41

ATOMIC STRUCTURE

41.1.

IDENTIFY: For a particle in a cubical box, different values of n X , nY and nZ can give the same energy.
SET UP: En

X

, nY , nZ

=

(n X2 + nY2 + nZ2 )π 2

2

.

2mL2

EXECUTE: (a) n X2 + nY2 + nZ2 = 3. This only occurs for n X = 1, nY = 1, nZ = 1 and the degeneracy is 1.
(b) n 2X + nY2 + nZ2 = 9. Occurs for n X = 2, nY = 1, nZ = 1, for n X = 1, nY = 2, nZ = 1 and for

n X = 1, nY = 1, nZ = 2. The degeneracy is 3.
41.2.

EVALUATE: In the second case, three different states all have the same energy.
IDENTIFY: Use an electron in a cubical box to model the hydrogen atom.

3π 2 2
6π 2 2
3π 2 2
4
SET UP: E1,1,1 =
. E2,1,1 =
. ΔE =
. L3 = π a 3.
2
2
3
2mL
2mL
2mL2
1/3

⎛ 4π ⎞
L=⎜
⎟
⎝ 3 ⎠

a = 8.527 × 10−11 m.

EXECUTE: Δ E =

E=−

41.3.

13.6 eV

n2

2(9.109 × 10−31 kg)(8.53 × 10−11 m)2

= 2.49 × 10−17 J = 155 eV. In the Bohr model,

. The energy separation between the n = 2 and n = 1 levels is

⎛1 1 ⎞ 3
Δ EBohr = (13.6 eV) ⎜ 2 − 2 ⎟ = (13.6 eV) = 10.2 eV.
⎝1 2 ⎠ 4
EVALUATE: A particle in a box is not a good model for a hydrogen atom.
IDENTIFY: The energy of the photon is equal to the energy difference between the states. We can use this
energy to calculate its wavelength.
hc
3π 2 2
9π 2 2
3π 2 2
SET UP: E1,1,1 =
=
Δ
=
.
E
.
E
. ΔE = .
2,2,1
2
2

2
λ
2mL
2mL
mL

EXECUTE: Δ E =

λ=
41.4.

3π 2 (1.055 × 10−34 J ⋅ s) 2

3π 2 (1.055 × 10−34 J ⋅ s) 2
(9.109 × 10

−31

kg)(8.00 × 10

−11

m)

2

= 5.653 × 10−17 J. Δ E =

hc


λ

gives

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 3.51 × 10−9 m = 3.51 nm.
ΔE
5.653 × 10−17 J

EVALUATE: This wavelength is much shorter than that of visible light.
IDENTIFY: Use the probability function for a particle in a three-dimensional box to find the points where
it is a maximum.
(a) SET UP: n X = 1, nY = 1, nZ = 1. ψ

2

3

π x ⎞⎛ 2 π y ⎞⎛ 2 π z ⎞
⎛L⎞ ⎛
= ⎜ ⎟ ⎜ sin 2
⎟⎜ sin
⎟⎜ sin
⎟.
L ⎠⎝
L ⎠⎝
L ⎠
⎝2⎠ ⎝


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41-1


41-2

Chapter 41

πy
πz
πx π
L
= ±1, sin
= ±1 , and sin
= ±1.
=
and x = .
L
L
L
L 2
2
π x 3π
3L
=
and x =
, but this is outside the box. Similar results obtain for y and z,
The next larger value is

L
2
2
EXECUTE:

so ψ

2

ψ

2

is maximum where sin

is maximum at the point x = y = z = L/2. This point is at the center of the box.
3

2π x ⎞⎛ 2 2π y ⎞⎛ 2 π z ⎞
⎛L⎞ ⎛
= ⎜ ⎟ ⎜ sin 2
⎟⎜ sin
⎟⎜ sin
⎟.
2
L ⎠⎝
L ⎠⎝
L ⎠
⎝ ⎠ ⎝
2π x

2π y
πz
2π x π
L
2
EXECUTE: ψ is maximum where sin
= ±1, sin
= ±1, and sin
= ±1.
=
and x = .
L
L
L
L
2
4
2π x 3π
3L
L
3L
L
2
and x = . Similarly, y = and
. As in part (a), z = . ψ is a maximum at the four
=
4
4
2
4

L
2
⎛ L L L ⎞ ⎛ L 3L L ⎞ ⎛ 3L L L ⎞
⎛ 3L 3 L L ⎞
points ⎜ , , ⎟ , ⎜ , , ⎟ , ⎜ , , ⎟ and ⎜ , , ⎟ .
4
4
2
4
4
2
4
4
2
⎝
⎠ ⎝
⎠ ⎝
⎠
⎝ 4 4 2⎠
EVLUATE: The points are located symmetrically relative to the center of the box.
IDENTIFY: A particle is in a three-dimensional box. At what planes is its probability function zero?
(b) SET UP: n X = 2, nY = 2, nZ = 1. ψ

41.5.

πx

2

3


2 ⎛L⎞ ⎛
2π x ⎞⎛ 2 2π y ⎞⎛ 2 π z ⎞
SET UP: ψ 2,2,1 = ⎜ ⎟ ⎜ sin 2
⎟⎜ sin
⎟⎜ sin
⎟.
L ⎠⎝
L ⎠⎝
L ⎠
⎝2⎠ ⎝
2
2π x
L
EXECUTE: ψ 2,2,1 = 0 for
= 0, π , 2π ,… . x = 0 and x = L correspond to walls of the box. x =
L
2
2
2
L
π
z
factor is
is the other plane where ψ 2,2,1 = 0. Similarly, ψ 2,2,1 = 0 on the plane y = . The sin 2
2
L

2


zero only on the walls of the box. Therefore, for this state ψ 2,2,1 = 0 on the following two planes other
than walls of the box: x =

L
L
and y = .
2
2

3

2 ⎛L⎞ ⎛
2π x ⎞⎛ 2 π y ⎞⎛ 2 π z ⎞
ψ 2,1,1 = ⎜ ⎟ ⎜ sin 2
⎟⎜ sin
⎟⎜ sin
⎟ is zero only on one plane ( x = L/2) other than the walls
L ⎠⎝
L ⎠⎝
L ⎠
⎝2⎠ ⎝

of the box.
3

2 ⎛L⎞ ⎛
π x ⎞⎛
π y ⎞⎛ 2 π z ⎞
ψ 1,1,1 = ⎜ ⎟ ⎜ sin 2 ⎟⎜ sin 2
⎟⎜ sin

⎟ is zero only on the walls of the box; for this state there are
L
L ⎠⎝
L ⎠
2
⎝ ⎠ ⎝
⎠⎝

41.6.

zero additional planes.
EVALUATE: For comparison, (2,1,1) has two nodal planes, (2,1,1) has one nodal and (1,1,1) has no nodal
planes. The number of nodal planes increases as the energy of the state increases.
IDENTIFY: A proton is in a cubical box approximately the size of the nucleus.
6π 2 2
3π 2 2
3π 2 2
SET UP: E1,1,1 =
=
Δ
=
E
.
.
E
.
2,1,1
2mL2
2mL2
2mL2


41.7.

3π 2 (1.055 × 10−34 J ⋅ s) 2

= 9.85 × 10−13 J = 6.15 MeV
2(1.673 × 10−27 kg)(1.00 × 10−14 m) 2
EVALUATE: This energy difference is much greater than the energy differences involving orbital electrons.
IDENTIFY: The possible values of the angular momentum are limited by the value of n.
SET UP: For the N shell n = 4, 0 ≤ l ≤ n – 1, m ≤ l , ms = ± 12 .
EXECUTE: Δ E =

EXECUTE: (a) The smallest l is l = 0. L = l (l + 1) , so Lmin = 0.
(b) The largest l is n − 1 = 3 so Lmax = 3(4) = 2 3 = 3.65 × 10−34 kg ⋅ m 2 /s.
(c) Let the chosen direction be the z-axis. The largest m is m = l = 3.

Lz ,max = m = 3 = 3.16 × 10−34 kg ⋅ m 2 /s.

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Atomic Structure

41-3

(d) S z = ± 12 . The maximum value is S z = /2 = 5.27 × 10−35 kg ⋅ m 2 /s.
1

Sz 2

1
=
= .
6
Lz 3
EVALUATE: The orbital and spin angular momenta are of comparable sizes.
IDENTIFY and SET UP: L = l (l + 1) . Lz = ml . l = 0, 1, 2, …, n − 1. ml = 0, ± 1, ± 2,..., ± l. cos θ = Lz /L.
(e)

41.8.

EXECUTE: (a) l = 0: L = 0, Lz = 0. l = 1: L = 2 , Lz = , 0, − . l = 2: L = 6 , Lz = 2 , , 0, − , −2 .
l = 3: L = 2 3 , Lz = 3 , 2 , , 0, − , −2 , −3 . l = 4: L = 2 5 , Lz = 4 , 3 , 2 , , 0, − , −2 , −3 , − 4 .

(b) L = 0: θ not defined. L = 2 : 45.0°, 90.0°, 135.0°. L = 6 : 35.3°, 65.9°, 90.0°, 114.1°, 144.7°.

L = 2 3 : 30.0°, 54.7°, 73.2°, 90.0°, 106.8°, 125.3°, 150.0°.
L = 2 5 : 26.6°, 47.9°, 63.4°, 77.1°, 90.0°, 102.9°, 116.6°, 132.1°, 153.4°.

(c) The minimum angle is 26.6° and occurs for l = 4, ml = +4. The maximum angle is 153.4° and occurs

for l = 4, ml = −4.
41.9.

EVALUATE: There is no state where L is totally aligned along the z-axis.
IDENTIFY and SET UP: The magnitude of the orbital angular momentum L is related to the quantum
number l by Eq. (41.22): L = l (l + 1) , l = 0, 1, 2,…
2

2

−34
kg ⋅ m 2 /s ⎞
⎛ L ⎞ ⎛ 4.716 × 10
EXECUTE: l (l + 1) = ⎜ ⎟ = ⎜
⎟ = 20
34
−
J ⋅ s ⎟⎠
⎝ ⎠ ⎜⎝ 1.055 × 10
And then l (l + 1) = 20 gives that l = 4.

41.10.

EVALUATE: l must be integer.
IDENTIFY and SET UP: L = l (l + 1) . Lz = ml . ml = 0, ±1, ± 2, …, ± l. cos θ = Lz /L.
EXECUTE: (a) (ml ) max = 2, so (Lz ) max = 2 .
(b) L = l (l + 1) = 6 = 2.45 . L is larger than (Lz )max .

⎛m ⎞
⎛L ⎞
(c) The angle is arccos ⎜ z ⎟ = arccos ⎜ l ⎟ , and the angles are, for ml = −2 to ml = 2, 144.7°,
⎝ L⎠
⎝ 6⎠
114.1°, 90.0°, 65.9°, 35.3°.

41.11.

EVALUATE: The minimum angle for a given l is for ml = l. The angle corresponding to ml = l will
always be smaller for larger l .
IDENTIFY and SET UP: The angular momentum L is related to the quantum number l by Eq. (41.22),

L = l (l + 1) . The maximum l, lmax , for a given n is lmax = n − 1.
EXECUTE: For n = 2, lmax = 1 and L = 2 = 1.414 .

For n = 20, lmax = 19 and L = (19)(20) = 19.49 .

41.12.

For n = 200, lmax = 199 and L = (199)(200) = 199.5 .
EVALUATE: As n increases, the maximum L gets closer to the value n postulated in the Bohr model.
IDENTIFY: l = 0, 1, 2, …, n − 1. ml = 0, ± 1, ± 2, …, ± l.
SET UP: En = −

13.60 eV

.
n2
EXECUTE: The (l , ml ) combinations are (0, 0), (1, 0), (1, ± 1), (2, 0), (2, ± 1), (2, ± 2), (3, 0),
(3, ± 1), (3, ± 2), (3, ± 3), (4, 0), (4, ± 1), (4, ± 2), (4, ± 3) and (4, ± 4) a total of 25.
(b) Each state has the same energy (n is the same), −

13.60 eV
= −0.544 eV.
25

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41-4


Chapter 41
EVALUATE: The number of l , ml combinations is n 2 . The energy depends only on n, so is the same for

41.13.

all l , ml states for a given n.
IDENTIFY: For the 5g state, l = 4, which limits the other quantum numbers.
SET UP: ml = 0, ±1, ± 2, … , ± l . g means l = 4. cosθ = Lz /L, with L = l (l + 1)

and Lz = ml .

EXECUTE: (a) There are eighteen 5g states: ml = 0, ±1, ± 2, ± 3, ± 4, with ms = ± 12 for each.
(b) The largest θ is for the most negative ml . L = 2 5 . The most negative Lz is Lz = −4 .

cosθ =

−4
2 5

and θ = 153.4°.

(c) The smallest θ is for the largest positive ml , which is ml = +4. cosθ =

4
2 5

and θ = 26.6°.

EVALUATE: The minimum angle between L and the z-axis is for ml = + l and for that ml , cos θ =
41.14.


IDENTIFY: The probability is P = ∫

a/2

0

l
.
l (l + 1)

2

ψ 1s 4π r 2dr.

SET UP: Use the expression for the integral given in Example 41.4.
a /2

EXECUTE: (a) P =

4 ⎡⎛ ar 2 a 2r a 3 ⎞ −2 r /a ⎤
−
− ⎟e
⎢⎜ −
⎥
2
4 ⎟⎠
a3 ⎢⎣⎜⎝ 2
⎦⎥ 0


= 1−

5e−1
= 0.0803.
2

(b) Example 41.4 calculates the probability that the electron will be found at a distance less than a from the
nucleus. The difference in the probabilities is (1 − 5e −2 ) − (1 − (5/2)e −1 ) = (5/2)(e −1 − 2e −2 ) = 0.243.
EVALUATE: The probability for distances from a/2 to a is about three times the probability for distances
between 0 and a/2. This agrees with Figure 41.8 in the textbook; P (r ) is maximum for r = a.
41.15.

a

2

a

1

0

π a3

IDENTIFY: P (a ) = ∫ ψ 1s dV = ∫
0

SET UP: From Example 41.4,

∫r


2 −2 r /a

e

e−2r/a (4π r 2 dr ).
⎛ −ar 2 a 2r a3 ⎞ −2 r /a
dr = ⎜
−
− ⎟e
.
⎜ 2
2
4 ⎟⎠
⎝

EXECUTE:
a

4 ⎡⎛ − ar 2 a 2r a 3 ⎞ −2r /a ⎤
4 ⎡⎛ − a 3 a 3 a3 ⎞ −2 a 3 0 ⎤
P(a ) = ∫
= 3 ⎢⎜
−
− ⎟e
= 3 ⎢⎜
− − ⎟ e + e ⎥ = 1 − 5e−2 .
⎥
⎜
⎟

⎜
2
2
4
2
2
4 ⎟⎠
4 ⎦⎥
a ⎢⎣⎝
⎠
⎦⎥ 0 a ⎣⎢⎝
EVALUATE: P (a ) < 1, as it must be.
IDENTIFY: Require that Φ (φ ) = Φ (φ + 2π )
4

a 2 −2 r /a
r e
dr
3 0
a

41.16.

SET UP: ei ( x1 + x2 ) = eix1 eix2
EXECUTE: Φ (φ + 2π ) = eiml (φ + 2π ) = eiml φ eiml 2π . eiml 2π = cos(ml 2π ) + i sin(ml 2π ). eiml 2π = 1 if ml is an

integer.
EVALUATE: If, for example, ml = 12 , eiml 2π = eiπ = cos(π ) + i sin(π ) = −1 and Φ (φ ) = −Φ (φ + 2π ). But if

ml = 1, eiml 2π = ei 2π = cos(2π ) + i sin(2π ) = +1 and Φ (φ ) = Φ (φ + 2π ), as required.

41.17.

IDENTIFY: Apply ΔU = μ B B.
SET UP: For a 3p state, l = 1 and ml = 0, ± 1.
EXECUTE: (a) B =

U

μB

=

(2.71 × 10−5 eV)
(5.79 × 10 −5 eV/T)

= 0.468 T.

(b) Three: ml = 0, ± 1.
EVALUATE: The ml = +1 level will be highest in energy and the ml = −1 level will be lowest. The

ml = 0 level is unaffected by the magnetic field.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Atomic Structure
41.18.

41-5


IDENTIFY: Apply Eq. (41.36).
SET UP: μ B = 5.788 × 10−5 eV/T
EXECUTE: (a) Δ E = μ B B = (5.79 × 10−5 eV/T)(0.400 T) = 2.32 × 10−5 eV.
(b) ml = −2 the lowest possible value of ml .
(c) The energy level diagram is sketched in Figure 41.18.
EVALUATE: The splitting between ml levels is independent of the n values for the state. The splitting is
much less than the energy difference between the n = 3 level and the n = 1 level.

Figure 41.18
41.19.

IDENTIFY and SET UP: The interaction energy between an external magnetic field and the orbital angular
momentum of the atom is given by Eq. (41.36). The energy depends on ml with the most negative ml

value having the lowest energy.
EXECUTE: (a) For the 5g level, l = 4 and there are 2l + 1 = 9 different ml states. The 5g level is split
into 9 levels by the magnetic field.
(b) Each ml level is shifted in energy an amount given by U = ml μB B. Adjacent levels differ in ml by
one, so ΔU = μ B B.

μB =

e
(1.602 × 10−19 C)(1.055 × 10−34 J ⋅ s)
=
= 9.277 × 10−24 A ⋅ m 2
2m
2(9.109 × 10−31 kg)

ΔU = μ B B = (9.277 × 10−24 A/m 2 )(0.600 T) = 5.566 × 10−24 J(1 eV/1.602 × 10−19 J) = 3.47 × 10−5 eV

(c) The level of highest energy is for the largest ml , which is ml = l = 4; U 4 = 4μ B B. The level of lowest

energy is for the smallest ml , which is ml = −l = −4; U −4 = −4μB B. The separation between these two
levels is U 4 − U −4 = 8μ B B = 8(3.47 × 10−5 eV) = 2.78 × 10−4 eV.
EVALUATE: The energy separations are proportional to the magnetic field. The energy of the n = 5 level
in the absence of the external magnetic field is (−13.6 eV)/52 = −0.544 eV, so the interaction energy with
41.20.

the magnetic field is much less than the binding energy of the state.
IDENTIFY: The effect of the magnetic field on the energy levels is described by Eq. (41.36). In a
transition ml must change by 0 or ±1.
SET UP: For a 2p state, ml can be 0, ± 1. For a 1s state, ml must be zero.
EXECUTE: (a) There are three different transitions that are consistent with the selection rules. The initial
ml values are 0, ±1; and the final ml value is 0.
(b) The transition from ml = 0 to ml = 0 produces the same wavelength (122 nm) that was seen without the

magnetic field.
(c) The larger wavelength (smaller energy) is produced from the ml = −1 to ml = 0 transition.
(d) The shorter wavelength (greater energy) is produced from the ml = +1 to ml = 0 transition.
EVALUATE: The magnetic field increases the energy of the ml = 1 state, decreases the energy for ml = −1
41.21.

and leaves the ml = 0 state unchanged.
IDENTIFY and SET UP: For a classical particle L = I ω . For a uniform sphere with mass m and radius R,
2
⎛2
⎞
I = mR 2 , so L = ⎜ mR 2 ⎟ ω. Solve for ω and then use v = rω to solve for v.
5
⎝5

⎠

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


41-6

Chapter 41

EXECUTE: (a) L =

ω=

5 3/4
2mR 2

=

3
4

so

2
3
mR 2ω =
5
4


5 3/4(1.055 × 10−34 J ⋅ s)
2(9.109 × 10−31 kg)(1.0 × 10−17 m)2

= 2.5 × 1030 rad/s

(b) v = rω = (1.0 × 10−17 m)(2.5 × 1030 rad/s) = 2.5 × 1013 m/s
EVALUATE: This is much greater than the speed of light c, so the model cannot be valid.
41.22.

IDENTIFY: Apply Eq. (41.40), with S z = − .
2
e
SET UP: μB =
= 5.788 × 10−5 eV/T.
2m
(2.00232)
⎛ e ⎞⎛ − ⎞
EXECUTE: (a) U = + (2.00232) ⎜
μ B B.
⎟⎜ ⎟ B = −
2
m
2
2
⎝
⎠⎝ ⎠

(2.00232)
(5.788 × 10−5 eV/T)(0.480 T) = −2.78 × 10−5 eV.
2

(b) Since n = 1, l = 0 so there is no orbital magnetic dipole interaction. But if n ≠ 1 there could be orbital
magnetic dipole interaction, since l < n would then allow for l ≠ 0.
U =−

EVALUATE: The energy of the ms = − 12 state is lowered in the magnetic field. The energy of the

ms = + 12 state is raised.
41.23.

IDENTIFY and SET UP: The interaction energy is U = − μ ⋅ B, with μ z given by Eq. (41.40).
EXECUTE: U = − μ ⋅ B = + μ z B, since the magnetic field is in the negative z-direction.

⎛ e ⎞
⎛ e ⎞
⎟ S z , so U = −(2.00232) ⎜
⎟ Sz B
⎝ 2m ⎠
⎝ 2m ⎠

μ z = −(2.00232) ⎜

⎛e ⎞
S z = ms , so U = −2.00232 ⎜
⎟ ms B
⎝ 2m ⎠
e
= μB = 5.788 × 10−5 eV/T
2m
U = −2.00232μBms B


1
level has lower energy.
2
1⎞
1⎞
⎛ 1 ⎛ 1 ⎞⎞
⎛
⎛
ΔU = U ⎜ ms = − ⎟ − U ⎜ ms = + ⎟ = −2.00232 μB B ⎜ − − ⎜ + ⎟ ⎟ = +2.00232 μB B
2
2
⎝
⎠
⎝
⎠
⎝ 2 ⎝ 2 ⎠⎠

The ms = +

ΔU = +2.00232(5.788 × 10−5 eV/T)(1.45 T) = 1.68 × 10−4 eV
EVALUATE: The interaction energy with the electron spin is the same order of magnitude as the
interaction energy with the orbital angular momentum for states with ml ≠ 0. But a 1s state has

41.24.

l = 0 and ml = 0, so there is no orbital magnetic interaction.
IDENTIFY: The transition energy ΔE of the atom is related to the wavelength λ of the photon by
hc
Δ E = . For an electron in a magnetic field the spin magnetic interaction energy is ± μB B. Therefore the


λ

effective magnetic field is given by ΔE = 2μB B when ΔE is produced by the hyperfine interaction.
SET UP: μB = 5.788 × 10−5 eV/T.
EXECUTE: (a) λ =

f =

c

λ

=

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
=
= 21 cm,
ΔE
(5.9 × 10−6 eV)

(3.00 × 108 m/s)
= 1.4 × 109 Hz, a short radio wave.
0.21 m

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Atomic Structure


41-7

(b) The effective field is B ≅ Δ E/2μB = 5.1 × 10−2 T, far smaller than that found in Example 41.7 for spin-

41.25.

41.26.

41.27.

orbit coupling.
EVALUATE: The level splitting due to the hyperfine interaction is much smaller than the level splittings
due to the spin-orbit interaction.
IDENTIFY and SET UP: j can have the values l + 1/2 and l − 1/2.
EXECUTE: If j takes the values 7/2 and 9/2 it must be that l − 1/2 = 7/2 and l = 8/2 = 4. The letter that
labels this l is g.
EVALUATE: l must be an integer.
IDENTIFY: Fill the subshells in the order of increasing energy. An s subshell holds 2 electrons, a
p subshell holds 6 and a d subshell holds 10 electrons.
SET UP: Germanium has 32 electrons.
EXECUTE: The electron configuration is 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 2 .
EVALUATE: The electron configuration is that of zinc (Z = 30) plus two electrons in the 4p subshell.
IDENTIFY: The ten lowest energy levels for electrons are in the n = 1 and n = 2 shells.
SET UP: l = 0, 1, 2, …, n − 1. ml = 0, ± 1, ± 2, …, ± l. ms = ± 12 .
EXECUTE: n = 1, l = 0, ml = 0, ms = ± 12 : 2 states. n = 2, l = 0, ml = 0, ms = ± 12 : 2 states.

n = 2, l = 1, ml = 0, ± 1, ms = ± 12 : 6 states.
EVALUATE: The ground state electron configuration for neon is 1s 2 2 s 2 2 p 6 . The electron configuration
41.28.


specifies the n and l quantum numbers for each electron.
IDENTIFY: Write out the electron configuration for ground-state carbon.
SET UP: Carbon has 6 electrons.
EXECUTE: (a) 1s 2 2s 2 2 p 2 .
(b) The element of next larger Z with a similar electron configuration has configuration
1s 2 2s 2 2 p 6 3s 2 3 p 2 . Z = 14 and the element is silicon.

41.29.

EVALUATE: Carbon and silicon are in the same column of the periodic table.
IDENTIFY: Write out the electron configuration for ground-state beryllium.
SET UP: Beryllium has 4 electrons.
EXECUTE: (a) 1s 2 2 s 2
(b) 1s 2 2 s 2 2 p 6 3s 2 . Z = 12 and the element is magnesium.
(c) 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 . Z = 20 and the element is calcium.

41.30.

EVALUATE: Beryllium, calcium and magnesium are all in the same column of the periodic table.
IDENTIFY and SET UP: Apply Eq. (41.45). The ionization potential is − En , where En is the level energy

for the least tightly bound electron.
EXECUTE: As electrons are removed, for the outermost electron the screening of the nucleus by the
remaining electrons decreases. The ground state electron configuration of magnesium is 1s 2 2 s 2 2 p 6 3s 2 .
For a 3s electron the other electrons screen the nucleus and Z eff ≈ 1. For Mg + the electron configuration
is 1s 2 2 s 2 2 p 6 3s and the 10 inner electrons screen the nucleus from the 3s electron. Z eff ≈ 2. For Mg 2+

41.31.

the electron configuration is 1s 2 2 s 2 2 p 6 . The screening for an outershell electron is further reduced and

now it is a n = 2 rather than an n = 3 electron that will be removed in ionization.
EVALUATE: Both screening and the shell structure of the atom determine the successive ionization
potentials.
IDENTIFY and SET UP: The energy of an atomic level is given in terms of n and Z eff by Eq. (41.45),

⎛ Z2
En = − ⎜ eff
⎜ n2
⎝

⎞
⎟⎟ (13.6 eV). The ionization energy for a level with energy − En is + En .
⎠

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41-8

Chapter 41

EXECUTE: n = 5 and Zeff = 2.771 gives E5 = −

(2.771)2
52

(13.6 eV) = −4.18 eV

The ionization energy is 4.18 eV.

2
EVALUATE: The energy of an atomic state is proportional to Z eff
.
41.32.

IDENTIFY and SET UP: Apply Eq. (41.45).
EXECUTE: For the 4s state, E = −4.339 eV and Z eff = 4 ( −4.339) /(−13.6) = 2.26. Similarly,

Z eff = 1.79 for the 4p state and 1.05 for the 4d state.

41.33.

EVALUATE: The electrons in the states with higher l tend to be farther away from the filled subshells and
the screening is more complete.
IDENTIFY and SET UP: Use the exclusion principle to determine the ground-state electron configuration,
as in Table 41.3. Estimate the energy by estimating Z eff , taking into account the electron screening of the

nucleus.
EXECUTE: (a) Z = 7 for nitrogen so a nitrogen atom has 7 electrons. N 2 + has 5 electrons: 1s 2 2 s 2 2 p.
(b) Z eff = 7 − 4 = 3 for the 2p level.

⎛ Z2 ⎞
32
En = − ⎜ eff
(13.6 eV) = − 2 (13.6 eV) = −30.6 eV
⎟
2
⎜ n ⎟
2
⎝

⎠
(c) Z = 15 for phosphorus so a phosphorus atom has 15 electrons.
P 2 + has 13 electrons: 1s 2 2 s 2 2 p 6 3s 2 3 p
(d) Z eff = 15 − 12 = 3 for the 3p level.

41.34.

⎛ Z2 ⎞
32
=
−
En = − ⎜ eff
(13.6
eV)
(13.6 eV) = −13.6 eV
⎟
⎜ n2 ⎟
32
⎝
⎠
EVALUATE: In these ions there is one electron outside filled subshells, so it is a reasonable approximation
to assume full screening by these inner-subshell electrons.
IDENTIFY and SET UP: Apply Eq. (41.45).
13.6 eV 2
EXECUTE: (a) E2 = −
Z eff , so Z eff = 1.26.
4
(b) Similarly, Z eff = 2.26.
EVALUATE: (c) Z eff becomes larger going down a column in the periodic table. Screening is less


41.35.

complete as n of the outermost electron increases.
IDENTIFY and SET UP: Estimate Z eff by considering electron screening and use Eq. (41.45) to calculate
the energy. Z eff is calculated as in Example 41.9.
EXECUTE: (a) The element Be has nuclear charge Z = 4. The ion Be + has 3 electrons. The outermost
electron sees the nuclear charge screened by the other two electrons so Z eff = 4 − 2 = 2.

⎛ Z2
En = − ⎜ eff
⎜ n2
⎝

⎞
22
⎟⎟ (13.6 eV) so E2 = − 2 (13.6 eV) = −13.6 eV
2
⎠

(b) The outermost electron in Ca + sees a Z eff = 2. E4 = −

22

(13.6 eV) = −3.4 eV
42
EVALUATE: For the electron in the highest l-state it is reasonable to assume full screening by the other
electrons, as in Example 41.9. The highest l-states of Be + , Mg + , Ca + , etc. all have a Z eff = 2. But the

41.36.


energies are different because for each ion the outermost sublevel has a different n quantum number.
IDENTIFY and SET UP: Apply Eq. (41.48) and solve for Z.
EXECUTE: EKα ≅ ( Z − 1) 2 (10.2 eV). Z ≈ 1 +

7.46 × 103 eV
= 28.0, which corresponds to the element
10.2 eV

Nickel (Ni).

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Atomic Structure

41.37.

41-9

EVALUATE: We use Z − 1 rather than Z in the expression for the transition energy, in order to account for
screening by the other K-shell electron.
IDENTIFY and SET UP: Apply Eq. (41.47). E = hf and c = f λ .
EXECUTE: (a) Z = 20: f = (2.48 × 1015 Hz)(20 − 1) 2 = 8.95 × 1017 Hz.

E = hf = (4.14 × 10−15 eV ⋅ s)(8.95 × 1017 Hz) = 3.71 keV. λ =

c 3.00 × 108 m/s
=
= 3.35 × 10−10 m.

f 8.95 × 1017 Hz

(b) Z = 27: f = 1.68 × 1018 Hz. E = 6.96 keV. λ = 1.79 × 10−10 m.

41.38.

(c) Z = 48: f = 5.48 × 1018 Hz, E = 22.7 keV, λ = 5.47 × 10−11 m.
EVALUATE: f and E increase and λ decreases as Z increases.
IDENTIFY: The energies of the x rays will be equal to the energy differences between the shells. From its
energy, we can calculate the wavelength of the x ray.
hc
SET UP: Δ E = . A Kα x ray is produced in a L → K transition and a K β x ray is produced in a

λ

M → K transition.
EXECUTE: Kα : Δ E = EL − EK = −12,000 eV − (−69,500 eV) = +57,500 eV.

λ=

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
=
= 0.0216 nm.
ΔE
57,500 eV

K β : Δ E = EM − EK = −2200 eV − (−69,500 eV) = +67,300 eV.
hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
=
= 0.0184 nm.

ΔE
67,300 eV
EVALUATE: These wavelengths are much shorter than the wavelengths in the visible spectrum of hydrogen.
IDENTIFY: The electrons cannot all be in the same state in a cubical box.
SET UP and EXECUTE: The ground state can hold 2 electrons, the first excited state can hold 6 electrons
and the second excited state can hold 6. Therefore, two electrons will be in the second excited state, which
has energy 3E1,1,1.

λ=

41.39.

EVALUATE: The second excited state is the third state, which has energy 3E1,1,1, as shown in Figure 41.4.
41.40.

IDENTIFY: Calculate the probability of finding a particle in certain regions of a three-dimensional box.
3

2 ⎛L⎞ ⎛
π x ⎞⎛ 2 π y ⎞⎛ 2 π z ⎞
SET UP: ψ 1,1,1 = ⎜ ⎟ ⎜ sin 2
⎟⎜ sin
⎟⎜ sin
⎟
L ⎠⎝
L ⎠⎝
L ⎠
⎝2⎠ ⎝
3


πx ⎤⎡ L 2 π y ⎤⎡ L 2 πz ⎤
⎛ 2 ⎞ ⎡ L/ 2
EXECUTE: (a) P = ⎜ ⎟ ⎢ ∫ sin 2
dx
sin
dy ⎥ ⎢ ∫ sin
dz .
0
L
L ⎦⎥ ⎣⎢ ∫0
L
L ⎦⎥
⎝ ⎠ ⎣
⎦⎣ 0
⎡ L 2 π y ⎤ ⎡ L 2 πz ⎤ L
⎢ ∫0 sin L dy ⎥ = ⎢ ∫0 sin L dz ⎥ = 2 .
⎣
⎦ ⎣
⎦
3

L /2

∫0

sin 2

πx

L/ 2


2π x ⎤
⎡x L
dx = ⎢ −
sin
L
L ⎥⎦ 0
⎣ 2 4π

⎛ L ⎞⎛ 1 ⎞
= ⎜ ⎟⎜ ⎟ .
⎝ 2 ⎠⎝ 2 ⎠

3

⎛ 2⎞ ⎛ L⎞ ⎛1⎞ 1
P = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = = 0.500.
⎝ L⎠ ⎝ 2 ⎠ ⎝2⎠ 2
3

πx ⎤⎡ L 2 π y ⎤⎡ L 2 πz ⎤
⎛ 2 ⎞ ⎡ L/ 2
(b) P = ⎜ ⎟ ⎢ ∫ sin 2
dx
sin
dy ⎥ ⎢ ∫ sin
dz .
L
/4
L ⎦⎥ ⎣⎢ ∫0

L
L ⎦⎥
⎝L⎠ ⎣
⎦⎣ 0
⎡ L 2 π y ⎤ ⎡ L 2 πz ⎤ L
⎢ ∫0 sin L dy ⎥ = ⎢ ∫0 sin L dz ⎥ = 2 .
⎣
⎦ ⎣
⎦
3

L/ 2

∫L /4

sin 2

πx

L/ 2

2π x ⎤
⎡x L
⎛ L ⎞⎛ 1 1 ⎞
dx = ⎢ −
sin
= ⎜ ⎟⎜ +
⎟.
⎥
L

L ⎦ L / 4 ⎝ 2 ⎠⎝ 4 2π ⎠
⎣ 2 4π

3

⎛ 2⎞ ⎛L⎞ ⎛1 1 ⎞ 1 1
P=⎜ ⎟ ⎜ ⎟ ⎜ +
= 0.409.
⎟= +
⎝ L ⎠ ⎝ 2 ⎠ ⎝ 4 2π ⎠ 4 2π
EVALUATE: In Example 41.1 for this state the probability for finding the particle between x = 0 and
x = L/4 is 0.091. The sum of this result and our result in part (b) is 0.091 + 0.409 = 0.500. This in turn
equals the probability of finding the particle in half the box, as calculated in part (a).

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41-10
41.41.

Chapter 41
IDENTIFY: Calculate the probability of finding a particle in a given region within a cubical box.
(a) SET UP and EXECUTE: The box has volume L3. The specified cubical space has volume ( L/4)3. Its

1
= 0.0156.
64

fraction of the total volume is


3

π x ⎤ ⎡ L / 4 2 π y ⎤ ⎡ L /4 2 π z ⎤
⎛ 2 ⎞ ⎡ L /4
(b) SET UP and EXECUTE: P = ⎜ ⎟ ⎢ ∫ sin 2
dx
sin
dy ⎥ ⎢ ∫ sin
dz .
0
L
L ⎦⎥ ⎣⎢ ∫0
L
L ⎦⎥
⎝ ⎠ ⎣
⎦⎣ 0
From Example 41.1, each of the three integrals equals
3

3

3

L L 1 ⎛ L ⎞⎛ 1 1 ⎞
−
= ⎜ ⎟⎜ − ⎟ .
8 4π 2 ⎝ 2 ⎠⎝ 2 π ⎠

3


⎛ 2⎞ ⎛L⎞ ⎛1⎞ ⎛1 1 ⎞
P = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ − ⎟ = 7.50 × 10−4.
⎝ L⎠ ⎝ 2 ⎠ ⎝ 2⎠ ⎝2 π ⎠
EVALUATE: Note that this is the cube of the probability of finding the particle anywhere between x = 0
and x = L/4. This probability is much less that the fraction of the total volume that this space represents. In
this quantum state the probability distribution function is much larger near the center of the box than near
its walls.
3

2 ⎛L⎞ ⎛
2π x ⎞⎛ 2 π y ⎞⎛ 2 π z ⎞
(c) SET UP and EXECUTE: ψ 2,1,1 = ⎜ ⎟ ⎜ sin 2
⎟⎜ sin
⎟⎜ sin
⎟.
L ⎠⎝
L ⎠⎝
L ⎠
⎝2⎠ ⎝

3

2π x ⎤ ⎡ L /4 2 π y ⎤ ⎡ L /4 2 π z ⎤
⎛ 2 ⎞ ⎡ L /4
P = ⎜ ⎟ ⎢ ∫ sin 2
dx ⎥ ⎢ ∫ sin
dy ⎥ ⎢ ∫ sin
dz .
0

L
L
L ⎦⎥
⎝L⎠ ⎣
⎦⎣ 0
⎦⎣ 0
⎡ L /4 2 π y ⎤ ⎡ L /4 2 π z ⎤ L ⎛ 1 ⎞⎛ 1 1 ⎞
⎢ ∫0 sin L dy ⎥ = ⎢ ∫0 sin L dz ⎥ = 2 ⎜ 2 ⎟⎜ 2 − π ⎟ .
⎣
⎦ ⎣
⎦
⎝ ⎠⎝
⎠
3

41.42.

2

2

L /4

∫0

sin 2

2π x
L
dx = .

L
8

2

⎛ 2⎞ ⎛ L⎞ ⎛1⎞ ⎛1 1 ⎞ ⎛ L⎞
P = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ − ⎟ ⎜ ⎟ = 2.06 × 10−3.
⎝ L⎠ ⎝ 2 ⎠ ⎝2⎠ ⎝2 π ⎠ ⎝ 8 ⎠
EVALUATE: This is about a factor of three larger than the probability when the particle is in the ground state.
∂
2
2
IDENTIFY: The probability is a maximum where ψ is a maximum, and this is where
ψ = 0. The
∂x
probability is zero where ψ
SET UP:

ψ 2 = A2 x 2e−2(α x

2
2

is zero.

+ β y 2 +γ z 2 )

. To save some algebra, let u = x 2 , so that ψ

2


= ue−2α u f ( y, z ).

∂
1
1
2
2
, x0 = ±
.
ψ = (1 − 2α u ) ψ ; the maximum occurs at u0 =
∂u
2α
2α
(b) ψ vanishes at x = 0, so the probability of finding the particle in the x = 0 plane is zero. The wave
function also vanishes for x = ±∞.
EXECUTE: (a)

EVALUATE:
41.43.

ψ

2

is a maximum at y0 = z0 = 0.
2

(a) IDENTIFY and SET UP: The probability is P = ψ dV with dV = 4π r 2dr.
EXECUTE:


ψ = A2e−2α r so P = 4π A2r 2e−2α r dr
2

2

2

(b) IDENTIFY and SET UP: P is maximum where
EXECUTE:

dP
= 0.
dr

d 2 −2α r 2
(r e
)=0
dr

2re −2α r − 4α r 3e −2α r = 0 and this reduces to 2r − 4α r 3 = 0
r = 0 is a solution of the equation but corresponds to a minimum not a maximum. Seek r not equal to 0 so
2

2

divide by r and get 2 − 4α r 2 = 0.
1
. (We took the positive square root since r must be positive.)
This gives r =

2α

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Atomic Structure
EVALUATE: This is different from the value of r, r = 0, where ψ

2

is a maximum. At r = 0, ψ

41-11
2

has a maximum but the volume element dV = 4π r 2dr is zero here so P does not have a maximum
at r = 0.
41.44.

IDENTIFY and SET UP: Evaluate ∂ 2ψ /∂x 2 , ∂ 2ψ /∂y 2 , and ∂ 2ψ /∂z 2 for the proposed ψ and put Eq.

(41.5). Use that ψ nx , ψ ny , and ψ nz are each solutions to Eq. (40.44).
EXECUTE: (a) −

⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ
+
+
⎜
2m ⎜⎝ ∂x 2 ∂y 2 ∂z 2

2

⎞
⎟⎟ + Uψ = Eψ
⎠

ψ nx , ψ ny , ψ nz are each solutions of Eq. (40.44), so −
−
−

2

d 2ψ ny

2m dy 2
2

d 2ψ nz

2m dz 2

2

d 2ψ nx

2m dx 2

1
+ k ′x 2ψ nx = Enxψ nx .
2


1
+ k ′y 2ψ ny = Eny ψ ny
2
1
+ k ′z 2ψ nz = Enz ψ nz
2
1
2

1
2

1
2

ψ = ψ nx ( x)ψ ny ( y )ψ nz ( z ), U = k ′x 2 + k ′y 2 + k ′z 2
⎛ d 2ψ n
x
⎜
=
∂x 2 ⎜⎝ dx 2

∂ 2ψ

So −

2
⎞
∂ 2ψ ⎛⎜ d ψ ny

⎟ψ n ψ n ,
=
⎟ y z ∂y 2 ⎜ dy 2
⎠
⎝

⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ
+
+
⎜
2m ⎜⎝ ∂x 2 ∂y 2 ∂z 2
2

⎞
2
⎛ d 2ψ n
z
⎟ψ n ψ n , ∂ ψ = ⎜
⎟ x z ∂z 2 ⎜ dz 2
⎝
⎠

⎞
⎟ψ n ψ n
⎟ x y
⎠

2 d 2ψ
⎛
⎞

⎞
1 2
nx
′x ψ n ⎟ψ n ψ n
k
+
⎟⎟ + Uψ = ⎜ −
x
⎜ 2m dx 2
⎟ y z
2
⎠
⎝
⎠

⎛
⎞
2 d 2ψ
2 d 2ψ
⎛
ny
1 2
1
nz
⎟ψ n ψ n + ⎜ −
′
+⎜ −
+
+ k ′z 2ψ nz
k

y
ψ
n
2
2
y
x
z
⎜ 2m dz
⎜ 2m dy
⎟
2
2
⎝
⎝
⎠

⎞
⎟ψ n ψ n
⎟ x y
⎠

⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ ⎞
+
+
⎜
⎟ + Uψ = ( Enx + Eny + Enz )ψ
2m ⎜⎝ ∂x 2 ∂y 2 ∂z 2 ⎟⎠
Therefore, we have shown that this ψ is a solution to Eq. (41.5), with energy
−


2

3⎞
⎛
Enx ny nz = Enx + Eny + Enz = ⎜ nx + n y + nz + ⎟ ω
2⎠
⎝
(b) and (c) The ground state has nx = n y = nz = 0, so the energy is E000 =

3
ω . There is only one set of
2

nx , n y and nz that give this energy.
First-excited state:
5
ω
2
There are three different sets of nx , n y , nz quantum numbers that give this energy, so there are three
nx = 1, n y = nz = 0 or n y = 1, nx = nz = 0 or nz = 1, nx = n y = 0 and E100 = E010 = E001 =

different quantum states that have this same energy.
EVALUATE: For the three-dimensional isotropic harmonic oscillator, the wave function is a product of
one-dimensional harmonic oscillator wavefunctions for each dimension. The energy is a sum of energies
for three one-dimensional oscillators. All the excited states are degenerate, with more than one state having
the same energy.

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41-12
41.45.

Chapter 41
IDENTIFY: Find solutions to Eq. (41.5).
SET UP: ω1 = k1′/m , ω 2 = k2′ /m . Let ψ nx ( x) be a solution of Eq. (40.44) with

1⎞
⎛
Enx = ⎜ nx + ⎟ ω1, ψ ny ( y ) be a similar solution, and let ψ nz ( z ) be a solution of Eq. (40.44) but with z as
2⎠
⎝
1⎞
⎛
the independent variable instead of x, and energy Enz = ⎜ nz + ⎟ ω2.
2⎠
⎝
EXECUTE: (a) As in Problem 41.44, look for a solution of the form ψ ( x, y , z ) = ψ nx ( x )ψ ny ( y )ψ nz ( z ).

Then, −
−

∂ 2ψ
∂ 2ψ
1
⎛
⎞
= ⎜ Enx − k1′x 2 ⎟ψ with similar relations for

and 2 . Adding,
2
2m ∂x
2
∂y
∂z
⎝
⎠
2

∂ 2ψ
2

⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ
+
+
⎜
2m ⎜⎝ ∂x 2 ∂y 2 ∂z 2
2

⎞ ⎛
1 2 1
2 1
2⎞
⎟⎟ = ⎜ Enx + Eny + Enz − k1′x − k1′ y − k2′ z ⎟ψ
2
2
2
⎝
⎠

⎠
= ( Enx + Eny + Enz − U )ψ = ( E − U )ψ

⎡
1⎞ ⎤
⎛
where the energy E is E = Enx + Eny + Enz = ⎢( nx + n y + 1)ω12 + ⎜ nz + ⎟ ω22 ⎥ , with nx , n y and nz all
2⎠ ⎦
⎝
⎣
nonnegative integers.
1
⎛
⎞
(b) The ground level corresponds to nx = n y = nz = 0, and E = ⎜ ω 21 + ω 22 ⎟ . The first excited level
2
⎝
⎠

41.46.

3
⎛
⎞
corresponds to nx = n y = 0 and nz = 1, since ω12 > ω 22 , and E = ⎜ ω 21 + ω 22 ⎟ .
2
⎝
⎠
(c) There is only one set of quantum numbers for both the ground state and the first excited state.
EVALUATE: For the isotropic oscillator of Problem 41.44 there are three states for the first excited level

but only one for the anisotropic oscillator.
IDENTIFY: An electron is in the 5f state in hydrogen. We want to find out about its angular mometum.
SET UP: For the 5f state, l = 3. Lz = ml . ml = 0, ±1, …, ± l. L = l (l + 1) .

EXECUTE: (a) The largest possible ml is ml = 3. Lz = 3 .
(b) L2x + L2y + L2z = L2 . L2 = 3(4)

L2x + L2x = L2 − L2z = 12

2

2

−9

= 12 2 .
2

= 3 .

EVALUATE: The restriction on Lz also places restrictions on Lx and Ly .
41.47.

IDENTIFY and SET UP: To calculate the total number of states for the n th principal quantum number shell
we must add up all the possibilities. The spin states multiply everything by 2. The maximum l value is
(n – 1), and each l value has (2l + 1) different ml values.
EXECUTE: The total number of states is
n −1

n −1


n −1

l =0

l =0

l =0

N = 2 ∑ (2l + 1) = 2∑1 + 4∑l = 2n +

41.48.

4(n − 1)( n)
= 2n + 2n 2 − 2n = 2n 2 .
2

(b) The n = 5 shell (O-shell) has 50 states.
EVALUATE: The n = 1 shell has 2 states, the n = 2 shell has 8 states, etc.
IDENTIFY: The orbital angular momentum is limited by the shell the electron is in.
SET UP: For an electron in the n shell, its orbital angular momentum quantum number l is limited by
0 ≤ l < n − 1, and its orbital angular momentum is given by L = l (l + 1) . The z-component of its angular

momentum is Lz = ml , where ml = 0, ± 1, … , ± l , and its spin angular momentum is S = 3/4 for all
electrons. Its energy in the n th shell is En = −(13.6 eV)/n 2 .

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Atomic Structure

41-13

EXECUTE: (a) L = l (l + 1) = 12 ⇒ l = 3. Therefore the smallest that n can be is 4, so

En = –(13.6eV)/n 2 = –(13.6 eV)/42 = –0.8500 eV.
(b) For l = 3, ml = ±3, ± 2, ± 1, 0. Since Lz = ml , the largest Lz can be is 3 and the smallest it can be
is −3 .
(c) S = 3/4 for all electrons.

41.49.

(d) In this case, n = 3, so l = 2, 1, 0. Therefore the maximum that L can be is Lmax = 2(2 + 1) = 6 .
The minimum L can be is zero when l = 0.
EVALUATE: At the quantum level, electrons in atoms can have only certain allowed values of their
angular momentum.
IDENTIFY: The total energy determines what shell the electron is in, which limits its angular momentum.
SET UP: The electron’s orbital angular momentum is given by L = l (l + 1) , and its total energy in the

n th shell is En = −(13.6 eV)/n 2 .
EXECUTE: (a) First find n: En = −(13.6eV)/n 2 = −0.5440 eV which gives n = 5, so l = 4, 3, 2, 1, 0.

Therefore the possible values of L are given by L = l (l + 1) , giving L = 0, 2 , 6 , 12 , 20 .
(b) E6 = − (13.6 eV)/62 = −0.3778 eV. ΔE = E6 − E5 = −0.3778 eV − ( −0.5440 eV) = +0.1662 eV
This must be the energy of the photon, so ΔE = hc/λ , which gives

λ = hc/Δ E = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)/(0.1662 eV) = 7.47 × 10−6 m = 7470 nm, which is in the

41.50.


infrared and hence not visible.
EVALUATE: The electron can have any of the five possible values for its angular momentum, but it cannot
have any others.
IDENTIFY: For the N shell, n = 4, which limits the values of the other quantum numbers.
SET UP: In the n th shell, 0 ≤ l < n − 1, ml = 0, ± 1, … , ± l , and ms = ±1/2. The orbital angular momentum

of the electron is L = l (l + 1) and its spin angular momentum is S = 3/4 .
EXECUTE: (a) For l = 3 we can have ml = ±3, ± 2± , ± 1, 0 and ms = ±1/2; for l = 2 we can have ml = ±2,
±1, 0 and ms = ±1/2; for l = 1, we can have ml = ±1, 0 and ms = ±1/2; for l = 0, we can have ml = 0 and
ms = ±1/2.

(b) For the N shell, n = 4, and for an f-electron, l = 3, giving L = l (l + 1) = 3(3 + 1) = 12 .

Lz = ml = ±3 , ± 2 , ± , 0, so the maximum value is 3 . S = 3/4 for all electrons.
(c) For a d-state electron, l = 2, giving L = 2(2 + 1) = 6 . Lz = ml , and the maximum value of ml is 2,

so the maximum value of Lz is 2 . The smallest angle occurs when Lz is most closely aligned along the
Lz
2
2
=
=
and
L
6
6
= 35.3°. The largest angle occurs when Lz is as far as possible from the L-vector, which is when Lz

angular momentum vector, which is when Lz is greatest. Therefore cos θ min =


θ min

−2
2
=−
and θ max = 144.7°.
6
6
(d) This is not possible since l = 3 for an f-electron, but in the M shell the maximum value of l is 2.
EVALUATE: The fact that the angle in part (c) cannot be zero tells us that the orbital angular momentum
of the electron cannot be totally aligned along any specified direction.
IDENTIFY: The inner electrons shield part of the nuclear charge from the outer electron.
Z2
SET UP: The electron’s energy in the n th shell, due to shielding, is En = − eff
(13.6 eV), where Z eff e is
n2
the effective charge that the electron “sees” for the nucleus.
is most negative. Therefore cos θ max =

41.51.

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41-14

Chapter 41


EXECUTE: (a) En = −

2
Z eff

n2

(13.6 eV) and n = 4 for the 4s state. Solving for Z eff gives

(42 )(−1.947 eV)
= 1.51. The nucleus contains a charge of +11e, so the average number of
13.6 eV
electrons that screen this nucleus must be 11 – 1.51 = 9.49 electrons.
(b) (i) The charge of the nucleus is +19e, but 17.2e is screened by the electrons, so the outer electron
“sees” 19e – 17.2e = 1.8e and Z eff = 1.8.
Z eff = −

(ii) En = −

2
Z eff

(1.8) 2

(13.6 eV) = −2.75 eV
42
n
EVALUATE: Sodium has 11 protons, so the inner 10 electrons shield a large portion of this charge from
the outer electron. But they don’t shield 10 of the protons, since the inner electrons are not totally
equivalent to a uniform spherical shell. (They are lumpy.)

2

(13.6 eV) = −

2

41.52.

IDENTIFY: At the r where P (r ) has its maximum value,
SET UP: From Example 41.4, r 2 ψ

2

d (r 2 ψ )
= 0.
dr

= Cr 2e −2 r/a .

2

d (r 2 ψ )
= Ce −2r/a (2r − (2r 2 /a )). This is zero for r = a. Therefore, P (r ) has its maximum
dr
value at r = a, the distance of the electron from the nucleus in the Bohr model.
EVALUATE: Our result agrees with Figure 41.8 in the textbook.
(a) IDENTIFY and SET UP: The energy is given by Eq. (39.14), which is identical to Eq. (41.21). The
potential energy is given by Eq. (23.9), with q = + Ze and q0 = − e.
EXECUTE:


41.53.

EXECUTE: E1s = −

E1s = U (r ) gives −
r=

(4π ⑀ 0 )2

me 4

1
2

2

(4π ⑀0 ) 2
1

me 4

(4π ⑀0 )2 2

2

; U (r ) = −

=−

1 e2

4π ⑀0 r

e2
4π ⑀0 r
1

2

= 2a
me 2
EVALUATE: The turning point is twice the Bohr radius.
(b) IDENTIFY and SET UP: For the 1s state the probability that the electron is in the classically forbidden
∞

2

2

∞

region is P (r > 2a ) = ∫ ψ 1s dV = 4π ∫ ψ 1s r 2 dr. The normalized wave function of the 1s state of
2a

2a

hydrogen is given in Example 41.4: ψ 1s ( r ) =

1

π a3


e − r/a . Evaluate the integral; the integrand is the same

as in Example 41.4.

⎛ 1 ⎞ ∞
EXECUTE: P (r > 2a ) = 4π ⎜ 3 ⎟ ∫ r 2e −2r /a dr
⎝ π a ⎠ 2a
⎛ r 2 2r
2 ⎞
Use the integral formula ∫ r 2e −α r dr = −e −α r ⎜ + 2 + 3 ⎟ , with α = 2/a.
⎜α α
α ⎟⎠
⎝
∞

⎛ ar 2 a 2r a 3 ⎞ ⎤
4 ⎡
4
P (r > 2a ) = − 3 ⎢e −2r /a ⎜
+
+ ⎟ ⎥ = + 3 e −4 (2a 3 + a 3 + a 3 /4)
⎜
⎟
2
4 ⎠ ⎥⎦
a ⎢⎣
a
⎝ 2
2a

P (r > 2a ) = 4e −4 (13/4) = 13e −4 = 0.238.

EVALUATE: These is a 23.8% probability of the electron being found in the classically forbidden region,
where classically its kinetic energy would be negative.

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Atomic Structure
41.54.

41-15

IDENTIFY and SET UP: Apply Eq. (41.45) and the concept of screening. For a level with quantum number
n the ionization energy is − En .
EXECUTE: (a) For large values of n, the inner electrons will completely shield the nucleus, so Z eff = 1

and the ionization energy would be
(b)

13.60 eV
3502

n2

.

= 1.11 × 10−4 eV, r350 = (350) 2 a0 = (350) 2 (0.529 × 10−10 m) = 6.48 × 10−6 m.
13.60 eV


= 3.22 × 10−5 eV, r650 = (650) 2 (0.529 × 10−10 m) = 2.24 × 10−5 m.
(650) 2
EVALUATE: For a Rydberg atom with large n the Bohr radius of the electron’s orbit is very large.
1
r ⎞ − r /2 a
⎛
ψ 2 s (r ) =
⎜ 2 − ⎟e
3⎝
a
⎠
32π a

(c) Similarly for n = 650,

41.55.

13.60 eV

∞

2

2

(a) IDENTIFY and SET UP: Let I = ∫ ψ 2 s dV = 4π ψ 2 s r 2dr. If ψ 2s is normalized then we will find
0

that I = 1.

2
r⎞
1 ∞⎛
4r 3 r 4 ⎞ − r /a
⎛ 1 ⎞ ∞⎛
EXECUTE: I = 4π ⎜
+ 2 ⎟e
2 − ⎟ e − r /a r 2dr = 3 ∫ ⎜ 4r 2 −
dr
3 ⎟ ∫0 ⎜
a⎠
a
8a 0 ⎜⎝
a ⎟⎠
⎝ 32π a ⎠ ⎝
∞
n!
Use the integral formula ∫ x ne−α x dx = n +1 , with α = 1/a.

α

0

1 ⎛
4
1
⎞ 1
I = 3 ⎜ 4(2!)( a 3 ) − (3!)(a ) 4 + 2 (4!)(a )5 ⎟ = (8 − 24 + 24) = 1; this ψ 2s is normalized.
a
8a ⎝

a
⎠ 8
(b) SET UP: For a spherically symmetric state such as the 2s, the probability that the electron will be

found at r < 4a is P (r < 4a ) = ∫

4a

0

EXECUTE: P (r < 4a ) =

Let P (r < 4a ) =
I1 = 4∫

4 a 2 − r/a

0

r e

1
8a 3

1

∫

4a


0

2

ψ 2 s r 2dr.

4a ⎛

8a 3 0

3
r4 ⎞
2 4r
+ 2 ⎟ e − r /a dr
⎜⎜ 4r −
a
a ⎟⎠
⎝

( I1 + I 2 + I3 ).

dr

Use the integral formula

2
2r
2 ⎞
−α r ⎛ r
2 −α r

=
−
+
+
with α = 1/a.
r
e
dr
e
⎜
∫
⎜ α α 2 α 3 ⎟⎟
⎝
⎠

I1 = −4 ⎡⎣e − r/a (r 2a + 2ra 2 + 2a 3 ) ⎤⎦
I2 = −

2

ψ 2 s dV = 4π ∫

4a
0

= ( −104e −4 + 8)a 3.

4 4a 3 − r/a
r e dr
a ∫0


Use the integral formula

3
3r 2 6r 6 ⎞
−α r ⎛ r
3 −α r
=
−
+
+
+
with α = 1/a.
r
e
dr
e
⎜
∫
⎜ α α 2 a 3 α 4 ⎟⎟
⎝
⎠

4a
4 ⎡ − r/a 3
e (r a + 3r 2a 2 + 6ra 3 + 6a 4 ) ⎤⎦ = (568e −4 − 24)a 3.
0
a⎣
1 4a
I 3 = 2 ∫ r 4e − r/a dr

a 0
⎛ r 4 4r 3 12r 2 24r 24 ⎞
Use the integral formula ∫ r 4e −α r dr = −e −α r ⎜ + 2 + 3 + 4 + 5 ⎟ with α = 1/a.
⎜α α
α
a
a ⎟⎠
⎝

I2 =

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41-16

Chapter 41

I3 = −

4a
1 ⎡ − r/a 4
e (r a + 4r 3a 2 + 12r 2a 3 + 24ra 4 + 24a 5 ) ⎤⎦ = ( −824e −4 + 24)a 3.
2⎣
0
a

Thus P (r < 4a ) =


1
8a

3

( I1 + I 2 + I3 ) =

1
8a 3

a 3 ([8 − 24 + 24] + e −4[−104 + 568 − 824])

1
P (r < 4a ) = (8 − 360e−4 ) = 1 − 45e −4 = 0.176.
8
EVALUATE: There is an 82.4% probability that the electron will be found at r > 4a. In the Bohr model the
electron is for certain at r = 4a; this is a poor description of the radial probability distribution for this state.
2

41.56.

d (r 2 ψ )
= 0.
dr
1
r ⎞ − r /2 a
⎛
SET UP: From Problem 41.55, ψ 2 s (r ) =
.
⎜ 2 − ⎟e

3⎝
a
⎠
32π a
IDENTIFY: P (r ) is a maximum or minimum when

EXECUTE: (a) Since the given ψ (r ) is real, r 2 ψ

2

= r 2ψ 2 . The probability density will be an extreme

d 2 2
dψ ⎞
dψ ⎞
⎛
⎛
(r ψ ) = 2 ⎜ rψ 2 + r 2ψ
⎟ = 2rψ ⎜ψ + r
⎟ = 0. This occurs at r = 0, a minimum, and when
dr
dr
dr ⎠
⎝
⎠
⎝
dψ
ψ = 0, also a minimum. A maximum must correspond to ψ + r
= 0. Within a multiplicative constant,
dr

dψ
1
ψ (r ) = (2 − r/a)e− r/ 2a ,
= − (2 − r/2a )e − r/2a , and the condition for a maximum is
dr
a

when

(2 − r/a ) = (r/a )(2 − r/2a ), or r 2 − 6ra + 4a 2 = 0 The solutions to the quadratic are r = a (3 ± 5). The ratio

of the probability densities at these radii is 3.68, with the larger density at r = a (3 + 5) = 5.24a and the
smaller density at r = a (3 − 5) = 0.76a. The maximum of P (r ) occurs at a value of r somewhat larger
than the Bohr radius of 4a.
(b) ψ = 0 at r = 2a

41.57.

EVALUATE: Parts (a) and (b) are consistent with Figure 41.8 in the textbook; note the two relative
maxima, one on each side of the minimum of zero at r = 2a.
L
⎛L
IDENTIFY: Use Figure 41.6 in the textbook to relate θ L to Lz and L: cosθ L = z so θ L = arccos ⎜ z
L
⎝ L
(a) SET UP: The smallest angle (θ L ) min

⎞
⎟.
⎠

is for the state with the largest L and the largest Lz . This is the

state with l = n − 1 and ml = l = n − 1.
EXECUTE: Lz = ml = (n − 1)

L = l (l + 1) = (n − 1)n
⎛ (n − 1)h ⎞
⎛ (n − 1) ⎞
⎛ n −1 ⎞
(θ L ) min = arccos ⎜
= arccos ⎜
= arccos ⎜⎜
⎟ = arccos( (1 − 1)/n).
⎜ ( n − 1) nh ⎟⎟
⎜ ( n − 1) n ⎟⎟
n ⎠⎟
⎝
⎝
⎠
⎝
⎠
EVALUATE: Note that (θ L ) min approaches 0° as n → ∞.
(b) SET UP: The largest angle (θ L )max is for l = n − 1 and ml = −l = − (n − 1).

(

EXECUTE: A similar calculation to part (a) yields (θ L )max = arccos − 1 − 1/n

)


EVALUATE: Note that (θ L )max approaches 180° as n → ∞.
41.58.

IDENTIFY and SET UP: L2x + L2y + L2z = L2 . L2 = l (l + 1) 2 . Lz = ml .
EXECUTE: (a) L2x + L2y = L2 − L2z = l (l + 1)

2

− ml2

2

so L2x + L2y = l (l + 1) − ml2 .

(b) This is the magnitude of the component of angular momentum perpendicular to the z-axis.

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Atomic Structure
(c) The maximum value is

41-17

l (l + 1) = L, when ml = 0. That is, if the electron is known to have no

z-component of angular momentum, the angular momentum must be perpendicular to the z-axis. The
minimum is l when ml = ± l.
EVALUATE: For l ≠ 0 the minimum value of L2x + L2y is not zero. The angular momentum vector


41.59.

cannot be totally aligned along the z-axis. For l ≠ 0, L must always have a component perpendicular to
the z-axis.
dP
IDENTIFY: At the value of r where P (r ) is a maximum,
= 0.
dr
⎛ 1 ⎞ 4 − r /a
SET UP: P (r ) = ⎜
.
⎟r e
⎝ 24a 5 ⎠
4
dP ⎛ 1 ⎞ ⎛ 3 r 4 ⎞ − r /a dP
3 r
=⎜
−
=
0
when
4
r
−
= 0; r = 4a. In the Bohr model,
4
r
e
.

⎜
⎟
⎟
dr
a
dr ⎝ 24a 5 ⎠ ⎜⎝
a ⎟⎠

EXECUTE:

41.60.

rn = n 2a so r2 = 4a, which agrees with the location of the maximum in P (r ).
EVALUATE: Our result agrees with Figure 41.8. The figure shows that P (r ) for the 2p state has a single
maximum and no zeros except at r = 0 and r → ∞.
IDENTIFY: Apply constant acceleration equations to relate Fz to the motion of an atom.

SET UP: According to Eq. (41.40), the magnitude of μ z is μ z = 9.28 × 10−24 A ⋅ m 2 . The atomic mass of

silver is 0.1079 kg/mol.
EXECUTE: The time required to transit the horizontal 50 cm region is t =

Δx 0.500 m
=
= 0.952 ms. The
vx 525 m/s

force required to deflect each spin component by 0.50 mm is
⎛
⎞ 2(0.50 × 10−3 m)

2Δz
0.1079 kg/mol
Fz = maz = ± m 2 = ± ⎜⎜
= ±1.98 × 10−22 N. Thus, the required
⎟⎟
23
−3 2
t
⎝ 6.022 × 10 atoms/mol ⎠ (0.952 × 10 s)
dBz
F
1.98 × 10−22 N
= z =
= 21.3 T/m.
μ z 9.28 × 10−24 J/T
dz
EVALUATE: The two spin components are deflected in opposite directions.
IDENTIFY: Apply Eq. (41.36).
SET UP: Decay from a 3d to 2 p state in hydrogen means that n = 3 → n = 2 and

magnetic-field gradient is
41.61.

ml = ±2, ± 1, 0 → ml = ±1, 0. However, selection rules limit the possibilities for decay. The emitted photon
carries off one unit of angular momentum so l must change by 1 and hence ml must change by 0 or ±1.
EXECUTE: The shift in the transition energy from the zero field value is
e B
(ml3 − ml2 ), where ml3 is the 3d ml value and ml2 is the 2 p ml value. Thus
U = ( ml3 − ml2 )μ B B =
2m

there are only three different energy shifts. The shifts and the transitions that have them, labeled by the ml

values, are:
e B
e B
: 2 → 1,1 → 0, 0 → −1. 0:1 → 1, 0 → 0, − 1 → −1. −
: 0 → 1, − 1 → 0, − 2 → −1.
2m
2m
41.62.

EVALUATE: Our results are consistent with Figure 41.15 in the textbook.
IDENTIFY: The presence of an external magnetic field shifts the energy levels up or down, depending
upon the value of ml .
SET UP: The selection rules tell us that for allowed transitions, Δl = 1 and Δml = 0 or ± 1.
EXECUTE: (a) E = hc/λ = (4.136 × 10 –15 eV ⋅ s)(3.00 × 108 m/s)/(475.082 nm) = 2.612 eV.

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41-18

Chapter 41
(b) For allowed transitions, Δl = 1 and Δml = 0 or ± 1. For the 3d state, n = 3, l = 2, and ml can have the

values 2, 1, 0, – 1, – 2. In the 2p state, n = 2, l = 1, and ml can be 1, 0, –1. Therefore the 9 allowed
transitions from the 3d state in the presence of a magnetic field are:
l = 2, ml = 2 → l = 1, ml = 1
l = 2, ml = 1 → l = 1, ml = 0

l = 2, ml = 1 → l = 1, ml = 1
l = 2, ml = 0 → l = 1, ml = 0
l = 2, ml = 0 → l = 1, ml = 1
l = 2, ml = 0 → l = 1, ml = −1
l = 2, ml = −1 → l = 1, ml = 0
l = 2, ml = −1 → l = 1, ml = −1
l = 2, ml = −2 → l = 1, ml = −1
(c) Δ E = µB B = (5.788 × 10−5 eV/T)(3.500 T) = 0.000203 eV
So the energies of the new states are –8.50000 eV + 0 and –8.50000 eV ± 0.000203 eV, giving energies
of: –8.50020 eV, –8.50000 eV and –8.49980 eV.
(d) The energy differences of the allowed transitions are equal to the energy differences if no magnetic
field were present (2.61176 eV, from part (a)), and that value ±ΔE (0.000203 eV, from part (c)).
Therefore we get the following:
For E = 2.61176 eV: λ = 475.082 nm (which was given)
For E = 2.61176 eV + 0.000203 eV = 2.611963 eV:

λ = hc/E = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)/(2.611963 eV) = 475.045 nm
For E = 2.61176 eV − 0.000203 eV = 2.61156 eV:

λ = hc/E = (4.136 × 10 –15 eV ⋅ s)(3.00 × 108 m/s)/(2.61156 eV) = 475.119 nm

41.63.

EVALUATE: Even a strong magnetic field produces small changes in the energy levels, and hence in the
wavelengths of the emitted light.
IDENTIFY: The presence of an external magnetic field shifts the energy levels up or down, depending
upon the value of ml .
SET UP: The energy difference due to the magnetic field is ΔE = µB B and the energy of a photon is
E = hc/λ .
EXECUTE: For the p state, ml = 0 or ± 1, and for the s state ml = 0. Between any two adjacent lines,


ΔE = µB B. Since the change in the wavelength (Δλ ) is very small, the energy change (ΔE ) is also very
small, so we can use differentials. E = hc/λ . dE =

μB B =

hcΔλ

λ

2

and B =

hcΔλ

μ Bλ 2

hc

λ

2

d λ and ΔE =

hcΔλ

λ2


. Since ΔE = µB B, we get

.

B = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)(0.0462 nm)/(5.788 × 10−5 eV/T)(575.050 nm)2 = 3.00 T

41.64.

EVALUATE: Even a strong magnetic field produces small changes in the energy levels, and hence in the
wavelengths of the emitted light.
IDENTIFY: Apply Eq. (41.36). Problem 39.86c says Δλ /λ = ΔE/E , when these quantities are small.
SET UP: μ B = 5.79 × 10−5 eV/T
EXECUTE: (a) The energy shift from zero field is ΔU 0 = ml μB B.

For ml = 2, ΔU 0 = (2)(5.79 × 10−5 eV/T)(1.40 T) = 1.62 × 10−4 eV.
For ml = 1, ΔU 0 = (1)(5.79 × 10−5 eV/T)(1.40 T) = 8.11 × 10 −5 eV.

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Atomic Structure
(b) Δλ = λ0

41-19

ΔE
⎛ 36 ⎞ 1
, where E0 = (13.6 eV)((1/4) − (1/9)), λ0 = ⎜ ⎟ = 6.563 × 10−7 m
E0

⎝ 5 ⎠R

and Δ E = 1.62 × 10−4 eV − 8.11 × 10−5 eV = 8.09 × 10−5 eV from part (a). Then,
Δλ = 2.81 × 10−11 m = 0.0281 nm. The wavelength corresponds to a larger energy change, and so the

wavelength is smaller.
EVALUATE: Δλ /λ = (0.0281 nm)/(656 nm) = 4.3 × 10−5. Δλ /λ is very small and the approximate
expression from Problem 39.86c is very accurate.
41.65.

IDENTIFY: The ratio according to the Boltzmann distribution is given by Eq. (39.18):

n1
= e− ( E1 − E0 )/kT ,
n0

where 1 is the higher energy state and 0 is the lower energy state.
⎛e ⎞
SET UP: The interaction energy with the magnetic field is U = − μ z B = 2.00232 ⎜
⎟ ms B (Example 41.6.).
⎝ 2m ⎠
1
1
The energy of the ms = + level is increased and the energy of the ms = − level is decreased.
2
2
n1/2
− (U1/ 2 −U −1/ 2 )/kT
=e
n−1/ 2

⎛ e ⎞ ⎛ 1 ⎛ 1 ⎞⎞
⎛e ⎞
EXECUTE: U1/2 − U −1/2 = 2.00232 ⎜
⎟ B ⎜ − ⎜ − ⎟ ⎟ = 2.00232 ⎜
⎟ B = 2.00232 μ B B
⎝ 2m ⎠ ⎝ 2 ⎝ 2 ⎠ ⎠
⎝ 2m ⎠

n1/2
= e− (2.00232) μB B /kT
n−1/ 2
(a) B = 5.00 × 10−5 T
−24
n1/2
= e−2.00232(9.274×10
n−1/2

A/m 2 )(5.00 ×10−5 T) / ([1.381×10−23 J/K][300 K])

−7
n1/2
= e−2.24 ×10 = 0.99999978 = 1 − 2.2 × 10−7
n−1/2

(b) B = 5.00 × 10−5 T,

−3
n1/ 2
= e−2.24 ×10 = 0.9978
n−1/2


−2
n1/2
= e−2.24 ×10 = 0.978
n−1/2
EVALUATE: For small fields the energy separation between the two spin states is much less than kT for
T = 300 K and the states are equally populated. For B = 5.00 T the energy spacing is large enough for
there to be a small excess of atoms in the lower state.
μ I
IDENTIFY: The magnetic field at the center of a current loop of radius r is B = 0 (Eq. 28.17).
2r
⎛ v ⎞
I = e⎜
⎟.
⎝ 2π r ⎠

(c) B = 5.00 × 10−5 T,

41.66.

SET UP: Using Eq. (41.22), L = mvr = l (l + 1) . The Bohr radius from Eq. (39.11) is n 2a0 .

l (l + 1)

2(6.63 × 10−34 J ⋅ s)

= 7.74 × 105 m/s. The magnetic field
m(n a0 ) 2π (9.11 × 10−31 kg)(4)(5.29 × 10−11 m)
generated by the “moving” proton at the electron’s position is
μ I μ ev

(1.60 × 10−19 C)(7.74 × 105 m/s)
B = 0 = 0 2 = (10−7 T ⋅ m/A)
= 0.277 T.
2r 4π r
(4) 2 (5.29 × 10−11 m) 2
EXECUTE: v =

2

=

EVALUATE: The effective magnetic field calculated in Example 41.7 for 3p electrons in sodium is much
larger than the value we calculated for 2p electrons in hydrogen.

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41-20

41.67.

Chapter 41

3
1
1 3
IDENTIFY and SET UP: ms can take on 4 different values: ms = − , − , + , + . Each nlml state can
2
2

2 2
have 4 electrons, each with one of the four different ms values. Apply the exclusion principle to determine
the electron configurations.
EXECUTE: (a) For a filled n = 1 shell, the electron configuration would be 1s 4 ; four electrons and Z = 4.
For a filled n = 2 shell, the electron configuration would be 1s 4 2 s 4 2 p12 ; twenty electrons and Z = 20.
(b) Sodium has Z = 11; 11 electrons. The ground-state electron configuration would be 1s 4 2 s 4 2 p 3.
EVALUATE: The chemical properties of each element would be very different.

41.68.

IDENTIFY: Apply Eq. (41.43) and Eq. (41.26), with e2 replaced by Ze2 . The photon wavelength λ is
hc
related to the transition energy ΔE for the atom by ΔE = .

λ

6+

SET UP: For N , Z = 7.
EXECUTE: (a) Z 2 ( −13.6 eV) = (7) 2 ( −13.6 eV) = −666 eV.
(b) The negative of the result of part (a), 666 eV.
(c) The radius of the ground state orbit is inversely proportional to the nuclear charge, and
a
= (0.529 × 10−10 m)/7 = 7.56 × 10−12 m.
Z
hc
hc
(d) λ =
=
, where E0 is the energy found in part (b), and λ = 2.49 nm.

ΔE
⎛1 1 ⎞
E0 ⎜ 2 − 2 ⎟
⎝1 2 ⎠

41.69.

EVALUATE: For hydrogen, the wavelength of the photon emitted in this transition is 122 nm (Section 39.3).
The wavelength for N 6 + is smaller by a factor of 7 2.
(a) IDENTIFY and SET UP: The energy of the photon equals the transition energy of the atom:
Δ E = hc/λ. The energies of the states are given by Eq. (41.21).
EXECUTE: En = −

13.60 eV
n

2

so E2 = −

13.60 eV
13.60 eV
and E1 = −
4
1

⎛ 1 ⎞ 3
ΔE = E2 − E1 = 13.60 eV⎜ − +1⎟ = (13.60 eV) = 10.20 eV = (10.20 eV)(1.602 × 10−19 J/eV) = 1.634 × 10−18 J
⎝ 4 ⎠ 4


λ=

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 1.22 × 10−7 m = 122 nm
ΔE
1.634 × 10−18 J

(b) IDENTIFY and SET UP: Calculate the change in ΔE due to the orbital magnetic interaction energy,
Eq. (41.36), and relate this to the shift Δλ in the photon wavelength.
EXECUTE: The shift of a level due to the energy of interaction with the magnetic field in the z-direction is
U = ml μB B. The ground state has ml = 0 so is unaffected by the magnetic field. The n = 2 initial state has

ml = −1 so its energy is shifted downward an amount U = ml μ B B = ( −1)(9.274 × 10−24 A/m 2 )(2.20 T) =
(−2.040 × 10−23 J)(1 eV/1.602 × 10 −19 J) = 1.273 × 10−4 eV.

Note that the shift in energy due to the magnetic field is a very small fraction of the 10.2 eV transition
energy. Problem 39.86c shows that in this situation Δλ /λ = Δ E/E . This gives

⎛ 1.273 × 10−4 eV ⎞
−3
Δλ = λ Δ E /E = 122 nm ⎜
⎟⎟ = 1.52 × 10 nm = 1.52 pm.
⎜
10.2 eV
⎝
⎠
EVALUATE: The upper level in the transition is lowered in energy so the transition energy is decreased.
A smaller ΔE means a larger λ ; the magnetic field increases the wavelength. The fractional shift in


wavelength, Δλ /λ is small, only 1.2 × 10−5.

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Atomic Structure

41.70.

IDENTIFY: Apply Eq. (41.36), where B is the effective magnetic field. ΔE =

hc

λ

41-21

.

e
eh
=
.
2m 4π m
EXECUTE: The effective field is that which gives rise to the observed difference in the energy level
Δ E hc ⎛ λ1 − λ2 ⎞ 4π mc ⎛ λ1 − λ2 ⎞
transition, B =
=
⎜

⎟=
⎜
⎟ . Substitution of numerical values gives
μB μB ⎝ λ1λ2 ⎠
e ⎝ λ1λ2 ⎠
SET UP: μ B =

B = 7.28 × 10−3 T.

41.71.

EVALUATE: The effective magnetic field we have calculated is much smaller than that calculated for
sodium in Example 41.7.
IDENTIFY: Estimate the atomic transition energy and use Eq. (39.5) to relate this to the photon
wavelength.
(a) SET UP: vanadium, Z = 23
minimum wavelength; corresponds to largest transition energy
EXECUTE: The highest occupied shell is the N shell (n = 4). The highest energy transition is N → K ,

with transition energy ΔE = EN − EK . Since the shell energies scale like 1/n 2 neglect E N relative to EK ,
so Δ E = EK = ( Z − 1) 2 (13.6 eV) = (23 − 1)2 (13.6 eV) = 6.582 × 103 eV = 1.055 × 10−15 J. The energy of the
emitted photon equals this transition energy, so the photon’s wavelength is given by
Δ E = hc/λ so λ = hc/Δ E.
(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)

= 1.88 × 10−10 m = 0.188 nm.
1.055 × 10−15 J
SET UP: maximum wavelength; corresponds to smallest transition energy, so for the Kα transition

λ=


EXECUTE: The frequency of the photon emitted in this transition is given by Moseley’s law (Eq. 41.47):
f = (2.48 × 1015 Hz)( Z − 1)2 = (2.48 × 1015 Hz)(23 − 1)2 = 1.200 × 1018 Hz

c 2.998 × 108 m/s
=
= 2.50 × 10−10 m = 0.250 nm
f 1.200 × 1018 Hz
(b) rhenium, Z = 45
Apply the analysis of part (a), just with this different value of Z.
minimum wavelength
Δ E = EK = ( Z − 1) 2 (13.6 eV) = (45 − 1) 2 (13.6 eV) = 2.633 × 104 eV = 4.218 × 10 −15 J.

λ=

λ = hc/Δ E =

(6.626 × 10−34 J . s)(2.998 × 108 m/s)
4.218 × 10−15 J

= 4.71 × 10−11 m = 0.0471 nm.

maximum wavelength
f = (2.48 × 1015 Hz)( Z − 1)2 = (2.48 × 1015 Hz)(45 − 1) 2 = 4.801 × 1018 Hz
c 2.998 × 108 m/s
=
= 6.24 × 10−11 m = 0.0624 nm
f 4.801 × 1018 Hz
EVALUATE: Our calculated wavelengths have values corresponding to x rays. The transition energies
increase when Z increases and the photon wavelengths decrease.

IDENTIFY: The interaction energy for an electron in a magnetic field is U = − μ z B , where μ z is given by

λ=

41.72.

Eq. (41.40).
SET UP: Δ S z =
EXECUTE: (a) Δ E = (2.00232)
(b) B =

e
e
hc
2π mc
⇒B=
BΔS z ≈ B =
.
λ
λe
2m
m

2π (9.11 × 10−31 kg)(3.00 × 108 m/s)
(0.0350 m)(1.60 × 10−19 C)

= 0.307 T.

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41-22

Chapter 41
EVALUATE: As shown in Figure 41.18 in the textbook, the lower state in the transition has ms = − 12 and

the upper state has ms = + 12 .
41.73.

IDENTIFY and SET UP: The potential U ( x) =

1 2
k ′x is that of a simple harmonic oscillator. Treated
2

(

)

quantum mechanically (see Section 40.5) each energy state has energy En = ω n + 12 . Since electrons
obey the exclusion principle, this allows us to put two electrons (one for each ms =

± 12

) for every value of

n—each quantum state is then defined by the ordered pair of quantum numbers (n, ms ).
EXECUTE: By placing two electrons in each energy level the lowest energy is
N −1 ⎤

⎛ N −1 ⎞
⎛ N −1 ⎛
⎡ N −1
1 ⎞⎞
1
⎡ ( N − 1)( N ) N ⎤
+ ⎥=
2 ⎜⎜ ∑ En ⎟⎟ = 2 ⎜⎜ ∑ ω ⎜ n + ⎟ ⎟⎟ = 2 ω ⎢ ∑ n + ∑ ⎥ = 2 ω ⎢
2 ⎠⎠
2
2⎦
⎣
⎥
n =0 2 ⎦
⎝ n =0 ⎠
⎝ n =0 ⎝
⎣⎢ n = 0
k′
. Here we realize that the first value of n is zero and the last value of
m
n is N – 1, giving us a total of N energy levels filled.

ω [N 2 − N + N ] = ω N 2 = N 2

EVALUATE: The minimum energy for one electron moving in this potential is

2N electrons the minimum energy is larger than (2 N )

1
2


ω , with ω =

( 12 ω ) , because only two electrons can be put into

each energy state. For example, for N = 2 (4 electrons), there are two electrons in the E =
state and two in the
41.74.

3
2

k′
. For
m

ω state, for a total energy of 2

1
2

ω energy

( 12 ω ) + 2( 32 ω ) = 4 ω , which is in agreement with

our general result.
IDENTIFY and SET UP: Apply Newton’s second law and Bohr’s quantization to one of the electrons.
EXECUTE: (a) Apply Coulomb’s law to the orbiting electron and set it equal to the centripetal force.
There is an attractive force with charge +2e a distance r away and a repulsive force a distance 2r away. So,
(+2e)(− e) ( −e)(− e)

− mv 2
+
=
. But, from the quantization of angular momentum in the first Bohr orbit,
r
4π ⑀0r 2
4π ⑀0 (2r ) 2
2

⎛
⎞
−m ⎜
2
⎟
−7 e 2
4π ⑀0 2
e
−2e
− mv
mr
⎝
⎠ =−
L = mvr = ⇒ v =
. So
+
=
=
⇒
=
−

.
4 r2
mr
r
r
mr 3
mr 3
4π ⑀0r 2 4π ⑀0 (2r ) 2
2

2

2

4 ⎛ 4π ⑀0 2 ⎞ 4
4
−10
−11
r= ⎜
⎟⎟ = a0 = (0.529 × 10 m) = 3.02 × 10 m. And
2
⎜
7 ⎝ me ⎠ 7
7
v=

mr

=


7
7
(1.054 × 10−34 J ⋅ s)
=
= 3.83 × 106 m/s.
4 ma0 4 (9.11 × 10−31 kg)(0.529 × 10−10 m)

⎛1
⎞
(b) K = 2 ⎜ mv 2 ⎟ = 9.11 × 10−31 kg (3.83 × 106 m/s) 2 = 1.34 × 10−17 J = 83.5 eV.
⎝2
⎠
⎛ −2e2 ⎞
e2
e2
−4e2
−7 ⎛ e2 ⎞
+
=
+
=
= −2.67 × 10−17 J = −166.9 eV
(c) U = 2 ⎜
⎟
⎜ 4π ⑀ r ⎟ 4π ⑀ (2r ) 4π ⑀ r 4π ⑀ (2r ) 2 ⎜⎜ 4π ⑀ r ⎟⎟
0
0
0
0
0

⎝
⎠
⎝
⎠
(d) E∞ = −[−166.9 eV + 83.5 eV] = 83.4 eV, which is only off by about 5% from the real value of 79.0 eV.
EVALUATE: The ground state energy of helium in this model is K + U = −83.4 eV. The ground state energy

of He+ is 4( −13.6 eV) = −54.4 eV. Therefore, the energy required to remove one electron from helium in
this model is − (−83.4 eV + 54.4 eV) = 29.0 eV. The experimental value for this quantity is 24.6 eV.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Atomic Structure
41.75.

41-23

IDENTIFY and SET UP: In the expression for the turning points and in the wave function replace a by a/Z
EXECUTE: (a) The radius is inversely proportional to Z, so the classical turning radius is 2a /Z .
1
(b) The normalized wave function is ψ 1s (r ) =
e− Zr/a and the probability of the electron being
3 3
π a /Z
∞
∞
4
2

ψ 4π r 2dr = 3 3 ∫ e−2 Zr/a r 2dr. Making the
found outside the classical turning point is P = ∫
2 a/Z 1s
2
a/z
a /Z
∞

change of variable u = Zr/a, dr = (a/Z )du changes the integral to P = 4∫ e−2uu 2du, which is independent
2

of Z. The probability is that found in Problem 41.53, 0.238, independent of Z.
EVALUATE: The probability of the electron being in the classically forbidden region is independent of Z.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.