chap36 - ATOMIC PHYSICS 36 Note For the problems in this...

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

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.