Atomic Structure
41-5
(d)
eff
15 12
3
Z
=−=
for the 3
p
level.
22
eff
3
(13.6 eV)
(13.6 eV)
13.6 eV
3
n
Z
E
n
⎛⎞
=−
⎜⎟
⎝⎠
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.
41.30.
(a)
2
2e
f
f
e
f
f
13.6 eV
, so
1.26.
4
EZ
Z
=
(b)
Similarly,
eff
2.26.
Z
=
(c)
eff
Z
becomes larger going down the columns in the periodic table.
41.31.
IDENTIFY
and
SET UP:
Estimate
eff
Z
by considering electron screening and use Eq.(41.27) to calculate the
energy.
eff
Z
is calculated as in Example 41.8.
EXECUTE: (a)
The element Be has nuclear charge
4.
Z
=
The ion
+
Be has 3 electrons. The outermost electron
sees the nuclear charge screened by the other two electrons so
eff
422
.
Z
2
eff
2
(13.6 eV)
n
Z
E
n
so
2
2
2
2
(13.6 eV)
13.6 eV
2
E
(b)
The outermost electron in
+
Ca sees a
eff
2.
Z
=
2
4
2
2
(13.6 eV)
3.4 eV
4
E
EVALUATE:
For the electron in the highest
l
-state it is reasonable to assume full screening by the other electrons,
as in Example 41.8. The highest
l
-states of
+
Be ,
++
Mg , Ca , etc. all have a
eff
2.
Z
=
But the energies are different
because for each ion the outermost sublevel has a different
n
quantum number.
41.32.
2
(1
)
(
1
0
.
2
e
V
)
kx
≅−
.
3
7.46 10 eV
12
8
.
0
,
10.2 eV
Z
×
≈+
=
which corresponds to the element Nickel (Ni).
41.33.
(a)
15
2
17
20:
(2.48 10 Hz)(20 1)
8.95 10 Hz
Zf
==
×
−
=
×
.
8
15
17
10
17
3.00 10 m s
(4.14 10
eV s) (8.95 10 Hz)
3.71 keV.
3.35 10
m.
8.95 10 Hz
c
Eh
f
f
λ
−
−
×
×
⋅
×
=
=
×
×
(b)
Z
= 27:
18
10
1.68 10 Hz.
6.96 keV.
1.79 10
m.
fE
−
=×
=
(c)
18
11
48:
5.48 10 Hz,
22.7 keV,
5.47 10
m.
E
−
×
=
=
×
41.34.
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
,
and its orbital angular momentum is given by
)
Ll
l
=+
=
. The
z-
component of its angular momentum is
,
zl
Lm
=
=
where
m
l
= 0, ±1, … , ±
l
, and its spin angular momentum is
3/4
S
=
=
for all electrons. Its energy in
the
n
th
shell is
2
(13.6 eV)/
n
En
.
EXECUTE: (a)
)
1
2
3
l
l
=
⇒
=
. Therefore the smallest that
n
can be is 4, so
E
n
= – (13.6 eV)/
n
2
=
– (13.6 eV)/4
2
= –0.8500 eV.
(b)
For
l
= 3,
m
l
= ±3, ±2, ±1, 0. Since
,
=
=
the largest
L
z
can be is 3
=
and the smallest it can be is –3
=
.
(c)
S
=
=
for
all
electrons.
(d)
In this case,
n
= 3, so
l
= 2, 1, 0. Therefore the maximum that
L
can be is
max
2(2 1)
6
L
=
. 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.
41.35.
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
=
, and its total energy in the
n
th
shell
is
2
(13.6
eV)/
n
.
EXECUTE: (a)
First find
n
:
2
(13.6 eV)/
n
=
−
0.5440 eV which gives
n
= 5, so
l
= 4, 3, 2, 1, 0. Therefore the
possible values of
L
are given by
)
l
=
, giving
L
= 0,
2,
=
6,
=
12 ,
=
20 .
=
(b)
E
6
= – (13.6 eV)/6
2
= –0.3778 eV.
Δ
E
=
E
6
–
E
5
= –0.3778 eV – (–0.5440 eV) = +0.1662 eV
This must be the energy of the photon, so
Δ
E
=
hc/
, which gives
=
hc
/
Δ
E
= (4.136
×
10
–15
eV s
⋅
)(3.00
×
10
8
m/s)/(0.1662 eV) = 7.47
×
10
–6
m = 7470 nm, which is in the infrared
and hence not visible.