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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture4
Hydrogen atomic spectrum consists of several series of lines
- It was known for some time that every atom when
subjected to high temperatures or an electrical discharge
emits electromagnetic radiation of characteristic
frequencies which called emission spectrum.
- Because the emission spectra of atoms consist of only
certain frequencies, the spectra called line spectra.
- Scientists show that the study of hydrogen atom spectrum
was a major step in the study of the electronic structure of
atoms.
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture4
Johann Balmer
- Balmer showed that


  8.22 1014 1 
4 

n2 
Hz
n= 3, 4, 5…
- He substituted   1


 
1 
 1
 109.680  2  2 

2 n 
1
cm-1 called wave number
Cm-1
(1- 14)
Where n= 3, 4, 5…………………..
This called Balmer formula.
Example (6)
Using Balmer’s formula, calculate the wave lengths of the
first few lines of the visible regions of the hydrogen atomic
spectrum n=3, n=4
Solution:
The first line is obtained by setting n= 3
 1 1
 2  Cm-1
2
2 3 
  109.680 

=1.523x104Cm-1
 
1

λ= 6.564x10-5Cm = 6564 Å
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture4
The next line is obtained by setting n=4
 1 1 
 2  Cm-1
2
2
4 

  109.680 

=2.056x104Cm-1
λ=4.863x10-5Cm=4863 Å
The Balmer series occurs in the visible and near ultraviolet
regions. There are lines in the hydrogen atomic spectrum in
other regions; in fact there are series of lines similar to the
Balmer series in the ultraviolet and in the infrared region.
The Rydberg formula accounts for all the lines in the
hydrogen atomic spectrum
Rydberg accounted for all lines in the hydrogen atomic
spectrum by generalizing the Balmer formula to
 1
1 
1

Cm

2
2
 n1 n 2 
  109.680 
(n 2  n1 ) (1-15)
This equation is called Rydberg formula.
The constant in Eq 1-15 is called Rydberg Constant
 1
1 

2
2 
n
n
 1
2 
  RH 
RH is the Rydberg constant = 109.67 Cm-1
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture4
[4]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture4
Example 7:
Calculate the wavelength of the second line in the Paschen series
and show that this line lies in the near infrared, that is, in the
infrared region near the visible.
Solution:
In the paschen series, n1=3 and n2= 4, 5, 6… According to figure
Thus the second line in the Paschen series is given by setting n1=3
and n2=5 in Eq. 1-15:
 1
1 
-1

Cm

2
2
 n1 n 2 
  RH 
1 1
-1

Cm
2
2 
3 5 
  109.680 
  7.799 103 Cm -1
=
1

 =1.282 104 Cm =12.820 Å
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture4
Angular momentum is the fundamental property of rotating
systems:
Linear momentum is given by m and is usually denoted by
symbol  . Now consider a particle rotating in a plane a bout fixed
centre as in the figure
Let rot be the frequency of rotations (Cycles/second)
The velocity of particle v=2πrrot= r ωrot where ωrot=2πrot has
units of radians/second and is called the angular velocity. The
kinetic energy of the revolving particle is:
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture4
k 
1
1
1
mv 2  mr 2 2  I  2
2
2
2
(1-17)
Where I =mr2 is the momentum of inertia where
v
L  I   mr 2 ( )  mvr
r
(1-18)
Where L is called the angular momentum
 Kinetic energy can be written in terms of momentum for
a linear system
mv 2 (mv )2 P 2
k 


2
2m
2m
(1-19)
And for a rotating system
I  2 (I  ) 2 L2
k 


2
2I
2I
mv 2
F
r
(1-20)
(1-21)
Where F : inward force to keep it moving in a circular orbit.
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