Chapter 28
Atomic Physics
Quick Quizzes
1.
(b). The allowed energy levels in a one-electron atom may be expressed as
En = − Z 2 ( 13.6 eV ) n2 , where Z is the atomic number. Thus, the ground state ( n = 1 level )
in helium, with Z = 2 , is lower than the ground state in hydrogen, with Z = 1 .
2.
(a). The energy of the photon emitted when the electron in a one-electron atom makes a
transition from a state having principal quantum number ni to one having principal
quantum number n f is
 1
1 
Eγ = Z 2 ( 13.6 eV )  2 − 2 
 n f ni 


Thus, for given values of ni and n f , the energy of the photon emitted by a helium atom,
with Z = 2 , is four times that of the photon emitted when an electron makes the
corresponding transition in a hydrogen atom, with Z = 1 .
3.
(a) For n = 5 , there are 5 allowed values of , namely = 0, 1, 2, 3, and 4.
(b) Since m ranges from – to + in integer steps, the largest allowed value of ( = 4 in
this case) permits the greatest range of values for m . For n = 5 , there are 9 possible
values for m : -4, -3, -2, –1, 0, +1, +2, +3, and +4.
(c) For each value of , there are 2 + 1 possible values of m . Thus, there is 1 distinct
pair with = 0 ; 3 distinct pairs possible with = 1 ; 5 distinct pairs with = 2 ; 7
distinct pairs with = 3 ; and 9 distinct pairs with = 4 . This yields a total of 25
distinct pairs of and m that are possible when n = 5 .
4.
(d). Krypton has a closed configuration consisting of filled n=1, n=2, and n=3 shells as
well as filled 4s and 4p subshells. The filled n = 3 shell (the next to outer shell in Krypton)
has a total of 18 electrons, 2 in the 3s subshell, 6 in the 3p subshell and 10 in the 3d
subshell.
423
424
CHAPTER 28
Answers to Even Numbered Conceptual Questions
2.
Neon signs do not emit a continuous spectrum. They emit many discrete wavelengths as
could be determined by observing the light from the sign through a spectrometer.
However, they do not emit all wavelengths. The specific wavelengths and intensities
account for the color of the sign.
4.
An atom does not have to be ionized to emit light. For example, hydrogen emits light
when a transition carries an electron from a higher state to the n = 2 state.
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 mv , 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.
12.
It replaced the simple circular orbits in the Bohr theory with electron clouds. More
important, quantum mechanics is consistent with Heisenberg’s uncertainty principle,
which tells us about the limits of accuracy in making measurements. In quantum
mechanics, we talk about the probabilistic nature of the outcome of a measurement of a
system, a concept which is incompatible with the Bohr theory. Finally, the Bohr theory of
the atom contains only one quantum number n, while quantum mechanics provides the
basis for additional quantum numbers to explain the finer details of atomic structure.
14.
Each of the given atoms has a single electron in an = 0 ( or s ) state outside a fully closedshell core, shielded from all but one unit of the nuclear charge. Since they reside in very
similar environments, one would expect these outer electrons to have nearly the same
electrical potential energies and hence nearly the same ionization energies. This is in
agreement with the given data values. Also, since the distance of the outer electron from
the nuclear charge should tend to increase with Z (to allow for greater numbers of
electrons in the core), one would expect the ionization energy to decrease somewhat as
atomic number increases. This is also in agreement with the given data.
16.
One assumption is natural from the standpoint of classical physics: The electron feels an
electric force of attraction which supplies the centripetal acceleration and holds it in orbit.
The other assumptions are in sharp contrast to the behavior of ordinary-size objects: The
electron’s angular momentum must be one of a set of certain special allowed values.
During the time when it is in one of these quantized orbits, the electron emits no
electromagnetic radiation. The atom radiates a photon when the electron makes a
quantum jump from one orbit to a lower one.
Atomic Physics
18.
(a) n, l, and ml are integers; ms is fractional
(b) n and l are always positive; ml and ms can be negative
(c) lmax = n − 1 = 1
(d) ml can have values of − 1, 0, or 1
425
426
CHAPTER 28
Answers to Even Numbered Problems
2.3 × 10 2 N
(b)
1.4 MeV
2.19 × 106 m s
(b)
13.6 eV
(c)
–27.2 eV
(a)
3.03 eV
(b)
410 nm
(c)
7.32 × 1014 Hz
12.
(a)
transition II
(b)
transition I
(c)
transitions II and III
14.
(a)
12.1 eV
(b)
12.1 eV, 10.2 eV, and 1.89 eV
20.
(a)
6
(b)
1.88 × 10 3 nm (in the Paschen series)
22.
(a) 1.52 × 10 −16 s
(b) 8.23 × 109 revolutions
(c) Yes, for 8.23 × 109 ª electron yearsº
(d) The electron moves so quickly that it can never meaningfully be said to be on any
particular side of the nucleus.
24.
4.43 × 10 4 m s
26.
(a)
2.89 × 10 34 kg ⋅ m 2 s
(b)
2.74 × 1068
28.
(a)
En = − 54.4 eV n 2
(b)
54.4 eV
30.
(a)
4.42 × 107 m −1
(b)
30.1 nm
34.
(a)
4
(b)
7
36.
(a)
1s2 2s2 2 p 4
(b)
( n = 1,
= 0, m = 0, ms = ± 12 ); ( n = 2,
= 0, m = 0, ms = ± 12 )
( n = 2,
= 1, m = 0, ms = ± 12 ); ( n = 2,
= 1, m = 1, ms = ± 12 )
4.
(a)
6.
45 fm
8.
(a)
10.
2
38.
(a)
(b)
40.
0.155 nm, 8.03 kV
42.
Z = 32 , germanium
44.
137
8
(c)
18
(d)
(c)
7.30 × 10 −69
(c)
ultraviolet
32
(e)
50
Atomic Physics
(b)
1.05 × 1019 photons
(b)
1 2πα
427
(a)
4.20 mm
(c)
8.82 × 1016 photons mm 3
48.
(a)
137
50.
The simplest diagram has 4 states with energies of -4.100 eV, -1.000 eV, -0.1000 eV, and 0.
52.
(a) 135 eV
(b) ≈ 10 times the magnitude of the ground state energy of hydrogen.
54.
when n → ∞ , f → fclassical = 4π 2 me ke2 e 4 h 3 n3
56.
(a)
2.56 × 10 −4 nm
(b)
− 2.82 × 10 3 eV, − 704 eV, − 313 eV
58.
(a)
nf = 1
(b)
ni = 3
46.
(c)
4π α
428
CHAPTER 28
Problem Solutions
28.1
The Balmer equation is
1
 1
= RH  2 − 2
λ
n
2
1
4  n2 

=
λ
,
or



RH  n2 − 4 

When n = 3 ,
λ=
4
 9 
−7
 = 6.56 × 10 m = 656 nm
7
−1 
1.097 37 × 10 m  9 − 4 
When n = 4 ,
λ=
4
 16 
−7
 = 4.86 × 10 m = 486 nm
7
−1 
1.097 37 × 10 m  16 − 4 
When n = 5 ,
λ=
28.2
4
 25 
−7
 = 4.34 × 10 m = 434 nm
7
−1 
1.097 37 × 10 m  25 − 4 
Start with Balmer’s equation,
or
λ=
 n2 − 4 
1 
 1
= RH  2 − 2  = RH 
2 
λ
n 
2
 4n 
1
4  n2 


RH  n 2 − 4 
Substituting RH =1.097 3732 × 107 m − 1 , we obtain
( 3.645 × 10
λ=
−7
m ) n2
n2 − 4
28.3
=
364.5 n2
nm where n = 3, 4, 5, . . . .
n2 − 4
(a) From Coulomb’s law,
F=
k e q1 q2
r2
( 8.99 × 10
=
9
N ⋅ m 2 C 2 )( 1.60 × 10 −19 C )
(1.0 × 10
−10
m)
2
2
= 2.3 × 10 −8 N
Atomic Physics
(b) The electrical potential energy is
9
2
2
−19
−19
k e q1 q2 ( 8.99 × 10 N ⋅ m C )( − 1.60 × 10 C )( 1.60 × 10 C )
=
PE =
1.0 × 10 −10 m
r

1 eV
= − 2.3 × 10 −18 J 
-19
 1.60 × 10
28.4

 = − 14 eV
J
(a) From Coulomb’s law,
9
2
2
−19
k e q1 q2 ( 8.99 × 10 N ⋅ m C )( 1.60 × 10 C )
=
= 2.3 × 10 2 N
F=
2
2
−
15
r
(1.0 × 10 m )
2
(b) The electrical potential energy is
9
2
2
−19
k e q1 q2 ( 8.99 × 10 N ⋅ m C )( 1.60 × 10 C )
=
PE =
1.0 × 10 −15 m
r
2
 1 MeV 
= 2.3 × 10 −13 J 
 = + 1.4 MeV
-13
 1.60 × 10 J 
28.5
(a) The electrical force supplies the centripetal acceleration of the electron, so
m
k e2
v 2 ke e 2
= 2 or v = e
r
r
mr
( 8.99 × 10 N ⋅ m C )(1.60 × 10 C )
( 9.11 × 10 kg )(1.0 × 10 m )
9
v=
(b)
No.
2
−31
−19
2
−10
2
= 1.6 × 106 m s
v 1.6 × 10 6 m s
=
= 5.3 × 10 −3 << 1 , so the electron is not relativistic.
c 3.00 × 108 m s
(c) The de Broglie wavelength for the electron is λ =
λ=
(d)
h
h
=
, or
p mv
6.63 × 10 −34 J ⋅ s
= 4.6 × 10 −10 m = 0.46 nm
−31
6
( 9.11 × 10 kg )(1.6 × 10 m s )
Yes. The wavelength and the atom are roughly the same size.
429
430
CHAPTER 28
28.6
Assuming a head-on collision, the α-particle comes to rest momentarily at the point of
closest approach. From conservation of energy,
KE f + PE f = KEi + PEi , or 0 +
k e ( 2 e )( 79 e )
rf
= KEi +
k e ( 2 e )( 79 e )
ri
With ri → ∞ , this gives the distance of closest approach as
−19
9
2
2
158 k e e 2 158 ( 8.99 × 10 N ⋅ m C )( 1.60 × 10 C )
rf =
=
KEi
5.0 MeV ( 1.60 × 10 -13 J MeV )
2
= 4.5 × 10 −14 m = 45 fm
28.7
(a)
rn = n2 a0 yields r2 = 4 ( 0.052 9 nm ) = 0.212 nm
(b) With the electrical force supplying the centripetal acceleration,
me vn2 ke e 2
ke e 2
me k e e 2
= 2 , giving vn =
and pn = me vn =
rn
rn
me rn
rn
Thus,
me k e e 2
=
p2 =
r2
( 9.11 × 10
−31
kg )( 8.99 × 10 9 N ⋅ m 2 C 2 )( 1.6 × 10 −19 C )
0.212 × 10 −9 m
= 9.95 × 10 −25 kg ⋅ m s
(c)
 h
Ln = n 
 2π

 6.63 × 10 −34 J ⋅ s 
−34
→
L
=
2


 = 2.11 × 10 J ⋅ s
2
2
π



9.95 × 10 −25 kg ⋅ m s )
(
p22
1
2
=
= 5.44 × 10 −19 J = 3.40 eV
(d) KE2 = me v2 =
2
2 me
2 ( 9.11 × 10 −31 kg )
2
(e)
( 8.99 × 109 N ⋅ m2 C2 )(1.60 × 10−19 C )
k ( −e ) e
=−
PE2 = e
r2
( 0.212 × 10−9 m )
= −1.09 × 10 −18 J = − 6.80 eV
(f)
E2 = KE2 + PE2 = 3.40 eV − 6.80 eV= − 3.40 eV
2
2
Atomic Physics
28.8
(a) With the electrical force supplying the centripetal acceleration,
me vn2 ke e 2
ke e 2
= 2 , giving vn =
rn
rn
me rn
where rn = n2 a0 = n 2 ( 0.052 9 nm )
Thus,
ke e 2
=
v1 =
me r1
( 8.99 × 10 N ⋅ m C )(1.60 × 10 C )
( 9.11 × 10 kg )( 0.052 9 × 10 m )
9
2
−19
2
−31
2
= 2.19 × 10 6 m s
−9
9
2
2
−19
k e 2 ( 8.99 × 10 N ⋅ m C )( 1.60 × 10 C )
1
(b) KE1 = me v12 = e =
2
2 r1
2 ( 0.052 9 × 10 −9 m )
= 2.18 × 10 −18 J = 13.6 eV
(c)
( 8.99 × 109 N ⋅ m2 C2 )(1.60 × 10−19 C )
k ( −e ) e
=−
PE1 = e
r1
( 0.052 9 × 10−9 m )
2
= − 4.35 × 10 −18 J = − 27.2 eV
28.9
Since the electrical force supplies the centripetal acceleration,
me vn2 ke e 2
k e2
= 2 or vn2 = e
rn
rn
me rn
From Ln = me rn vn = n , we have rn =
vn2 =
n
, so
me vn
k e e 2  me vn 
ke e 2
which
reduces
to
v
=


n
me  n 
n
2
431
432
CHAPTER 28
28.10
(b) From
 1
1 
= RH  2 − 2 
 n f ni 
λ


1
2 2
1  ni n f
 2
RH  ni − n2f
or λ =
λ=
28.11
(a)
E=
hc
(c)
f=
c
λ
λ

 with ni = 6 and n f = 2


 ( 36 )( 4 ) 
1
−7
 = 4.10 × 10 m = 410 nm
7
−1 
1.097 37 × 10 m  36 − 4 
( 6.63 × 10
=
=
−34
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 Hz
−9
410 × 10 m
The energy of the emitted photon is
Eγ =
hc
λ
=
( 6.626 × 10
−34
J ⋅ s )( 2.998 × 108 m s )  1 nm  
1 ev
 −9 
656 nm
10
m
1.602
× 10 −19



 = 1.89 eV
J
This photon energy is also the difference in the electron’s energy in its initial and final
orbits. The energies of the electron in the various allowed orbits within the hydrogen
atom are
En = −
13.6
eV
n2
where
n = 1, 2, , 3, …
giving E1 = −13.6 eV, E2 = −3.40 eV, E3 = −1.51 eV, E4 = −0.850 eV, …
Observe that Eγ = E3 − E2 . Thus, the transition was from
the n = 3 orbit to the n = 2 orbit
Atomic Physics
28.12
The change in the energy of the electron is
 1
1 
∆E = E f − Ei = 13.6 eV  2 − 2 
 ni n f 


Transition I:
1 1 
∆E = 13.6 eV  −  = 2.86 eV (absorption)
 4 25 
Transition II:
 1 1
∆E = 13.6 eV  −  = − 0.967 eV (emission)
 25 9 
Transition III:
1 
 1
∆E = 13.6 eV 
−  = − 0.572 eV (emission)
 49 16 
Transition IV:
1 
 1
∆E = 13.6 eV  −  = 0.572 eV (absorption)
 16 49 
(a) Since λ =
hc
hc
=
, transition II emits the shortest wavelength photon.
Eγ − ∆E
(b) The atom gains the most energy in transition I
(c) The atom loses energy in transitions II and III
28.13
The energy absorbed by the atom is
 1
1 
Eγ = E f − Ei = 13.6 eV  2 − 2 
 ni n f 


(a)
1 1 
Eγ = 13.6 eV  −  = 0.967 eV
 9 25 
1 
 1
(b) Eγ = 13.6 eV  −  = 0.266 eV
 25 49 
28.14
(a) The energy absorbed is
 1
1 
1 1
∆E = E f − Ei = 13.6 eV  2 − 2  = 13.6 eV  −  = 12.1 eV
 ni n f 
1 9


433
434
CHAPTER 28
(b) Three transitions are possible as the electron returns to the ground state. These
transitions and the emitted photon energies are
28.15
ni = 3 → n f = 1 :
 1 1
∆E = 13.6 eV  2 − 2
1 3

 = 12.1 eV

ni = 3 → n f = 2 :
1
 1
∆E = 13.6 eV  2 − 2
3
2

 = 1.89 eV

ni = 2 → n f = 1 :
 1 1
∆E = 13.6 eV  2 − 2
1 2

 = 10.2 eV

 1
1 
= RH  2 − 2  , it is seen that (for a fixed value of n f ) λmax occurs when
 n f ni 
λ


ni = n f + 1 and λmin occurs when ni → ∞ .
From
1
(
)
(a) For the Lyman series n f = 1 ,
1
λmax
1 1 
= ( 1.097 37 × 107 m −1 )  2 − 2  → λmax = 1.22 × 10 −7 m = 122 nm
1 2 
and
1
λmin
1 1
= ( 1.097 37 × 107 m −1 )  2 −  → λmin = 9.11 × 10 −8 m = 91.1 nm
1 ∞
(
)
(b) For the Paschen series n f = 3 ,
1
λmax
1
 1
= ( 1.097 37 × 107 m −1 )  2 − 2
4
3

−6
3
 → λmax = 1.87 × 10 m = 1.87 × 10 nm

and
1
λmin
 1 1
= ( 1.097 37 × 107 m −1 )  2 −  → λmin = 8.20 × 10 −7 m = 820 nm
∞
3
Atomic Physics
28.16
435
The electron is held in orbit by the electrical force the proton exerts on it. Thus,
me
v 2 ke e 2
= 2
r
r
v=
or
ke e 2
me r
If we divide by the speed of light, and recognize that in the first Bohr orbit r = a0 where
a0 = 0.052 9 nm , this becomes
ke e 2
ke e 2
v1
=
=
=
c
me c 2 r1
me c 2 a0
or
28.17
( 8.99 × 10 N ⋅ m C )(1.60 × 10 C )
( 9.11 × 10 kg ) ( 3.00 × 10 m s ) ( 0.052 9 × 10
v1
1
= 7.28 × 10 −3 =
c
137
9
2
−31
Thus,
−19
2
8
2
2
−9
m)
v1 = ( 1 137 ) c
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
1
1
= RH  2 − 2
λ
4
1
1
28.18
16
16
 15 RH
or λ =
=
= 97.2 nm
=
15 RH 15 ( 1.097 37 × 107 m −1 )
16

The magnetic force supplies the centripetal acceleration, so
mv
mv 2
= qvB , or r =
qB
r
If angular momentum is quantized according to
Ln = mvn rn = 2n , then mvn =
2n
rn
and the allowed radii of the path are given by
rn =
1  2n 

 or
qB  rn 
rn =
2n
qB
436
CHAPTER 28
28.19
(a) The energy emitted by the atom is
1 
 1
∆E = E4 − E2 = − 13.6 eV  2 − 2  = 2.55 eV
2 
4
The wavelength of the photon produced is then
−34
8
hc hc ( 6.63 × 10 J ⋅ s )( 3.00 × 10 m s )
=
=
Eγ ∆E
( 2.55 eV ) (1.60 × 10−19 J eV )
λ=
= 4.88 × 10 −7 m = 488 nm
(b) Since momentum must be conserved, the photon and the atom go in opposite
directions with equal magnitude momenta. Thus, p = matom v = h λ or
v=
28.20
h
matom λ
=
6.63 × 10 −34 J ⋅ s
= 0.814 m s
(1.67 × 10−27 kg )( 4.88 × 10−7 m )
(a) Starting from the n = 4 state, there are 6 possible transitions as the electron returns
to the ground ( n = 1) state. These transitions are: n = 4 → n = 1, n = 4 → n = 2,
n = 4 → n = 3, n = 3 → n = 1, n = 3 → n = 2, and n = 2 → n = 1 . Since there is a
different change in energy associated with each of these transitions there will be
6 different wavelengths observed in the emission spectrum of these atoms.
(b) The longest observed wavelength is produced by the transition involving the
smallest change in energy. This is the n = 4 → n = 3 transition, and the wavelength
is
λmax
6.626 × 10 −34 J ⋅ s ) ( 2.998 × 108 m s ) 
(
1 eV
hc
=
=

-19
1 
E4 − E3
 1
 1.602 × 10
−13.6 eV  2 − 2 
3 
4
  1 nm 


J   10 -9 m 
or λmax = 1.88 × 10 3 nm
Since this transition terminates on the n = 3 level, this is part of the
Paschen series
28.21
When the centripetal acceleration is supplied by the gravitational force,
GM
mv 2 GMm
=
or v 2 =
2
r
r
r
Atomic Physics
(a) With PE = − GMm r , the total energy is
E = KE + PE =
1
GMm m  GM  GMm
GMm
mv 2 −
= 
= −
−
2
r
2 r 
r
2r
(b) Using the Bohr quantization rule, Ln = mvn rn = n , so vn =
n
and
mrn
2
 n  GM
GM
v =
becomes 
 =
r
rn
 mrn 
2
which reduces to rn =
r0 =
n2 2
= n2 r0 with
2
GMm
2
GMm2
( 6.63 × 10 J ⋅ s )
N ⋅ m kg )(1.99 × 10
−34
=
4 π 2 ( 6.67 × 10 −11
2
2
2
30
kg )( 5.98 × 10 24 kg )
2
r0 = 2.32 × 10 −138 m
(c) The energy in the nth orbit is En = −
E0 =
GMm
GMm  GMm2
=−

2 rn
2  n2 2
G2 M 2 m3
2 2
4 π 2 ( 6.67 × 10 −11 ) ( 1.99 × 10 30 ) ( 5.98 × 10 24 )
2
=

E0
 = − 2 , where
n

(d) rn = n 2 r0 , so n2 =
2 ( 6.63 × 10
2
)
−34 2
3
= 1.71 × 10182 J
rn
1.49 × 1011 m
=
= 6.42 × 10148
-138
m
r0 2.32 × 10
or n = 2.53 × 1074
(e) No, the quantum numbers are too large, and the allowed energies are essentially
continuous in this region.
437
438
CHAPTER 28
28.22
(a) The time for one complete orbit is
From Bohr’s quantization postulate,
we see that v =
T=
2π r
v
L = me vr = n
n
Thus, the orbital period becomes:
me r
2π me r 2 2π me ( a0 n
T=
=
n
n
)
2 2
=
2π me a 02
n3
or T = t0 n3 where
t0 =
2π me a 02
=
2π ( 9.11 × 10 − 31 kg )( 0.052 9 × 10 − 9 m )
1.055 × 10 − 34 J ⋅ s
2
= 1.52 × 10 −16 s
(b) With n = 2 , we have T =8 t0 =8 ( 1.52 × 10 − 16 s ) =1.21 × 10 − 15 s
Thus, if the electron stays in the n = 2 state for 10 µ s , it will make
10.0 × 10 − 6 s
= 8.23 × 10 9 revolutions of the nucleus
1.21 × 10 − 15 s rev
(c)
Yes, for 8.23 × 109 “electron years”
(d) The electron moves so quickly that it can never meaningfully be said to be on any
particular side of the nucleus.
28.23
(a) The wavelength emitted in the ni = 2 → n f = 1 transition is
2 2
1  ni n f
λ=

RH  ni2 − n 2f

 ( 4 )( 1) 
1
= 1.22 × 10 −7 m
=
 ( 1.097 37 × 107 m −1 )  4 − 1 

and the frequency is f =
c
λ
=
3.00 × 10 8 m s
= 2.47 × 1015 Hz
1.22 × 10 −7 m
From Ln = me vn rn = n , the speed of the electron is vn = n
me rn
Atomic Physics
439
Therefore, with rn = n2 a0 , the orbital frequency is
f orb =

n
h
1 vn
=
=
=
2
T rn 2 π me rn  4 π 2 me a02
For the n = 2 orbit, f orb =
6.59 × 1015 Hz
( 2)
3
 1 6.59 × 1015 Hz
 3 =
n3
n
= 8.23 × 1014 Hz
(b) For the ni = 10 000 → n f = 9 999 transition,
 ( 10 000 )2 ( 9 999 )2 
1

 = 4.56 × 10 4 m
λ=
2
2
−1
7
(1.0977 37 × 10 m )  (10 000 ) − ( 9 999 ) 
and
f=
c
λ
=
3.00 × 108 m s
= 6.59 × 10 3 Hz
4.56 × 10 4 m
For the n = 10 000 orbit, f orb =
6.59 × 1015 Hz
(10 000 )
3
= 6.59 × 10 3 Hz
For small n, significant differences between classical and quantum results appear.
However, as n becomes large, classical theory and quantum theory approach one
another in their results. This is in agreement with the correspondence principle.
28.24
Each atom gives up its kinetic energy in emitting a photon, so
KE =
−34
8
hc ( 6.63 × 10 J ⋅ s )( 3.00 × 10 m s )
1
=
= 1.64 × 10 −18 J
mv 2 =
λ
2
121.6 × 10 −9 m
2 ( 1.64 × 10 −18 J )
2 ( KE )
v=
=
= 4.43 × 10 4 m s
−27
matom
1.67 × 10 kg
440
CHAPTER 28
28.25
For minimum initial kinetic energy, KEtotal = 0 after collision. Hence, the two atoms must
have equal and opposite momenta before impact. The atoms then have the same initial
kinetic energy, and that energy is converted into excitation energy of the atom during
the collision. Therefore,
KEatom =
or
28.26
(a)
v=
1
matom v 2 = E2 − E1 = 10.2 eV
2
2 ( 10.2 eV ) ( 1.60 × 10 −19 J eV )
2 ( 10.2 eV )
=
= 4.42 × 10 4 m s
matom
1.67 × 10 −27 kg
 2π r 
L = mvr = m 
r
 T 
=
(b) n =
2 π ( 7.36 × 10 22 kg )( 3.84 × 108 m )
2
= 2.89 × 10 34 kg ⋅ m 2 s
2.36 × 106 s
L
34
2
2π L 2π ( 2.89 × 10 kg ⋅ m s )
=
=
= 2.74 × 1068
h
6.63 × 10 −34 J ⋅ s
(c) The gravitational force supplies the centripetal acceleration so
mv 2 GME m
=
, or rv 2 = GME
2
r
r
Then, from Ln = mvn rn = n or vn =
n
,
mrn
2
2
 n 
2
we have rn 
which
gives
r
n
=
GM
=


E
n
2
 GME m
 mrn 

2
 = n r1

Therefore, when n increases by 1, the fractional change in the radius is
2
∆r rn +1 − rn ( n + 1) r1 − n r1 2n + 1 2
=
=
=
≈
r
rn
n 2 r1
n2
n
2
∆r
2
≈
= 7.30 × 10 −69
r
2.74 × 1068
Z 2 ( 13.6 eV )
( 3 ) ( 13.6 eV )
, E1 = −
(a) From En = −
= − 122 eV
2
2
n
( 1)
2
28.27
Atomic Physics
441
n2 a0
( 1) a0 0.052 9 × 10 −9 m
gives r1 =
(b) Using rn =
=
= 1.76 × 10 −11 m
Z
3
3
2
28.28
(a) The energy levels of a hydrogen-like ion
whose charge number is Z are given by
En = ( −13.6 eV )
2
Z
n2
For Helium, Z = 2 and the energy levels
are
En = −
54.4 eV
n = 1, 2, 3, . . .
n2
n=¥
___________________
0
n=5
n=4
___________________
___________________
–2.18 eV
–3.40 eV
n=3
___________________
–6.04 eV
n=2
___________________
–13.6 eV
n=1
___________________
–54.4 eV
(b) For He + , Z = 2 , so we see that the ionization energy (the energy required to take
the electron from the n = 1 to the n = ∞ state) is
( −13.6 eV )( 2 )
= 54.4 eV
2
( 1)
2
E = E∞ − E1 = 0 −
28.29
28.30
rn =
2
n2 

Z  me k e e 2
 n 2 a0
0.052 9 nm
a
, so r1 = 0 =
=
Z
Z
Z

(a) For He+ ,
Z = 2 and r =
0.052 9 nm
= 0.026 5 nm
2
(b) For Li 2+ ,
Z = 3 and r =
0.052 9 nm
= 0.017 6 nm
3
(c) For Be 3+ ,
Z = 4 and r =
0.052 9 nm
= 0.013 2 nm
4
(a) For hydrogen-like atoms having atomic number Z, the Rydberg constant is
R=
me k e2 Z 2 e 4
4π c 3
Thus, for singly ionized helium with Z = 2 , we have
( 9.11 × 10
R=
−31
kg )( 8.99 × 10 9 N ⋅ m 2 C 2 ) ( 2 ) ( 1.60 × 10 −19 C )
4π ( 3.00 × 10
8
) (1.05 × 10
2
2
−34
J ⋅ s)
3
4
= 4.42 × 107 m −1
442
CHAPTER 28
(b)
 1
1 
 1 1
= RHe  2 − 2  = ( 4.42 × 107 m −1 )  2 − 2
 n f ni 
λ
1 2


1

 1 nm 
m  −9  = 30.1 nm
so λ = 
7
 3.32 × 10
 10 m 
1

7
−1
 = 3.32 × 10 m

(c) This wavelength is in the deep ultraviolet portion of the electromagnetic
spectrum.
28.31
From L = me vn rn = n and rn = n2 a0
we find that pn = mvn =
n
h
=
rn ( 2π a0 ) n
Thus, the de Broglie wavelength of the electron in the nth orbit is
λ = h pn = ( 2π a0 ) n . For n = 4 , this yields
λ = 8π a0 = 8π ( 0.052 9 nm ) = 1.33 nm
28.32
(a) For standing waves in a string fixed at both ends, L =
or λ =
2L
h
. According to the de Broglie hypothesis, p =
n
λ
Combining these expressions gives p = mv =
(b) Using E =
En =
28.33
nλ
2
nh
2L
p2
1
mv 2 =
, with p as found in (a) above:
2
2m
n2 h 2
h2
2
=
where
=
n
E
E
0
0
4 L2 ( 2m )
8 mL2
In the 3p subshell, n = 3 and
n=3
n=3
n=3
=1
=1
=1
= 1 . The 6 possible quantum states are
m = +1
m =0
m = −1
ms = ± 12
ms = ± 12
ms = ± 12
443
Atomic Physics
28.34
(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
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 ml 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 ml . These values are:
ml = −3, − 2, − 1, 0, + 1, + 2, and + 3
28.35
The 3d subshell has n = 3 and = 2 . For ρ-mesons, we also have s = 1 . Thus, there are 15
possible quantum states as summarized in the table below.
n
m
ms
28.36
3
2
+2
+1
3
2
+2
0
3
2
+2
-1
3
2
+1
+1
3
2
+1
0
3
2
+1
-1
3
2
0
+1
3
2
0
0
3
2
0
-1
3
2
-1
+1
3
2
-1
0
3
2
-1
-1
3
2
-2
+1
3
2
-2
0
3
2
-2
-1
(a) The electronic configuration for oxygen ( Z = 8 ) is 1s2 2s2 2 p 4
(b) The quantum numbers for the 8 electrons can be:
n=1
n=2
1s states
2s states
m
m
m
m
=0
n=2
2 p states
28.37
=0
=1
=0
ms
ms
ms
ms
=0
=0
=1
= ± 12
= ± 12
= ± 12
= ± 12
(a) For Electron #1 and also for Electron #2, n = 3 and = 1 . The other quantum
numbers for each of the 30 allowed states are listed in the tables below.
Electron #1
m
+1
ms
+ 12
m
+1
ms
+ 12
m
+1
ms
+ 12
m
+1
ms
− 12
m
+1
ms
− 12
m
+1
ms
− 12
Electron #2
+1
− 12
0
± 12
-1
± 12
+1
+ 12
0
± 12
-1
± 12
Electron #1
m
0
ms
+ 12
m
0
ms
+ 12
m
0
ms
+ 12
m
0
ms
− 12
m
0
ms
− 12
m
0
ms
− 12
Electron #2
+1
± 12
0
− 12
-1
± 12
+1
± 12
0
+ 12
-1
± 12
444
CHAPTER 28
Electron #1
m
-1
ms
+ 12
m
-1
ms
+ 12
m
-1
ms
+ 12
m
-1
ms
− 12
m
-1
ms
− 12
m
-1
ms
− 12
Electron #2
+1
± 12
0
± 12
-1
− 12
+1
± 12
0
± 12
-1
+ 12
There are 30 allowed states , since Electron #1 can have any of three possible
values of m 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.
28.38
(a) For n = 1,
= 0 and there are 2(2 + 1) states = 2(1) = 2 sets of quantum numbers
(b) For n = 2, = 0 for 2 (2 + 1) states = 2(0 + 1) = 2 sets
and
= 1 for 2(2 + 1) states = 2(2 + 1) = 6 sets
total number of sets = 8
(c) For n = 3,
and
and
= 0 for 2 (2 + 1) states = 2(0 + 1) = 2 sets
= 1 for 2(2 + 1) states = 2(2 + 1) = 6 sets
= 2 for 2(2 + 1) states = 2(4 + 1) = 10 sets
total number of sets = 18
(d) For n = 4,
and
and
and
=0
=1
=2
=3
for
for
for
for
2 (2
2(2
2(2
2(2
+ 1)
+ 1)
+ 1)
+ 1)
states
states
states
states
= 2(0 + 1) = 2 sets
= 2(2 + 1) = 6 sets
= 2(4 + 1) = 10 sets
= 2(6 + 1) = 14 sets
total number of sets = 32
(e) For n = 5,
and
and
and
and
=0
=1
=2
=3
=4
for
for
for
for
for
2 (2
2(2
2(2
2(2
2(2
+ 1)
+ 1)
+ 1)
+ 1)
+ 1)
states
states
states
states
states
= 2(0 + 1) = 2 sets
= 2(2 + 1) = 6 sets
= 2(4 + 1) = 10 sets
= 2(6 + 1) = 14 sets
= 2(8 + 1) = 18 sets
total number of sets = 50
For n = 1: 2n 2 = 2
For n = 2: 2n2 = 8
For n = 3: 2n2 = 18
For n = 4: 2n2 = 32
For n = 5: 2n2 = 50
Atomic Physics
445
Thus, the number of sets of quantum states agrees with the 2n2 rule.
28.39
(a) Zirconium, with 40 electrons, has 4 electrons outside a closed Krypton core. The
Krypton core, with 36 electrons, has all states up through the 4p subshell filled.
Normally, one would expect the next 4 electrons to go into the 4d subshell.
However, an exception to the rule occurs at this point, and the 5s subshell fills
(with 2 electrons) before the 4d subshell starts filling. The two remaining electrons
in Zirconium are in an incomplete 4d subshell. Thus, n = 4, and = 2 for each of
these electrons.
(b) For electrons in the 4d subshell, with
= 2 , the possible values of m are
m = 0, ± 1, ± 2 and those for ms are ms = ± 1 2
(c) We have 40 electrons, so the electron configuration is:
1s2 2s2 2 p6 3s2 3 p 6 3d10 4 s2 4 p 6 4 d 2 5s2 = [Kr]4 d 2 5s2
28.40
The photon energy is Eγ = EL − EK = −951 eV − ( −8 979 eV ) = 8 028 eV , and the
wavelength is
−34
8
hc ( 6.63 × 10 J ⋅ s )( 3.00 × 10 m s )
=
= 1.55 × 10 −10 m = 0.155 nm
λ=
Eγ
( 8 028 eV ) (1.60 × 10−19 J eV )
To produce the Kα line, an electron from the K shell must be excited to the L shell or
higher. Thus, a minimum energy of 8 028 eV must be given to the atom. A minimum
accelerating voltage of ∆V = 8 028 V = 8.03 kV is required.
446
CHAPTER 28
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 × 10 3 eV
2
( 1)
13.6 eV
2 ( 13.6 eV )
= − ( 25 )
= − 2.13 × 10 3 eV
2
4
( 2)
Thus, Eγ = EL − EK = −2.13 keV − ( −9.91 keV ) = 7.78 keV
and
28.42
−34
8
hc ( 6.63 × 10 J ⋅ s )( 3.00 × 10 m s )
=
= 1.60 × 10 −10 m = 0.160 nm
λ=
Eγ
7.78 keV ( 1.60 × 10 −16 J keV )
The energies in the K and M shells are
EK ≈ − ( Z − 1)
2
13.6 eV
2 13.6 eV
and EM ≈ − ( Z − 9 )
2
2
( 1)
( 3)
 ( Z − 9 )2
2
8

Thus, Eγ = EM − EK ≈ ( 13.6 eV )  −
+ ( Z − 1)  = ( 13.6 eV )  Z 2 − 8 
9
9



and Eγ =
hc
λ
gives Z 2 =
Z≈ 9+

9
hc
8 +
 , or
8
( 13.6 eV ) λ 
9 ( 6.63 × 10 −34 J ⋅ s ) ( 3.00 × 108 m s ) 

1 eV

 = 32.0
−19
−9
8 ( 13.6 eV ) ( 0.101 × 10 m )
 1.60 × 10 J 
The element is Germanium
Atomic Physics
28.43
The transitions that produce the three
longest wavelengths in the K series are
shown at the right. The energy of the K
shell is EK = − 69.5 keV .
EN
N shell
EM
M shell
EL
L shell
Thus, the energy of the L shell is
EL = EK +
or
l1
hc
EL = − 69.5 keV +
l2
( 6.63 × 10
J ⋅ s )( 3.00 × 108 m s )
−34
0.021 5 × 10 −9 m


1 keV
= − 69.5 keV + 9.25 × 10 −15 J 

−16
 1.60 × 10 J 
= − 69.5 keV + 57.8 keV = − 11.7 keV
Similarly, the energies of the M and N shells are
EM = EK +
hc
λ2
= − 69.5 keV +
( 6.63 × 10
−34
( 0.020 9 × 10
−9
J ⋅ s )( 3.00 × 10 8 m s )
m )( 1.60 × 10 −16 J keV )
= − 10.0 keV
and
EN = EK +
hc
λ1
= − 69.5 keV +
( 6.63 × 10
−34
( 0.018 5 × 10
−9
J ⋅ s )( 3.00 × 108 m s )
m )( 1.60 × 10 −16 J keV )
The ionization energies of the L, M, and N shells are
11.7 keV, 10.0 keV, and 2.30 keV respectively
l3
K shell
EK
λ3
447
= − 2.30 keV
448
CHAPTER 28
28.44
According to the Bohr model, the radii of the electron orbits in hydrogen are given
by
rn = n2 a0 with a0 = 0.052 9 nm = 5.29 × 10 −11 m
Then, if rn ≈ 1.00 µ m = 1.00 × 10 −6 m , the quantum number is
rn
1.00 × 10 −6 m
=
≈ 137
a0
5.29 × 10 −11 m
n=
28.45
(a)
∆E = E2 − E1 = − 13.6 eV (2)2 − ( − 13.6 eV (1)2 ) = 10.2 eV
(b) The average kinetic energy of the atoms must equal or exceed the needed
excitation energy, or 32 k B T ≥ ∆E which gives
T≥
28.46
(a)
L = c ( ∆t ) = ( 3.00 × 10 8 m s )( 14.0 × 10 −12 s ) = 4.20 × 10 −3 m = 4.20 mm
(b) N =
=
(c)
−19
J eV )
2 ( ∆E ) 2 ( 10.2 eV ) ( 1.60 × 10
=
= 7.88 × 10 4 K
−23
3k B
3 ( 1.38 × 10
J K)
n=
=
Epulse
Eγ
=
Epulse
hc λ
( 694.3 × 10
( 6.63 × 10
−34
−9
m ) ( 3.00 J )
J ⋅ s )( 3.00 × 10 m s )
8
= 1.05 × 1019 photons
N
N
=
V L (π d 2 4 )
4 ( 1.05 × 1019 photons )
( 4.20 mm ) π ( 6.00 mm )
2
= 8.82 × 1016 photons mm 3
Atomic Physics
28.47
(a)
hc
E1 = E∞ −
E2 = E1 +
λlimit
( 6.63 × 10
=0−
−34
J ⋅ s )( 3.00 × 10 8 m s )
152.0 × 10 −9 m ( 1.60 × 10 −19 J eV )
λ1
−34
J ⋅ s )( 3.00 × 10 8 m s )
202.6 × 10 −9 m ( 1.60 × 10 −19 J eV )
λ2
−34
J ⋅ s )( 3.00 × 108 m s )
170.9 × 10 −9 m ( 1.60 × 10 −19 J eV )
= − 0.904 eV
hc
λ3
( 6.63 × 10
= − 8.18 eV +
−34
J ⋅ s )( 3.00 × 10 8 m s )
162.1 × 10 −9 m ( 1.60 × 10 −19 J eV )
E5 = E1 +
= − 2.04 eV
hc
( 6.63 × 10
= − 8.18 eV +
E4 = E1 +
= − 8.18 eV
hc
( 6.63 × 10
= − 8.18 eV +
E3 = E1 +
= − 0.510 eV
hc
λ4
= − 8.18 eV +
(b) From λ =
( 6.63 × 10
−34
J ⋅ s )( 3.00 × 108 m s )
158.3 × 10 −9 m ( 1.60 × 10 −19 J eV )
= − 0.325 eV
hc
hc
=
, the longest and shortest wavelengths in the Balmer series for
Eγ Ei − E f
this atom are
λlong =
and
449
( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m s )
hc
=
= 1.09 × 10 3 nm
E3 − E2  − 0.904 eV − ( − 2.04 eV )  ( 1.60 × 10 −19 J eV )
λshort =
( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m s ) = 609 nm
hc
=
E∞ − E2  0 − ( − 2.04 eV )  ( 1.60 × 10 −19 J eV )
(b)
28.49
a0
λC
2
=
me k e e 2
1  c 
1
=
 2 =
h me c
2 π  ke e  2 π α
(c)
1 RH 4 π
=
a0
(a)
Eγ =
hc
λ
3
2
 c
c me k e2 e 4
= 4π  2
2
me k e e
 ke e
( 6.626 × 10
λ ( 1.602 × 10
−34
=
 4π
=
α

J ⋅ s )( 2.998 × 108 m s )
−19
J eV )( 10
−9
m nm )
=
1 240 eV ⋅ nm
λ
For:
λ = 253.7 nm, Eγ = 4.888 eV ;
λ = 185.0 nm, Eγ = 6.703 eV ;
λ = 158.5 nm, Eγ = 7.823 eV
Thus, the energies of the first three excited states are:
E1 = − 10.39 eV + 4.888 eV= − 5.502 eV
E2 = − 10.39 eV + 6.703 eV= − 3.687 eV
and
E3 = − 10.39 eV + 7.823 eV= − 2.567 eV
Third Excited State
Second Excited State
First Excited State
Ground State
2
253.7 nm
(a)
6.626 × 10 −34 J ⋅ s )( 2.998 × 108 m s )
(
c
hc
=
=
=
= 137
α ke e 2 2 π ke e 2 2 π ( 8.987 × 10 9 N ⋅ m 2 C2 )( 1.602 × 10 −19 C )2
1
185.0 nm
28.48
CHAPTER 28
158.5 nm
450
– 2.567 eV
– 3.687 eV
3
5
– 5.502 eV
4
1
6
– 10.39 eV
Atomic Physics
(
451
)
(b) From λ = ( 1 240 eV ⋅ nm ) Ei − E f , the wavelengths of the emission lines shown are
λ1 = 158.5 nm , λ2 = 422.5 nm , λ3 = 1 107 nm , λ4 = 185.0 nm ,
λ5 = 683.2 nm , and λ6 = 253.7 nm
(c) To have an inelastic collision, we must excite the atom from the ground state to the
first excited state, so the incident electron must have a kinetic energy of at least
KE = 10.39 eV − 5.502 eV = 4.888 eV ,
2 ( 4.888 eV ) ( 1.602 × 10 −19 J eV )
2 ( KE )
so v =
=
= 1.31 × 10 6 m s
−31
9.11 × 10 kg
me
28.50
Eγ =
hc
λ
( 6.626 × 10
=
λ ( 1.602 × 10
−34
J ⋅ s )( 2.998 × 108 m s )
−19
J eV )( 10
−9
m nm )
=
1 240 eV ⋅ nm
λ
= ∆E
For:
λ = 310.0 nm , ∆E = 4.000 eV
λ = 400.0 nm , ∆E = 3.100 eV
λ = 1 378 nm , ∆E = 0.900 0 eV
and
The ionization energy is 4.100 eV. The energy level diagram having the fewest number
of levels and consistent with these energy differences is shown below.
n=¥
Second Excited State
Ground State
28.51
(a)
I=
400 nm
310 nm
First Excited State
1 378 nm
0
–0.100 0 eV
–1.000 eV
–4.100 eV
−3
−9
P ( ∆E ∆t ) 4 ( 3.00 × 10 J 1.00 × 10 s )
=
=
= 4.24 × 1015 W m 2
2
2
−
6
A πd 4
π ( 30.0 × 10 m )
(b) E = I A ( ∆t )
2
W  π

=  4.24 × 1015
0.600 × 10 −9 m )  ( 1.00 × 10 −9 s ) = 1.20 × 10 −12 J
(
2 
m  4


452
CHAPTER 28
28.52
(a) Given that the de Broglie wavelength is λ = 2 a0 , the momentum is p = h λ = h 2 a 0 .
The kinetic energy of this non-relativistic electron is
KE =
p2
h2
=
2 me 8 me a 20
( 6.63 × 10
=
8 ( 9.11 × 10
−34
J ⋅ s ) ( 1 eV 1.60 × 10 −19 J )
−31
2
kg )( 0.052 9 × 10 −9 m )
2
= 135 eV
(b) The kinetic energy of this electron is ≈ 10 times the magnitude of the ground state
energy of the hydrogen atom which is –13.6 eV.
28.53
In the Bohr model,
f =
=
∆E En − En −1
=
h
h
1  − me k e2 e 4  1
1

 2 −
2
2
h  2
( n − 1)
n
which reduces to
28.54
f=
  4π 2 me k e2 e 4
  =
2 h3
 
2π 2 me k e2 e 4  2n − 1

2 2
h3
 ( n − 1) n
 1
1
−


2
n2 
 ( n − 1)



As n → ∞ , 2n − 1 → 2n and n − 1 → n . In this limit, the result of Problem 28.53 reduces
to
f=
2π 2 me k e2 e 4  2n  4π 2 me k e2 e 4
 4 =
h3
h3 n3
n 
Since the electrical force supplies the centripetal acceleration,
me v 2 ke e 2
k e2
= 2 or v = e
r
r
me r
Atomic Physics
The classical frequency is then
f =
v
2π r
=
f=
This gives
ke e 2
n2 h 2
2
where
=
=
r
n
a
0
4π 2 me k e e 2
me r 3
1
2π
v
2π r
=
k e e 2  64 π 6 me3 k e3 e 6 
4π 2 me k e2 e 4
=


4π 2 me 
n6 h 6
h 3 n3

Thus, the frequency from the Bohr model is the same as the classical frequency in the
limit n → ∞ .
28.55
(a) The energy levels in this atom are
En = −
mπ Z 2 k e2 e 4
2 2 n2
273 ( 2 )  me ke2 e 4

n2  2 2
2
=−

− 1.485 × 10 4 eV
273  2

2
13.60
eV
=
−
=
( ) (
)

n2 
n2

The energies of the first six levels are:
E1 = −1.485 × 10 4 eV
E2 = −3.71 × 10 3 eV
E3 = −1.65 × 10 3 eV
E4 = − 928 eV
E5 = −594 eV
E6 = − 413 eV
(b) From the Compton shift formula, the emitted wavelength was
λ0 = λ ′ − λC ( 1 − cosθ ) = 0.089 929 3 nm − ( 0.002 43 nm ) ( 1 − cos 42.68° )
= 0.089 289 nm
The energy radiated by the atom is then
∆E = Ei − E f =
=
hc
λ0
( 6.63 × 10
−34
J ⋅ s )( 3.00 × 108 m s )
( 0.089 289 × 10−9 m )(1.60 × 10−19 J eV )
= 1.392 × 10 4 eV
453
454
CHAPTER 28
Since ∆E > E2 , the final state must be the ground state E1 . The energy of the initial
state was
Ei = E f + ∆E = − 1.485 × 10 4 eV + 1.392 × 10 4 eV = −928 eV
This is seen to be E4 . Thus, the transition made by the pi meson was n = 4 → n = 1
28.56
(a) Using a0 =
2
mµ k e e 2
, with mµ = 207 me , gives the Bohr radius for the “muonic atom”
as
a0 =
2
1 

207  me k e e 2

1
( 0.052 9 nm ) = 2.56 × 10−4 nm
=
207

(b) The energy levels in this atom are
En = −
mµ Z 2 k e2 e 4
2
2
n2
207 ( 1)  me k e2 e 4 
−2.82 × 10 3 eV
207
=−
=
−
=
13.6
eV
(
)


n2  2 2 
n2
n2
2
The energies of the three lowest levels are:
E1 = − 2.82 × 10 3 eV
28.57
E2 = − 704 eV
E3 = − 313 eV
(a) From Newton’s second law, F = k e q1q2 r 2 = me a , and the acceleration is
a=
k e2
F
= e 2
me me a0
( 8.99 × 10 N ⋅ m C )(1.60 × 10 C )
=
( 9.11 × 10 kg )( 0.052 9 × 10 m )
9
2
−31
(b) √ = −
−19
2
−9
2
= 9.03 × 10 22 m s 2
2
2 ke e 2 a2
3 c3
2 ( 8.99 × 10 9 N ⋅ m 2 C 2 )(1.60 × 10 −19 C ) ( 9.03 × 10 22 m s 2 )
2
=−
3 ( 3.00 × 108 m s )
√ = − 4.63 × 10 −8 J s = − 4.63 × 10 −8 W
3
2
Atomic Physics
455
(c) With the electrical force supplying the centripetal acceleration,
me v 2 ke e 2
k e2
k e2
1
= 2 or me v 2 = e and KE = me v 2 = e
r
r
r
2
2r
Thus,
−19
9
2
2
k e 2 ( 8.99 × 10 N ⋅ m C )( 1.60 × 10 C )
= 2.17 × 10 −18 J
KE = e =
−9
2 a0
2 ( 0.052 9 × 10 m )
2
The time required to radiate all this energy, and the estimated lifetime is
∆t =
28.58
KE 2.17 × 10 −18 J
=
= 4.69 × 10 −11 s or ∆t ~ 10 −11 s
P 4.63 × 10 −8 J s
(a) The photon emitted by the hydrogen atom must have an energy Eγ ≥ 4.58 eV if it is
to eject a photoelectron from tungsten (φ = 4.58 eV ) . Thus, the electron in the
hydrogen atom must give up at least 4.58 eV of energy, meaning that the energy of
the final state must be E f ≤ −4.58 eV . The only state in the hydrogen atom satisfying
this condition is the ground ( n = 1) state, so it is necessary that n f = 1
(b) If the stopping potential of the ejected photoelectron is Vs = 7.51 V , the kinetic
energy of this electron as it leaves the tungsten is
KEmax = eVs = e ( 7.51 V ) = 7.51 eV
and the photoelectric effect equation gives the photon energy as
Eγ = hf = φ + KEmax = 4.58 eV + 7.51 eV = 12.09 eV
But, the photon energy equals the energy given up by the electron in the hydrogen
atom. That is, Eγ = Ei − E f . Since we determined in Part (a) that n f = 1 , then
E f = −13.6 eV and we have
Eγ = Ei − E f = Ei + 13.6 eV=12.09 eV
Thus, from En =
−13.6 eV
,
n2
ni =
or
Ei = 12.09 eV − 13.6 eV = −1.51 eV
−13.6 eV
−13.6 eV
=
= 3
Ei
−1.51 eV
456
CHAPTER 28