 2,− 1,0,+ 1,+ 2, and + 3
28.36
(a) The electronic configuration for oxygen ( Z = 8) is 1s2 2s2 2p4
(b) The quantum numbers for the 8 electrons can be:
1s states
2s states
n=1
n= 2
l=0
l=0
2p states
n= 2
l =1
ml =0
ml =0
ml =0
ml =1
m s = ± 12
m s = ± 12
m s = ± 12
m s = ± 12
427
Atomic Physics
28.37
(a) For Electron #1 and also for Electron #2, n = 3 and l = 1 . The other quantum
numbers for each of the 30 allowed states are listed in the tables below.
Electron #1
ml
+1
ms
+ 12
ml
+1
ms
+ 12
ml
+1
ms
+ 12
ml
+1
ms
− 12
ml
+1
ms
− 12
ml
+1
ms
− 12
Electron #2
+1
− 12
0
± 12
-1
± 12
+1
+ 12
0
± 12
-1
± 12
Electron #1
ml
0
ms
+ 12
ml
0
ms
+ 12
ml
0
ms
+ 12
ml
0
ms
− 12
ml
0
ms
− 12
ml
0
ms
− 12
Electron #2
+1
± 12
0
− 12
-1
± 12
+1
± 12
0
+ 12
-1
± 12
Electron #1
ml
-1
ms
+ 12
ml
-1
ms
+ 12
ml
-1
ms
+ 12
ml
-1
ms
− 12
ml
-1
ms
− 12
ml
-1
ms
− 12
Electron #2
+1
± 12
0
± 12
-1
− 12
+1
± 12
0
± 12
-1
+ 12
There are 30 allow ed states , since Electron #1 can have any of three possible
values of m l for both spin up and spin down, totaling six possible states. For each
of these states, Electron #2 can be in either of the remaining five states.
(b) Were it not for the exclusion principle, there would be 36 possible states, six
for each electron independently.
28.41
For nickel, Z = 28 and
EK ≈ − ( Z − 1)
2
EL ≈ − ( Z − 3)
2
13.6 eV
2
= − ( 27) ( 13.6 eV ) = − 9.91× 103 eV
2
( 1)
6 eV )
13.6 eV
2 ( 13.
= − ( 25)
= − 2.13× 103 eV
2
4
( 2)
Thus, Eγ = EL − EK = −2.13 keV − ( −9.91 keV ) = 7.78 keV
and
λ=
−34
8
hc ( 6.63× 10 J⋅ s)( 3.00 × 10 m s)
=
= 1.60× 10−10 m = 0.160 nm
Eγ
7.78 keV ( 1.60 × 10−16 J keV )
428
CHAPTER 28