Announcements
•Exam 1 is next time. Covers everything up
through today. Solutions to today’s
homework will be posted Friday afternoon.
•First project is due in two weeks. In addition
to a short (~10 minute) presentation you
must turn in a written report on your project
•Homework Set 5: Chapter 4 # 41 & 42 +
Supplemental Problems
Light!
Frequency = n
Wavelength = l
Energy E = hn
c = 299792458 m/s ≈ 3.00x108 m/s
Speed = c = ln
h = 6.626x10-34 J-s
Some simple examples
•Determine the frequency of the red emission line of
hydrogen whose wavelength is 656.3 nm
•Determine the energy of a sodium D1 photon whose
wavelength is 589.592 nm
•What is the wavelength of an gamma-ray photon
whose energy is 5.0 MeV? 1 eV = 1.602x10-19 Joules?
Example Solution
Given a wavelength of 656.3 nm,
find the frequency.
2.9979 108 m s
14
c  nl  n  

4.568

10
Hz
9
l 656.3 10 m
c
Example Solution 2
Given a wavelength of
589.592 nm, find the energy.
c  ln  n 
so E 
hc
l
c
l


and
E  hn

6.626 1034 Js 2.99792 108 m s
9
589.592 10 m
 3.370 1019 J
A useful unit is the eV.
One ev is 1.60217657 1019 Joules
So, E  2.103eV
  3.36914 10
19
J
Example 3 Solution
Given an energy of 5.0 MeV, find the wavelength.
First convert energy from MeV to Joules since h
has units of J-s
E   5.0 106 eV   1.60218  1019 J eV   8.0109  1019 J
Now use energy formula to find wavelength
hc
hc  6.626 10
E  hn 
l 

l
34

J  s 3.0 108 m s
E
8.0109 10
 2.5 1013 m
19
J
  2.485 10
13
m
Light comes from electron
transitions within the atom
For the hydrogen atom these transitions are named
after the scientist who studied them. The Lyman lines
are in the UV range, the Balmer are visible and the
Paschen and Brackett are in the IR
The energy of an emitted
photon is just the difference
in energy of two energy levels
For the hydrogen atom
E  E final  Einitial  hn
1
1
 13.6eV  2  2 
 n f ni 
n is the quantum number for
the initial or final energy
level. Note that there is no
n = 0 energy level. The lowest
energy level, n = 1, is referred
to as the ground state.
Examples
The Balmer emission lines of hydrogen are
transitions whose final energy level is the n = 2
level. Determine the wavelength, in nm, of the first
four Balmer lines.
Example Solution
The Balmer lines all have nf = 2 so the first four lines
will have ni = 3, 4, 5 and 6. First do some algebra to
get a formula for wavelength in terms of quantum
numbers. Convert eV to J while we are at it.
1
1
1
1
19 J
E  13.6eV  2  2   13.6eV 1.60218 10
eV  2  2 
 n f ni 
 n f ni 
hc
hc
hc
hc


 l 
E
E
l
1 1 
1
1
18
18
2.17896 10 J   2 
2.17896 10 J  2  2 
 4 ni 
 n f ni 
so



l
hc
 2.17896 10
18
1 1 
J   2
 4 ni 




Example Solution 2
Lump all the constants together since we have to
do the same calculation several times


6.62607 1034 J  s 2.99792  108 m s
hc
l

1 1 
1 1 
13.6eV   2 
2.17896 1018 J   2 
 4 ni 
 4 ni 

9.116452  10 m 

l
8
1 1 
 4  n2 
i 



Example Solution 3
Now just plug
in values for ni
9.116452 108 m
l
1 1 
  2
 4 ni 
9.116452 108 m
l1 
 6.5634 107 m  656nm
1 1 
 4  32 
9.116452 10 8 m
l2 
 4.8621107 m  486nm
1 1 
 4  42 
9.116452 10 8 m
l3 
 4.341107 m  434nm
1 1 
 4  52 
9.116452 10 8 m
l4 
 4.1024 107 m  410nm
1 1 
 4  62 