The document discusses concepts related to the quantum mechanical model of the hydrogen atom. It provides answers to conceptual problems involving energy levels, quantum numbers, and properties of atomic orbitals. Key points include: - As the principal quantum number n increases, the spacing between adjacent energy levels decreases. - For n=4, the orbital quantum number l can take on values from 0 to 3. - In sodium, the 3s state is at a lower energy than the 3p state due to penetration of the 3s orbital closer to the nucleus. In hydrogen, the 3s and 3p states have similar energies. - The Ritz combination principle, where 1/λ1 + 1/λ2

1. Chapter 36
Atoms
Conceptual Problems
1 • For the hydrogen atom, as n increases, does the spacing of adjacent
energy levels increase or decrease?
Determine the Concept Examination of Figure 35-4 indicates that as n increases,
the spacing of adjacent energy levels decreases.
7 • For the principal quantum number n = 4, how many different values
can the orbital quantum number l take on? (a) 4, (b) 3, (c) 7, (d) 16, (e) 25
Determine the Concept We can find the possible values of by using the
constraints on the quantum numbers n and .
l
l
The allowed values for the
orbital quantum number forl
n = 1, 2, 3, and 4 are summarized
in table shown to the right:
n l
1 0
2 0, 1
3 0, 1, 2
4 0, 1, 2, 3
From the table it is clear that l can have 4 values. )(a is correct.
9 •• Why is the energy of the 3s state considerably lower than the energy of
the 3p state for sodium, whereas in hydrogen 3s and 3p states have essentially the
same energy?
Determine the Concept The s state, with l = 0, is a ″penetrating″ state in which
the probability density near the nucleus is significant. Consequently, the 3s
electron in sodium is in a region of low potential energy for a significant portion
of the time. In the state = 1, the probability density at the nucleus is zero, so the
2p electron of sodium is shielded from the nuclear charge by the 1s electrons. In
hydrogen, the 3s and 2p electrons experience the same nuclear potential.
l
19 •• The Ritz combination principle states that for any atom, one can find
different spectral lines λ1, λ2, λ3, and λ4, so that 1/λ1 + 1/λ2 = 1/λ3 + 1/λ4. Show
why this is true using an energy-level diagram.
Determine the Concept The Ritz combination principle is due to the quantization
of energy levels in the atom. We can use the relationship between the wavelength
of the emitted photon and the difference in energy levels within the atom that
results in the emission of the photon to express each of the wavelengths and then
16

2. Atoms 17
the sum of the reciprocals of the first and second wavelengths and the sum of the
reciprocals of the third and fourth wavelengths.
Express the wavelengths of the
spectral lines λ1, λ2, λ3, and λ4 in
terms of the corresponding energy
transitions:
23
1
EE
hc
−
=λ ,
02
2
EE
hc
−
=λ
13
3
EE
hc
−
=λ and
01
4
EE
hc
−
=λ
Add the reciprocals of λ1 and λ2 to
obtain:
hc
EE
hc
EE
hc
EE
03
0223
21
11
−
=
−
+
−
=+
λλ
(1)
Add the reciprocals of λ3 and λ4 to
obtain:
hc
EE
hc
EE
hc
EE
03
0113
43
11
−
=
−
+
−
=+
λλ
(2)
Because the right-hand sides of
equations (1) and (2) are equal:
4321
1111
λλλλ
+=+
One possible set of energy levels is
shown to the right:
E3
E2
E1
E0
λ1
2
3
4
λ
λ
λ
Estimation and Approximation
21 •• In laser cooling and trapping, a beam of atoms traveling in one
direction are slowed by interaction with an intense laser beam in the opposite
direction. The photons scatter off the atoms by resonance absorption, a process by
which the incident photon is absorbed by the atom, and a short time later a photon
of equal energy is emitted in a random direction. The net result of a single such
scattering event is a transfer of momentum to the atom in a direction opposite to
the motion of the atom, followed by a second transfer of momentum to the atom
in a random direction. Thus, during photon absorption the atom loses speed, but
during photon emission the change in speed of the atom is, on average, zero
(because the directions of the emitted photons are random). An analogy often
made to this process is that of slowing down a bowling ball by bouncing ping-
pong balls off of it. (a) Given that the typical photon energy used in these

3. Chapter 3618
experiments is about 1 eV, and that the typical kinetic energy of an atom in the
beam is the typical kinetic energy of the atoms in a gas that has a temperature of
about 500 K (a typical temperature for an oven that produces an atomic beam),
estimate the number of photon-atom collisions that are required to bring an atom
to rest. (The average kinetic energy of an atom is equal to 3
2
kT, where k is the
Boltzmann constant and T is the temperature. Use this to estimate the speed of the
atoms.) (b) Compare the Part (a) result with the number of ping-pong ball–
bowling ball collisions that are required to bring the bowling ball to rest. Assume
the typical speed of the incident ping-pong balls are all equal to the initial speed
of the bowling ball.) (c) 85
Rb is a type of atom often used in cooling experiments.
The wavelength of the light resonant with the cooling transition of these atoms is
λ = 780.24 nm. Estimate the number of photons needed to slow down an 85
Rb
atom from a typical thermal velocity of 300 m/s to a stop.
Picture the Problem The number of photons need to stop a 85
Rb atom traveling
at 300 m/s is the ratio of its momentum to that of a typical photon. In Part (b),
assume the mass of a bowling ball to be 6.0 kg and the mass of a ping pong ball to
be 4.0 g.
(a) The number N of photon-atom
collisions needed to bring an atom
to rest is the ratio of the change in
the momentum of the atom as it
stops to the momentum brought to
the collision by each photon:
E
mvc
c
E
mv
p
p
N ==
Δ
=
photon
atom
where m is the mass of the atom.
The kinetic energy of an atom
whose temperature is T is: kTmv 2
32
2
1
= ⇒
m
kT
v
3
=
Substituting for v gives:
mkT
E
c
m
kT
E
mc
N 3
3
==
For an atom whose mass is 50 u:
( )( )
5
23
27
19
8
10
K500J/K10381.1
u
kg101.661
u503
eV
J101.602
eV1
m/s10998.2
≈
×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
×
×
×
×
= −
−
−
N

4. Atoms 19
(b) The number N of ping-pong
ball-bowling ball collisions needed
to bring the bowling ball to rest is
the ratio of the change in the
momentum of the bowling ball as it
stops to the momentum brought to
the collision by each ping-pong ball:
ppbppb
bbbb
ballpong-ping
ballbowling
vm
vm
p
p
N =
Δ
=
Provided the speeds of the
approaching bowling ball and ping-
pong ball are approximately the
same:
3
ppb
bb
ballpong-ping
ballbowling
10
g4.0
kg0.6
≈≈≈
Δ
=
m
m
p
p
N
(c) The number of photons N needed
to stop a 85
Rb atom is the ratio of the
change in the momentum of the
atom to the momentum brought to
the collision by each photon:
h
mv
h
mv
p
p
N
λ
λ
==
Δ
=
photon
atom
Substitute numerical values and evaluate N:
( )( )( ) 4
34
27
1008.5
sJ10626.6
nm24.780m/s300kg10661.185
×=
⋅×
×
= −
−
N
The Bohr Model of the Hydrogen Atom
27 ••• In the center-of-mass reference frame of a hydrogen atom, the electron
and nucleus have momenta that have equal magnitudes p and opposite directions.
(a) Using the Bohr model, show that the total kinetic energy of the electron and
nucleus can be written K = p2
/(2μ) where μ = meM/(M + me) is called the reduced
mass, me is the mass of the electron, and M is the mass of the nucleus. (b) For the
equations for the Bohr model of the atom, the motion of the nucleus can be taken
into account by replacing the mass of the electron with the reduced mass. Use
Equation 36-14 to calculate the Rydberg constant for a hydrogen atom that has a
nucleus of mass M = mp. Find the approximate value of the Rydberg constant by
letting M go to infinity in the reduced mass formula. To how many figures does
this approximate value agree with the actual value? (c) Find the percentage
correction for the ground-state energy of the hydrogen atom by using the reduced
mass in Equation 36-16. Remark: In general, the reduced mass for a two-body
problem with masses m1 and m2 is given by
μ =
m1
m2
m1
+ m2
.

5. Chapter 3620
Picture the Problem We can express the total kinetic energy of the electron-
nucleus system as the sum of the kinetic energies of the electron and the nucleus.
Rewriting these kinetic energies in terms of the momenta of the electron and
nucleus will lead to K = p2
/2mr.
(a) Express the total kinetic energy
of the electron-nucleus system:
ne KKK +=
Express the kinetic energies of the
electron and the nucleus in terms of
their momenta:
e
2
e
2m
p
K = and
M
p
K
2
2
n =
Substitute and simplify to obtain:
μ2
2
2
11
222
2
e
e
2
e
e
2
e
22
e
2
p
mM
Mm
p
Mm
mMp
Mm
p
M
p
m
p
K
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+=+=
provided we let μ = meM/(M + me).
(b) From Equation 36-14 we have:
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
+
==
M
m
m
C
c
ekm
R
e
e
3
42
r
14 hπ
(1)
where
3
42
4 hc
ek
C
π
=
Use the Table of Physical Constants
at the end of the text to obtain:
kg/m10204663.1 137 −
×=C
For H:
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
+
=
p
e
e
H
1
m
m
m
CR

6. Atoms 21
Substitute numerical values and evaluate :HR
( ) 17
27
31
31
137
H m10096850.1
kg10673.1
kg10109.9
1
kg10109.9
kg/m10204663.1 −
−
−
−
−
×=
⎟⎟
⎟
⎟
⎟
⎠
⎞
⎜⎜
⎜
⎜
⎜
⎝
⎛
×
×
+
×
×=R
Let M → ∞ in equation (1) to obtain
:approxH,R
eapproxH, CmR =
Substitute numerical values and evaluate :approxH,R
( )( ) 1731137
approxH, m10097448.1kg10109.9kg/m10204663.1 −−−
×=××=R
HR and agree to three significant figures.approxH,R
(c) Express the ratio of the kinetic
energy K of the electron in its orbit
about a stationary nucleus to the
kinetic energy of the reduced-mass
system K′:
p
eep
p
ep
ep
ee
2
e
2
1
1
1
2
2
m
mmm
m
mm
mm
mmp
m
p
K'
K
+
=
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
===
μ
μ
Substitute numerical values and
evaluate the ratio of the kinetic
energies:
999455.0
kg10673.1
kg10109.9
1
1
27
31
=
×
×
+
=
−
−
K'
K
or
K'K 999455.0=
and the correction factor is the ratio of
the masses or %0545.0
Remarks: The correct energy is slightly less than that calculated neglecting
the motion of the nucleus.

7. Chapter 3622
Quantum Numbers in Spherical Coordinates
33 •• Find the minimum value of the angle θ between L
r
and the +z
direction for an electron in an atom that has (a) l = 1, (b) l = 4, and (c) l = 50.
Picture the Problem The minimum angle between the z axis and L
r
is the angle
between the L
r
vector for m = and the z axis.l
Express the angle θ as a function of
Lz and L:
⎟
⎠
⎞
⎜
⎝
⎛
= −
L
Lz1
cosθ
Relate the z component of L
r
to ml
and :l
lhhl == mLz
Express the angular momentum L: ( )hll 1+=L
Substitute to obtain:
( ) ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
= −−
1
cos
1
cos 11
l
l
hll
lh
θ
(a) For l = 1:
°=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
= −
0.45
11
1
cos 1
θ
(b) For l = 4:
°=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
= −
6.26
14
4
cos 1
θ
(c) For l = 50:
°=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
= −
05.8
150
50
cos 1
θ
Quantum Theory of the Hydrogen Atom
37 • (a) If electron spin is not included, how many different wave functions
are there corresponding to the first excited energy level n = 2 for a hydrogen
atom? (b) List these functions by giving the quantum numbers for each state.
Picture the Problem We can use the constraints on n, l and m to determine the
number of different wave functions, excluding spin, corresponding to the first
excited energy state of hydrogen.

8. Atoms 23
For n = 2: 1or0=l
(a) For l = 0, ml= 0 and we have: 1 state
For l = 1, ml = −1, 0, +1 and we
have:
3 states
Hence, for n = 2 we have: states4
(b) The four wave functions are
summarized to the right.
n l ml (n, l, ml)
2 0 0 (2,0,0)
2 1 −1 (2,1,−1)
2 1 0 (2,1,0)
2 1 1 (2,1,1)
General Problems
69 •• We are often interested in finding the quantity ke2
/r in electron volts
when r is given in nanometers. Show that ke2
= 1.44 eV⋅nm.
Picture the Problem We can show that ke2
= 1.44 eV⋅nm by solving the equation
for the ground state energy of an atom for ke2
.
Express the ground state energy of an
atom as a function of k, e, and a0: 0
2
0
2a
ke
E = ⇒ 00
2
2 aEke =
Substitute numerical values and
evaluate ke2
:
( )( )
nmeV44.1
nm0529.0eV6.1322
⋅=
=ke