Chapter 40
p. 1
CHAPTER 40 – Quantum Mechanics of Atoms
Note:
At the atomic scale, it is most convenient to have energies in electronvolts and wavelengths in
nanometers.
A useful expression for the energy of a photon in terms of its wavelength is
E
=
hf
=
hc
/
λ
= (6.63
×
10
–34
J
·
s)(3
×
10
8
m/s)(10
9
nm/m)/(1.60
×
10
–19
J/eV)
;
E
= (1.24
×
10
3
eV
·
nm)/
.
1.
The value of
¬
can range from 0 to
n
– 1.
Thus for
n
= 6, we have
¬
= 0, 1, 2, 3, 4, 5
.
2.
The value of
m
¬
can range from –
¬
to +
¬
.
Thus for
¬
= 3, we have
m
¬
= – 3, – 2, – 1, 0, 1, 2, 3
.
The possible values of
m
s
are
–
!
, +
!
.
3.
The value of
¬
can range from 0 to
n
– 1.
Thus for
n
= 4, we have
¬
= 0, 1, 2, 3.
For each
¬
the value of
m
¬
can range from –
¬
to +
¬
, or 2
¬
+ 1 values.
For each of these there are two
values of
m
s
.
Thus the total number for each
¬
is 2(2
¬
+ 1).
The number of states is
N
= 2(0 + 1) + 2(2 + 1) + 2(4 + 1) + 2(6 + 1) =
32 states
.
We start with
¬
= 0, and list the quantum numbers in the order (
n
,
¬
,
m
¬
,
m
s
);
(4, 0, 0, –
!
),
(4, 0, 0, +
!
),
(4, 1, –1, –
!
),
(4, 1, –1, +
!
),
(4, 1, 0, –
!
),
(4, 1, 0, +
!
),
(4, 1, 1, –
!
),
(4, 1, 1, +
!
),
(4, 2, – 2, –
!
),
(4, 2, – 2, +
!
),
(4, 2, –1, –
!
),
(4, 2, –1, +
!
),
(4, 2, 0, –
!
),
(4, 2, 0, +
!
),
(4, 2, 1, –
!
),
(4, 2, 1, +
!
),
(4, 2, 2, –
!
),
(4, 2, 2, +
!
),
(4, 3, – 3, –
!
),
(4, 3, – 3, +
!
),
(4, 3, – 2, –
!
),
(4, 3, – 2, +
!
),
(4, 3, – 1, –
!
),
(4, 3, – 1, +
!
),
(4, 3, 0, –
!
),
(4, 3, 0, +
!
),
(4, 3, 1, –
!
),
(4, 3, 1, +
!
),
(4, 3, 2, –
!
),
(4, 3, 2, +
!
),
(4, 3, 3, –
!
),
(4, 3, 3, +
!
).
4.
The value of
m
¬
can range from –
¬
to +
¬
, so we have
¬
= 3
.
The value of
¬
can range from 0 to
n
– 1.
Thus we have
n
=
¬
+ 1 (minimum 4).
There are two values of
m
s
:
m
s
= –
!
, +
!
.
5.
The value of
¬
can range from 0 to
n
– 1.
Thus for
¬
= 4, we have
n
= 5
.
For each
¬
the value of
m
¬
can range from –
¬
to +
¬
:
m
¬
= – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4
.
There are two values of
m
s
:
m
s
= –
!
, +
!
.
6.
The magnitude of the angular momentum depends on
¬
only:
L
=
˙
[
¬
(
¬
+ 1)]
1/2
= (1.055
×
10
–34
J
·
s)[(2)(2 + 1)]
1/2
=
v6
˙
= 2.58
×
10
–34
kg
·
m
2
/s
.
7. (
a
) The principal quantum number is
n
=
6
.
(
b
) The energy of the state is
E
6
= – (13.6 eV)/
n
2
= – (13.6 eV)/6
2
=
– 0.378 eV
.
(
c
) From spdfgh, we see that the “g” subshell has
¬
= 4
.
The magnitude of the angular momentum
d
e
p
e
n
d
s
o
n
¬
only:
L
=
˙
[
¬
(
¬
+ 1)]
1/2
= (1.055
×
10
–34
J
·
s)[(4)(4 + 1)]
1/2
=
v20
˙
= 4.72
×
10
–34
kg
·
m
2
/s
.
(
d
) For each
¬
the value of
m
¬
can range from –
¬
to +
¬
:
m
¬
= – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4
.
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View Full DocumentChapter 40
p. 2
8. (
a
) For each
¬
the value of
m
¬
can range from –
¬
to +
¬
, or 2
¬
+ 1 values.
For each of these there are
two values of
m
s
.
Thus the total number of electrons allowed in a subshell is
N
= 2(2
¬
+ 1).
(
b
) For the values of
¬
we have
¬
= 0,
N
= 2[2(0) + 1] = 2;
¬
= 1,
N
= 2[2(1) + 1] = 6;
¬
= 2,
N
= 2[2(2) + 1] = 10;
¬
= 3,
N
= 2[2(3) + 1] = 14;
¬
= 4,
N
= 2[2(4) + 1] = 18;
¬
= 5,
N
= 2[2(5) + 1] = 22;
¬
= 6,
N
= 2[2(6) + 1] = 26
.
9.
From Problem 8 the total number of electrons allowed in a subshell with quantum number
¬
is 2(2
¬
+ 1).
For a state with quantum number
n
the value of
¬
can range from 0 to
n
– 1.
Thus we have
N
=2
(2
+1)
Σ
=0
n
– 1
=4
Σ
n
– 1
+2
1
Σ
n
– 1
(
n
–1)
n
2
n
n
2
.
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 Spring '09
 Fielding
 Physics, Atom, Angular Momentum, Energy, Photon, Atomic orbital, radial probability density

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