36.1
ATOMIC PHYSICS
Note:
For the problems in this chapter, useful numerical values of Planck’s constant, in SI and atomic units,
are:
34
15
6.626
10
J
s
4.136
10
eV
s
1240 eV
nm/c,
h
−
−
=
×
⋅ =
×
⋅ =
⋅
and
34
/2
1.055
10
J s
h
π
−
=
=
×
⋅ =
h
16
6.582
10
eV
s
197.3 MeV
fm/ .
c
−
×
⋅ =
⋅
Useful constants and combinations, in SI and atomic units, are:
34
15
34
16
8
2
27
2
23
5
6.626
10
J
s
4.136
10
eV
s
1240 eV
nm/
/2
1.055
10
J
s
6.582
10
eV
s
197.3 MeV
fm/
2.998
10
m/s,
1.440 eV
nm,
1
1.661
10
kg
931.5 MeV/
1.381
10
J/K
8.617
10
eV/K
B
h
c
h
c
c
ke
u
c
k
π
−
−
−
−
−
−
−
=
×
⋅ =
×
⋅ =
⋅
=
=
×
⋅ =
×
⋅ =
⋅
=
×
=
⋅
=
×
=
=
×
=
×
h
E
XERCISES
Section 36.1 The Hydrogen Atom
14.
2
34
2
11
0
0
2
31
19
2
4
4
(8.854 pF/m)(6.626
10
J S/2 )
5.293
10
m
(9.109
10
kg)(1.602
10
C)
a
me
πε
π
π
−
−
−
−
×
⋅
=
=
=
×
×
×
h
while
15.
I
NTERPRET
Our system consists of a group of hydrogen atoms in the same excited state characterized by the
quantum number
n
. The 1.5 eV minimum energy is the ionization energy.
D
EVELOP
The hydrogen energy levels are given by Equation 36.6:
1
2
2
13.6 eV
n
E
E
n
n
−
=
=
where
n
is the principal quantum number. The ionization energy for this state is the difference between the zero of
energy and the energy of the state.
E
VALUATE
So
2
1.5 eV
13.6 eV/
,
n
E
n
= −
=
which yields
13.6/1.5
3.
n
=
=
A
SSESS
The result makes sense since principal quantum must be a positive integer.
16.
The quantum number
l
can take integer values from 0 to
1,
n
−
so its maximum value is 6. From Equation 36.9,
6
7
42 .
L
=
×
=
h
h
17.
I
NTERPRET
This problem is about the allowed values of the magnitude of angular momentum.
D
EVELOP
The magnitude of orbital angular momentum is given by Equation 36.9:
(
1) ,
0,1,2, ...
L
l l
l
=
+
=
h
E
VALUATE
Successive values of
(
1),
l l
+
starting with
3,
l
=
are:
3
4
12,
4
5
20,
5
6
30,
× =
× =
× =
36
2
34
2
18
1
2
31
11
2
0
(6.626
10
J s/2 )
2.180
10
J
13.61 eV
2
2(9.109
10
kg)(5.292
10
m)
E
ma
π
−
−
−
−
−
×
⋅
= −
=
= −
×
= −
×
×
h
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36.2
Chapter 36
6
7
42,
× =
etc. Evidently,
(d)
is an erroneous possibility.
A
SSESS
Since orbital angular momentum is quantized, only certain discrete values are allowed.
18.
From Equation 36.9,
2
2
(
1)
( / )
(2.585/1.054)
6.01
6.
l l
L
+
=
=
=
≈
h
Therefore
2.
l
=
Since
1,
l
n
≤ −
the smallest
value of
n
is 3; thus the minimum energy is
2
13.6 eV/(3)
1.51 eV
−
= −
(from Equation 36.6).
19.
I
NTERPRET
We’re given the magnitude of the orbital angular momentum of an electron and asked to find its
corresponding orbital quantum number.
D
EVELOP
The magnitude of orbital angular momentum is given by Equation 36.9:
(
1) ,
0,1,2, ...
L
l l
l
=
+
=
h
E
VALUATE
From the above equation, we have
(
1)
30,
l l
+
=
or
5.
l
=
A
SSESS
In this case, the
value
l
is easily found by inspection. In general,
l
is a nonnegative integer solution of the
quadratic formula.
20.
A 6
f
state has
6
n
=
and
3
l
=
(see the explanation following Example 36.2). Thus,
13.6 eV/36
0.378 eV,
E
= −
= −
and
34
34
12(1.054
10
J
s)
3.65
10
J s
L
−
−
=
×
⋅
=
×
⋅
(see Equations 36.6 and 35.9).
21.
I
NTERPRET
We’re given the energy and orbital angular momentum of an electron and asked about its state. We
need to find both principal and orbital quantum numbers.
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 Fall '10
 Pheong
 Atom, Angular Momentum, Atomic physics

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